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Printed: September 1, 2011




Authors
Brenda Meery, Richard Parsons


Contributors
Jonathan Edge, Lillian Gu


Editor
Shonna Robinson


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Contents


1 Measurement in Chemistry 1
1.1 Making Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Measurement Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 The SI System of Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.4 Significant Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5 Using Algebra in Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.6 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.7 Evaluating Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.8 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31


2 Independent and Dependent
Events 40
2.1 Independent Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
2.2 Dependent Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
2.3 Mutually Inclusive and Mutually Exclusive Events . . . . . . . . . . . . . . . . . . . . . . . 48
2.4 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54


3 The Next Step...Conditional
Probability 56
3.1 What are Tree Diagrams? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.2 Order and Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
3.3 Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.4 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73


4 Introduction to Discrete
Random Variables 77
4.1 What are Variables? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
4.2 The Probability Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
4.3 A Glimpse at Binomial and Multinomial Distributions . . . . . . . . . . . . . . . . . . . . . 82
4.4 Using Technology to Find Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . 86


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4.5 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97


5 Probability Distributions 101
5.1 Normal Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.2 Binomial Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
5.3 Exponential Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
5.4 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124


6 Measures of Central Tendency 128
6.1 The Mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
6.2 The Median . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
6.3 The Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
6.4 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152


7 The Shape, Center and Spread
of a Normal Distribution 162
7.1 Estimating the Mean and Standard Deviation of a Normal Distribution . . . . . . . . . . . 164
7.2 Calculating the Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
7.3 Connecting the Standard Deviation and Normal Distribution . . . . . . . . . . . . . . . . . 177
7.4 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180


8 Organizing and Displaying
Distributions of Data 185
8.1 Line Graphs and Scatter Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
8.2 Circle Graphs, Bar Graphs, Histograms, and Stem-and-Leaf Plots . . . . . . . . . . . . . . . 197
8.3 Box-and-Whisker Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
8.4 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225


9 Organizing and Displaying
Data for Comparison 237
9.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
9.2 Double Line Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
9.3 Two-sided Stem-and-Leaf Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
9.4 Double Bar Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248
9.5 Double Box-and-Whisker Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
9.6 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254


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Chapter 1


Measurement in Chemistry


1.1 Making Observations
Lesson Objectives
The student will:


• explain the importance of observations.
• define qualitative and quantitative observations.
• distinguish between qualitative and quantitative observations.
• explain the importance of using quantitative observations in measurements where possible.


Vocabulary
• observation
• qualitative observation
• quantitative observation


Introduction
Observation is very important when using scientific methods to investigate phenomena. Observation
involves using the senses to gather information about the natural world. Science depends on keeping
records of observations for later interpretations. These interpretations may lead to the development of
scientific theories or laws. Without accurate observations, scientists cannot make any interpretations and
therefore cannot draw conclusions.
Take out a piece of paper and record a chart similar to the one shown in Table 1.1. A chart is a useful
tool that can help us record and organize our observations. Look up from this text and scan the room.
Write down what you see around you in as much detail as you feel necessary in order for you or someone
else to picture the room.


Table 1.1: Record of Observations


Item Observation
1.


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Table 1.1: (continued)


Item Observation
2.
3.


One summer evening, Scott and Brenda came home from work to find their house in shambles. Neighbors,
friends, and colleagues were baffled by the strange occurrence. Inside the house, they found the television
turned on. Food on the table was ready to be eaten. All of Scott’s coin collections, his precious metals,
and Brenda’s prized possession – her statue of Galileo – were gone. Foul play is suspected.
Here is a simple test for you. Pretend you are visiting a forensic scientist, hired to investigate the scene of
the crime. You are asked to only analyze the observations gathered by the other investigators at the scene.
You must try to make as few assumptions as possible and make your decision based on the data at hand.
The lead investigator gives you the following observations gathered from the scene and the suspects:
Observations at Scene


1. Blood type = B
2. Fiber sample = polyester
3. Powder found = white
4. Shoe print found = work boot


Table 1.2: Suspect Information


Suspect 1: 180 lb male Suspect 2: 220 lb male Suspect 3: 120 lb female
Blood type = B Blood type = B Would not comply
Sweater = polyester Blazer = wool knit Pants = polyester
Works in sugar factory Pastry chef Automobile sales woman


From Table 1.2, can you deduce who might have been involved in the alleged crime? Do you need more
information? How detailed do the observations need to be in order for you, the scientist, to make accurate
conclusions? What will you base your decision on? What other information do you need? Remember that
if you guess randomly, an innocent person could be convicted.


Quantitative and Qualitative Observations Defined
There are two types of observations: quantitative and qualitative. Quantitative observations involve
measurements or estimates that yield meaningful, numerical results. Qualitative observations yield
descriptive, nonnumerical results. Although all the observations we can make on a phenomenon are valu-
able, quantitative observations are often more helpful than qualitative ones. Qualitative observations are
somewhat vague because they involve comparative terms. Quantitative observations, on the other hand,
have numbers and units associated with them and therefore convey more information. Even an estimated
number is more valuable than no number at all.
A qualitative observation would be, for example, ‘‘The attendance clerk is a small woman.” If the observer
is 6 feet 4 inches tall, he might find a woman who is 5 feet 8 inches tall to be ‘‘small.” If the observer
reported this to someone who is 5 feet 2 inches tall, however, the listener would not acquire a good idea


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of the attendance clerk’s height because he would not think a woman who is 5 feet 8 inches tall is small.
The description ‘‘a small woman” could refer to any woman whose height was between 3 feet and 6 feet,
depending on who made the observation, as illustrated in the image below.


Similarly, ‘‘a small car” could refer to anything from a compact car to a child’s toy car. The word ‘‘small”
is a comparative term. The same is true for all words like tall, short, fast, slow, hot, cold, and so forth.
These words do not have exact meanings. Under various circumstances, temperatures of 90◦F, 110◦F,
212◦F, and 5000◦F could all be described as ‘‘hot.” The word ‘‘hot” does not convey as much information
as the numerical description. Even observations of colors are not exact because there are many shades of
each color. Two people may both be wearing red shirts, but the colors of the shirts may not be exactly the
same. Exact descriptions of colors would require reporting the frequency or wavelength of the color.


Quantitative and Qualitative Observations Compared


Table 1.3: Comparison of Qualitative and Quantitative Observations


Qualitative (words only) Quantitative (words and numbers)
The girl has very little money. The girl has 85 cents.
The man is short. The man is 5 feet 2 inches tall.
Use a small test tube. Use a test tube that is 12 centimeters long.
It is a short walk to my house. It is about 1 mile to my house.


You can see from the last example in Table 1.3 that even if the number is an estimate, a quantitative
observation contains more information because of the number associated with the observation. Some
people might think that a 1-mile walk is short, while others may not. If an actual measuring device is not
available, the observer should always try to estimate a measurement so that the observation will have a
number associated with it.
While estimated measurements may not be accurate, they are valuable because they establish an approxi-
mate numerical description for the observation. ‘‘The car is small” is an observation that provides us with
certain information. We know that the object is some kind of car (perhaps real or perhaps a toy), and we
know that it is probably smaller than a limousine because almost no one would describe a limousine as
small. Suppose instead that the observation is: ‘‘The car is about 3 feet tall, 3 feet long, and 2 feet wide.”


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While these estimated measurements are not accurate, we now know that are not dealing with a compact
automobile, nor are we dealing with a toy car. With these estimated measurements, we know that we
are dealing with a car about the size of a tricycle. It is not a problem if we discover later that the car
was actually 2 feet tall instead of 3 feet tall, because we knew the original measurement was an estimate.
Estimates are excellent observations if we do not have the ability to measure the object accurately and still
qualify as quantitative observations.
Example Questions:
Pick out the quantitative and qualitative observations from each phrase.


1. 3.0 grams of NaCl dissolve in 10 milliliters of H2O to produce a clear solution.
2. The spider on the wall has only seven legs remaining but is still big and hairy.
3. When 0.50 milliliter of a solution is put into a flame, the flame turns a brilliant green.


Solutions:


1. Quantitative: 3.0 grams and 10 milliliters; Qualitative: clear solution
2. Quantitative: seven legs; Qualitative: big and hairy
3. Quantitative: 0.50 milliliter; Qualitative: brilliant green


Lesson Summary
• Observation involves using the senses to gather information about the natural world.
• There are two types of observations: qualitative and quantitative.
• Scientists gather information by making both qualitative and quantitative observations.
• Qualitative observations yield descriptive, nonnumerical results.
• Quantitative observations yield meaningful, numerical results.
• Observations, either qualitative or quantitative, are used by scientists as tools to make representations
and interpretations about the surroundings.


Further Reading / Supplemental Links
This website helps to build your observation skills.


• http://www.mrsoshouse.com/pbl/observe/indexobserve.htm


Review Questions
Label each observation as qualitative or quantitative.


1. The temperature of this room is 25◦C.
2. It is comfortably warm in this room.
3. Most people have removed their coats.
4. The building is 25 stories high.
5. It is a very tall building.
6. The building is taller than any nearby trees.
7. The bottle is green.
8. The bottle contains 250 milliliters of liquid.


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9. Robert bought his son a small car.
10. The car is smaller than his hand.
11. The car is about three inches long.
12. The race is about 27 miles long.


1.2 Measurement Systems
Lesson Objectives
The student will:


• state an advantage of using the metric system over the United States customary system.
• state the different prefixes used in the metric system.


Vocabulary
• base unit
• metric system


Introduction
Even in ancient times, humans needed measurement systems for commerce. Land ownership required
measurements of length, and the sale of food and other commodities required measurements of mass. The
first elementary efforts in measurement required convenient objects to be used as standards, such as the
human body. Inch and foot are examples of measurement units that are based on parts of the human
body. The inch is based on the width of a man’s thumb, and the foot speaks for itself. The grain is a unit
of mass measurement that is based upon the mass of a single grain of wheat. Because grains of wheat are
fairly consistent in mass, the quantity of meat purchased could be balanced against some number of grains
of wheat on a merchant’s balance.
It should be apparent that measuring the foot of two different people would lead to different results. One
way to achieve greater consistency was for everyone to use the foot of one person, such as the king, as the
standard. The length of the king’s foot could be marked on pieces of wood, and everyone who needed to
measure length could have a copy. Of course, this standard would change when a new king was crowned.
What were needed were objects that could be safely stored without changing over time to serve as standards
of measurement. Copies of these objects could then be made and distributed so that everyone was using the
exact same units of measure. This was especially important when the requirements of science necessitated
accurate, reproducible measurements.


The Metric System
The metric system is an international decimal-based system of measurement. Because the metric system
is a decimal system, making conversions between different units of the metric system are always done with
factors of ten. To understand why this makes the metric system so useful and easy to manipulate, let’s
consider the United States customary system – that is, the measurement system commonly used in the US.
For instance, if you need to know how many inches are in a foot, you need to remember: 12 inches = 1 foot.
Now imagine that you now need to know how many feet are in a mile. What happens if you have never


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memorized this fact before? Of course, you can find this conversion online or elsewhere, but the point is
that this information must be given to you, as there is no way for you to derive it by yourself. This is true
about all parts of the United States customary system: you have to memorize all the facts that are needed
for different measurements.


Metric Prefixes and Equivalents
The metric system uses a number of prefixes along with the base units. A base unit is one that cannot
be expressed in terms of other units. The base unit of mass is the gram (g), that of length is the meter
(m), and that of volume is the liter (L). Each base unit can be combined with different prefixes to define
smaller and larger quantities. When the prefix ‘‘centi-” is placed in front of gram, as in centigram, the unit
is now 1100 of a gram. When ‘‘milli-” is placed in front of meter, as in millimeter, the unit is now 11,000 of a
meter. Common prefixes are shown in Table 1.4.


Table 1.4: Common Prefixes


Prefix Meaning Symbol
pico- 10−12 p
nano- 10−9 n
micro- 10−6 µ (pronounced mu)
milli- 10−3 m
centi- 10−2 c
deci- 10−1 d
kilo- 103 k


Common metric units, their symbols, and their relationships to a base unit are shown below:


1.00 picogram = 1.00 pg = 1.00 × 10−12 g
1.00 nanosecond = 1.00 ns = 1.00 × 10−9 g
1.00 micrometer = 1.00 µm = 1.00 × 10−6 m
1.00 centimeter = 1.00 cm = 1.00 × 10−2 m


1.00 deciliter = 1.00 dL = 1.00 × 10−1 L
1.00 kilogram = 1.00 kg = 1.00 × 103 g


You can express a given measurement in more than one unit. If you express a measured quantity in two
different metric units, then the two measurements are metric equivalents. Common metric equivalents are
shown below.


• Length:


1, 000 millimeters = 1 meter
100 centimeters = 1 meter
10 millimeters = 1 centimeter


• Mass:


1, 000 milligrams = 1 gram
1, 000 grams = 1 kilogram


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• Volume:


1 liter = 1, 000 milliliters


Lesson Summary
• The metric system is an international decimal-based system of measurement.
• The metric system uses a number of prefixes along with the base units.
• The prefixes in the metric system are multiples of 10.
• A base unit is one that cannot be expressed in terms of other units
• If you express a measured quantity in two different metric units, then the two measurements are
metric equivalents.


Further Reading / Supplemental Links
The following website provides more information about the metric system and measurements in chemistry.


• http://www.chemistry24.com/teach_chemistry/measurement-and-math-in-chemistry.html


Review Questions
Fill in the blanks in Table 1.5.


Table 1.5: Table for Review Question


Prefix Meaning Symbol
pico- 10−12 p
nano- ? n
? 10−6 µ
milli- 10−3 ?
centi- ? c
deci- 10−1 ?
? 103 k


1.3 The SI System of Measurement
Lesson Objectives
The student will:


• explain the difference between mass and weight.
• identify SI units of mass, distance (length), volume, temperature, and time.
• define derived unit.
• describe absolute zero.


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Vocabulary
• absolute zero
• cubic meter
• derived units
• heat
• International System of Units
• Kelvin temperature scale
• kilogram
• length
• mass
• meter
• second
• temperature
• volume
• weight


Introduction
The International System of Units, abbreviated SI from the French Le Système International d’ Unites,
is the main system of measurement units used in science. Since the 1960s, the International System of
Units has been agreed upon internationally as the standard metric system. The SI base units are based
on physical standards. The definitions of the SI base units have been and continue to be modified, and
new base units are added as advancements in science are made. Each SI base unit, except the kilogram, is
described by stable properties of the universe.


Mass and Its SI Unit
Mass and weight are not the same thing. Although we often use these terms interchangeably, each one
has a specific definition and usage. The mass of an object is a measure of the amount of matter in it
and remains the same regardless of where the object is placed. For example, moving a brick to the moon
does not cause matter in the brick to disappear or to be removed. The weight of an object is the force
of attraction between the object and the Earth (or whatever large, gravity-producing body the object is
located on). This attraction is due to the force of gravity. Since the force of gravity is not the same at
every point on the Earth’s surface, the weight of an object is not constant. The gravitational pull on the
object varies and depends on where the object is with respect to the Earth or other gravity-producing
object. For example, a man who weighs 180 pounds on Earth would weigh only 45 pounds if he were in
a stationary position 4,000 miles above the Earth’s surface. This same man would weigh only 30 pounds
on the moon, because the moon’s gravitational pull is one-sixth that of Earth’s. The mass of this man,
however, would be the same in each situation because the amount of matter in the man is constant.
We measure weight with a scale, which contains a spring that compresses when an object is placed on it.
An illustration of a scale is depicted on the left in the diagram below. If the gravitational pull is less, the
spring compresses less and the scale shows less weight. We measure mass with a balance, depicted on the
right in the diagram below. A balance compares the unknown mass to known masses by balancing them
on a lever. If we take our balance and known masses to the moon, an object will have the same measured
mass that it had on the Earth. The weight, of course, would be different on the moon. Consistency requires
that scientists use mass and not weight when measuring the amount of matter.


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The basic unit of mass in the International System of Units is the kilogram. A kilogram is equal to 1,000
grams. A gram is a relatively small amount of mass, so larger masses are often expressed in kilograms.
When very tiny amounts of matter are measured, we often use milligrams, with one milligram equal to
0.001 gram. Other larger, smaller, or intermediate mass units may also be appropriate.
At the end of the 18th century, a kilogram was the mass of a cubic decimeter of water. In 1889, a new
international prototype of the kilogram was made from a platinum-iridium alloy. The kilogram is equal to
the mass of this international prototype, which is held in Paris, France. A copy of the standard kilogram
is shown in Figure 1.1.


Figure 1.1: This image shows a copy of the standard kilogram stored and used in Denmark.


Length and Its SI Unit
Length is the measurement of anything from end to end. In science, length usually refers to how long an
object is. There are many units and sets of standards used in the world for measuring length. The ones
familiar to you are probably inches, feet, yards, and miles. Most of the world, however, measure distances
in meters and kilometers for longer distances, and in centimeters and millimeters for shorter distances. For
consistency and ease of communication, scientists around the world have agreed to use the SI standards,
regardless of the length standards used by the general public.


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Figure 1.2: This image shows the standard meter used in France in the 18th century.


The SI unit of length is the meter. In 1889, the definition of the meter was the length of a bar made
of platinum-iridium alloy stored under conditions specified by the International Bureau of Standards. In
1960, this definition of the standard meter was replaced by a definition based on a wavelength of krypton-
86 radiation. In 1983, that definition was replaced by the following: the meter is the length of the path
traveled by light in a vacuum during a time interval of 1299,792,458 of a second.


Volume: A Derived Unit
The volume of an object is the amount of space it takes up. In the International System of Units, volume
is a derived unit, meaning that it is based on another SI unit. Consider a cube with each side measuring
1.00 meter. The volume of this cube is 1.00 m× 1.00 m× 1.00 m = 1.00 m3, or one cubic meter. The cubic
meter is the SI unit of volume and is based on the meter, the SI unit of length. The cubic meter is a very
large unit and is not very convenient for most measurements in chemistry. A more common unit is the liter
(L), which is 11,000 of a cubic meter. One liter is slightly larger than one quart: 1.000 liter = 1.057 quart.
Another commonly used volume measurement is the milliliter, which is equal to 11,000 of a liter. Since 11,000
of a liter is also equal to 1.00 cubic centimeter, then 1.00 mL = 1.00 cm3.


As seen in the illustration above, the volume of 1,000 blocks, each with a volume of 1 cubic centimeter, is
equivalent to 1 liter.


Measuring Temperature
When used in a scientific context, the words heat and temperature do not mean the same thing. Tem-
perature represents the average kinetic energy of the particles making up a material. Increasing the
temperature of a material increases its thermal energy; objects contain thermal energy, not ‘‘heat.” Heat


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is the movement of thermal energy from a warmer object to a cooler object. When thermal energy moves
from one object to another, the temperature of both objects change. These different types of energies will
be re-examined in more detail in later chapters, but the key concept to remember here is that temperature
reflects the thermal energy contained in an object, while heat is the movement of thermal energy between
two objects.
Consider a small beaker of boiling water (100◦C) and a bathtub of water at a temperature of 50◦C. The
amount of thermal energy contained in the bathtub is 40,000,000 joules (a measure of energy), while the
amount of thermal energy in the beaker is 4,000 joules. Although the temperature of the beaker of water
is only twice the temperature of the bathtub of water, the amount of thermal energy contained in the
bathtub is many times greater than that in the beaker of water. The important thing to note here is that
the amount of thermal energy contained in an object increases greatly with an increase in temperature.
A thermometer is a device that measures temperature. The name is made up of thermo, which means
heat, and meter, which means to measure. One of the earliest inventors of a thermometer was Galileo. He
is said to have used a device called a thermoscope around the year 1600. The thermometers we typically
use today, however, are different from the one Galileo used.
The type of thermometers most people are familiar with operates on the principle that the volume of
most liquids increases when heated and decreases when cooled. If a liquid is trapped inside an evacuated
tube with an attached bulb reservoir, like that shown in the diagram below, the liquid in the tube will
move higher in the tube when the liquid is heated and lower when the liquid is cooled. After a short
period of time, the temperature of the liquid in the bulb will be the same temperature as the surrounding
material. The liquid in the tube reflects the temperature of the surrounding because the molecules of
material surrounding the bulb will collide with the tube and transfer heat during the process. If heat is
transferred to the liquid in the bulb, the liquid will rise and indicate an increase in temperature. If heat is
transferred to the surrounding material, the height of the liquid in the tube will fall and indicate a decrease
in temperature.


Each thermometer is calibrated by placing it in a liquid whose exact temperature is known. Most ther-
mometers are calibrated using consistent known temperatures that are easy to reproduce. At normal sea
level and atmospheric pressure, a stable mixture of ice and water will be at the freezing point of water,
and a container of boiling water will be at the boiling point of water. When the height of the liquid
inside the thermometer reflects the temperature of the surrounding liquid, a mark (scratch) is made on
the tube to indicate that temperature. Once the freezing and boiling temperatures have been marked on


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the thermometer, the distance between the marks can be marked up into equal divisions called degrees.
Daniel Fahrenheit established the Fahrenheit scale. On his temperature scale, Fahrenheit designated the
freezing point of water as 32◦F and the boiling point of water as 212◦F. Therefore, the distance between
these two points would be divided into 180 degrees. The Fahrenheit temperature scale is used in the
United States for most daily expressions of temperature. In another temperature scale established by
Anders Celsius, Celsius designated the freezing point of water as 0◦C and the boiling point of water as
100◦C. Therefore, the temperatures between these two points on the Celsius scale are divided into 100
degrees. Clearly, the size of a Celsius degree and the size of a Fahrenheit degree are not the same.
Earlier in the lesson, the temperature of a substance is defined to be directly proportional to the average
kinetic energy it contains. In order for the average kinetic energy and temperature of a substance to be
directly proportional, it is necessary for the average kinetic energy to be zero when the temperature is
zero. This is not true with either the Fahrenheit or Celsius temperature scales. Most of us are familiar
with temperatures that are below the freezing point of water. It should be apparent that even though the
air temperature may be −5◦C, the molecules of air are still moving. Substances like oxygen and nitrogen
have already become vapor at temperatures below −150◦C, indicating that molecules are still in motion at
over a hundred degrees below zero.
A third temperature scale was established to address this issue. This temperature scale was designed
by Lord Kelvin. Lord Kelvin stated that there is no upper limit to how hot things can get, but there
is a limit as to how cold things can get. Kelvin developed the idea of absolute zero, which is the
temperature that molecules stop moving and have zero kinetic energy. The Kelvin temperature scale
has its zero at absolute zero (determined to be −273.15◦C) and uses the same degree size as a degree on the
Celsius scale. As a result, the mathematical relationship between the Celsius scale and the Kelvin scale is:
K = ◦C+273.15. On the Kelvin scale, water freezes at 273.15 K and boils at 373.15 K. In the case of the
Kelvin scale, the degree sign is not used. Temperatures are expressed, for example, simply as 450 K.
It should be noted that many mathematical calculations in chemistry involve the difference between two
temperatures, symbolized by ∆T (pronounced delta T). Since the size of a degree is the same in Celsius
and in Kelvin, the ∆T will be the same for either scale. For example, 20◦C = 293 K and 50◦C = 323 K; the
difference between the Celsius temperatures is 30◦C, and the difference between the Kelvin temperatures
is 30 K. When the calculations involve ∆T , it is not necessary to convert Celsius to Kelvin, but when the
temperature is used directly in an equation, it is necessary to convert Celsius to Kelvin.


This video is an explanation of how to convert among the Celsius, Kelvin, and Fahrenheit temperature
scales and includes a sample problem (4e): http://www.youtube.com/watch?v=SASnMMGp5mo (4:37).


Figure 1.3: (Watch Youtube Video)
http://www.ck12.org/flexbook/embed/view/353


This video is an explanation of particle temperature, average temperature, heat flow, pressure, and volume
(7a): http://www.youtube.com/watch?v=tfE2y_7LqA4 (4:00).


www.ck12.org 12




Figure 1.4: (Watch Youtube Video)
http://www.ck12.org/flexbook/embed/view/354


Time and Its SI Unit
The SI unit for time is the second. The second was originally defined as a tiny fraction of the time
required for the Earth to orbit the Sun. It has since been redefined several times. The definition of a
second (established in 1967 and reaffirmed in 1997) is: the duration of 9,192,631,770 periods of radiation
corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133
atom.


Lesson Summary
• The International System of Units, abbreviated SI from the French Le Système International d’


Unites, is internationally agreed upon since the 1960s as the standard metric system.
• Mass and weight:


– The mass of an object is a measure of the amount of matter in it.
– The mass of an object remains the same regardless of where the object is placed.
– The basic unit of mass in the International System of Units is the kilogram.
– The weight of an object is the force of attraction between the object and the earth (or whatever
large, gravity-producing body the object is located on).


• Length:
– Length is the measurement of anything from end to end.
– The SI unit of length is the meter.


• Volume:
– The volume of an object is the amount of space it takes up.
– The cubic meter is the SI unit of volume.


• Temperature and heat:
– Temperature represents the average kinetic energy of the particles that make up a material.
– Increasing the temperature of a material increases its thermal energy.
– Heat is the movement of thermal energy from a warmer object to a cooler object.
– When thermal energy moves from one object to another, the temperature of both objects change.
– Absolute zero is the temperature at which molecules stop moving and therefore have zero kinetic
energy.


– The Kelvin temperature scale has its zero at absolute zero (determined to be −273.15◦C) and
uses the same degree size as the Celsius scale.


– The mathematical relationship between the Celsius scale and the Kelvin scale is K = ◦C+273.15.
• Time:


– The SI unit for time is the second.


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Further Reading / Supplemental Links
This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 1-3 is on
measuring matter, lesson 1-7 is on temperature conversion, and lesson 2-1 is on the International System
of measurements.


• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lessindex.htm


Review Questions
1. What is the basic unit of measurement in the metric system for length?
2. What is the basic unit of measurement in the metric system for mass?
3. What unit is used in the metric system to measure volume? How is this unit related to the measure-
ment of length?


4. Give the temperatures in Celsius for the freezing and boiling points of water.
5. Give the temperatures in Kelvin for the freezing and boiling points of water.
6. Would it be comfortable to swim in a swimming pool whose water temperature was 275 K? Why or
why not?


1.4 Significant Figures
Lesson Objectives
The student will:


• explain the necessity for significant figures.
• determine significant figures of the equipment pieces chosen.
• identify the number of significant figures in a measurement.
• use significant figures properly in measurements and calculations.
• determine the number of significant figures in the result of a calculation.
• round calculated values to the correct number of significant figures.


Vocabulary
• significant figures


Introduction
The numbers you use in math class are considered to be exact numbers. When you are given the number
2 in a math problem, it does not mean 1.999 rounded up to 2, nor does it mean 2.00001 rounded down
to 2. In math class, the number 2 means exactly 2.000000. . . with an infinite number of zeros – a perfect
2! Such numbers are produced only by definition, not by measurement. We can define 1 foot to contain
exactly 12 inches with both numbers being perfect numbers, but we cannot measure an object to be exactly
12 inches long. In the case of measurements, we can only read our measuring instruments to a limited
number of subdivisions. We are limited by our ability to see smaller and smaller subdivisions, and we are
limited by our ability to construct smaller and smaller subdivisions on our measuring devices. Even with
the use of powerful microscopes to construct and read our measuring devices, we eventually reach a limit.


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Therefore, although the actual measurement of an object may be a perfect 12 inches, we cannot prove it
to be so. Measurements do not produce perfect numbers; the only perfect numbers in science are defined
numbers, such as conversion factors. Since measurements are fundamental to science, science does not
produce perfect measurements.
It is very important to recognize and report the limitations of a measurement along with the magnitude
and unit of the measurement. Many times, the measurements made in an experiment are analyzed for
regularities. If the numbers reported show the limits of the measurements, the regularity, or lack thereof,
becomes visible.
Table 1.6: Comparison of Observations with the Proper Number of Significant Figures


Observation List A Observation List B
22.41359 m 22.4 m
22.37899 m 22.4 m
22.42333 m 22.4 m
22.39414 m 22.4 m


In the lists of observations shown in Table 1.6, List A shows measurements without including the limits
of the measuring device. In comparison, List B has the measurements rounded to reflect the limits of the
measuring device. It is difficult to perceive regularity in List A, but the regularity stands out in List B.


Rules for Determining Significant Figures
Significant figures, also known as significant digits, are all of the digits that can be known with certainty
in a measurement plus an estimated last digit. Significant figures provide a system to keep track of the
limits of the original measurement. To record a measurement, you must write down all the digits actually
measured, including measurements of zero, and you must not write down any digit not measured. The
only real difficulty with this system is that zeros are sometimes used as measured digits, while other times
they are used to locate the decimal point.


In the sketch shown above, the correct measurement is greater than 1.2 inches but less than 1.3 inches. It is
proper to estimate one place beyond the calibrations of the measuring instrument. This ruler is calibrated
to 0.1 inches, so we can estimate the hundredths place. This reading should be reported as 1.25 or 1.26
inches.


In this second case (sketch above), it is apparent that the object is, as nearly as we can read, 1 inch. Since


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we know the tenths place is zero and can estimate the hundredths place to be zero, the measurement should
be reported as 1.00 inch. It is vital that you include the zeros in your reported measurement because these
are measured places and are significant figures.


This measurement is read as 1.15 inches, 1.16 inches, or perhaps even 1.17 inches.


This measurement is read as 1.50 inches.
In all of these examples, the measurements indicate that the measuring instrument had subdivisions of a
tenth of an inch and that the hundredths place is estimated. There is some uncertainty about the last,
and only the last, digit.
In our system of writing measurements to show significant figures, we must distinguish between measured
zeros and place-holding zeros. Here are the rules for determining the number of significant figures in a
measurement.
Rules for Determining the Number of Significant Figures:


1. All non-zero digits are significant.
2. All zeros between non-zero digits are significant.
3. All beginning zeros are not significant.
4. Ending zeros are significant if the decimal point is actually written in but not significant if the decimal
point is an understood decimal (the decimal point is not written in).


Examples of the Significant Figure Rules:


1. All non-zero digits are significant.


543 has 3 significant figures.
22.437 has 5 significant figures.
1.321754 has 7 significant figures.


2. All zeros between non-zero digits are significant.


7, 004 has 4 significant figures.
10.3002 has 6 significant figures.
103 has 3 significant figures.


3. All beginning zeros are not significant.


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0.00000075 has 2 significant figures.
0.02 has 1 significant figure.
0.003003 has 4 significant figures.


4. Ending zeros are significant if the decimal point is actually written in but not significant if the decimal
point is an understood decimal.


37.300 has 5 significant figures.
33.00000 has 7 significant figures.
100. has 3 significant figures.
100 has 1 significant figure.
302, 000 has 3 significant figures.
1, 050 has 3 significant figures.


Equipment Determines Significant Figures
Quality measuring instruments are made with as much consistency as possible and are individually cali-
brated after construction. In a graduated cylinder, for example, it is desirable for the sides to be perfectly
vertical and for the inside diameter to be the same all the way up the tube. After the graduated cylinder is
completed, exact volumes of liquids are placed in the cylinder and the calibration marks are then scribed
onto the side of the tube.
The choice of measuring instrument determines the unit of measure and the number of significant figures
in the measurement. Consider the two graduated cylinders shown below.


Both cylinders are marked to measure milliliters, but the cylinder on the left only shows graduations for
whole milliliters. In comparison, the cylinder on the right has calibrations for tenths of milliliters. The
measurer reads the volume from the calibrations and estimates one place beyond the calibrations. For the
cylinder on the left, a reasonable reading is 4.5 mL. For the cylinder on the right, the measurer estimates
one place beyond the graduations and obtains a reasonable reading of 4.65 mL. The choice of the measuring
instrument determines both the units and the number of significant figures. If you were mixing up some
hot chocolate at home, the cylinder on the left would be adequate. If you were measuring out a chemical
solution for a very delicate reaction in the lab, however, you would need the cylinder on the right.
Similarly, the equipment chosen for measuring mass will also affect the number of significant figures. For
example, if you use a pan balance (illustrated on the left in the image below) that can only measure to ±
0.1 g, you could only measure out 3.3 g of NaCl rather than 3.25 g. In comparison, the digital balance
(illustrated on the right in the image below) might be able to measure to ± 0.01 g. With this instrument,


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you could measure what you need more exactly. The difference between these two balances has to do with
the number of significant figures that the balances are able to measure. Whenever you need to make a
measurement, make sure to check the number of significant figures a measuring instrument can measure
before choosing an appropriate instrument.


Significant Figures in Calculations
In addition to using significant figures to report measurements, we also use them to report the results of
computations made with measurements. The results of mathematical operations on measurements must
indicate the number of significant figures in the original measurements. There are two rules for determining
the number of significant figures after performing a mathematical operation. Most of the errors that occur
in this area result from using the wrong rule, so always double check that you are using the correct rule
for the mathematical operation involved.


Addition and Subtraction
The answer to an addition or subtraction operation must not have any digits further to the right than the
shortest addend. In other words, the answer should have as many decimal places as the addend with the
smallest number of decimal places.
Example:


Notice that the top addend has a 3 in the last column on the right, but neither of the other two addends
have a number in that column. In elementary math classes, you were taught that these blank spaces can
be filled in with zeros and the answer would be 17.6163 cm. In the sciences, however, these blank spaces
are unknown numbers, not zeros. Since they are unknown numbers, you cannot substitute any numbers
into the blank spaces. As a result, you cannot know the sum of adding (or subtracting) any column of
numbers that contain an unknown number. When you add the columns of numbers in the example above,
you can only be certain of the sums for the columns with known numbers in each space in the column.
In science, the process is to add the numbers in the normal mathematical process and then round off all
columns that contain an unknown number (a blank space). Therefore, the correct answer for the example
above is 17.62 cm and has only four significant figures.
Example:


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In this case, the addend 12 has no digits beyond the decimal. Therefore, all columns past the decimal
point must be rounded off in the final answer. We get the seemingly odd result that the answer is still 12,
even after adding a number to 12. This is a common occurrence in science and is absolutely correct.
Example:


Multiplication and Division
The answer for a multiplication or division operation must have the same number of significant figures as
the factor with the least number of significant figures.
Example:


(3.556 cm) · (2.4 cm) = 8.5344 cm2 = 8.5 cm2


The factor 3.556 cm has four significant figures, and the factor 2.4 cm has two significant figures. Therefore
the answer must have two significant figures. The mathematical answer of 8.5344 cm2 must be rounded
back to 8.5 cm2 in order for the answer to have two significant figures.
Example:


(20.0 cm) · (5.0000 cm) = 100.00000 cm2 = 100. cm2


The factor 20.0 cm has three significant figures, and the factor 5.0000 cm has five significant figures. The
answer must be rounded to three significant figures. Therefore, the decimal must be written in to show
that the two ending zeros are significant. If the decimal is omitted (left as an understood decimal), the
two zeros will not be significant and the answer will be wrong.
Example:


(5.444 cm) · (22 cm) = 119.768 cm2 = 120 cm2


In this case, the answer must be rounded back to two significant figures. We cannot have a decimal after
the zero in 120 cm2 because that would indicate the zero is significant, whereas this answer must have
exactly two significant figures.


Lesson Summary
• Significant figures are all of the digits that can be known with certainty in a measurement plus an
estimated last digit.


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• Significant figures provide a system to keep track of the limits of a measurement.
• Rules for determining the number of significant figures:


1. All non-zero digits are significant.
2. All zeros between non-zero digits are significant.
3. All beginning zeros are not significant.
4. Ending zeros are significant if the decimal point is actually written in but not significant if the
decimal point is an understood decimal.


• The choice of measuring instrument is what determines the unit of measure and the number of
significant figures in the measurement.


• The results of mathematical operations must include an indication of the number of significant figures
in the original measurements.


• The answer for an addition or subtraction operation must not have any digits further to the right
than the shortest addend.


• The answer for a multiplication or division operation must have the same number of significant figures
as the factor with the least number of significant figures.


Further Reading / Supplemental Links
A problem set on unit conversions and significant figures.


• http://science.widener.edu/svb/pset/convert1.html


This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 2-3 is
on significant figures.


• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson23.htm


Review Questions
1. How many significant figures are in the following numbers?


(a) 2.3
(b) 17.95
(c) 9.89 × 103
(d) 170
(e) 22.1
(f) 1.02
(g) 19.84


2. Perform the following calculations and give your answer with the correct number of significant figures:
(a) 10.5 + 11.62
(b) 0.01223 + 1.01
(c) 19.85 − 0.0113


3. Perform the following calculations and give your answer with the correct number of significant figures:
(a) 0.1886 × 12
(b) 2.995 ÷ 0.16685
(c) 1210 ÷ 0.1223
(d) 910 × 0.18945


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1.5 Using Algebra in Chemistry
Lesson Objectives
The student will:


• perform algebraic manipulations to solve equations.
• use the density equation to solve for the density, mass, or volume when two of the quantities in the
equation are known.


• construct conversion factors from equivalent measurements.
• apply the techniques of dimensional analysis to solving problems.
• perform metric conversions using dimensional analysis.


Vocabulary
• conversion factor
• dimensional analysis


Introduction
During your studies of chemistry (and physics as well), you will note that mathematical equations are
used in a number of different applications. Many of these equations have a number of different variables
that you will need to work with. You should also note that these equations will often require you to use
measurements with their units. Algebra skills become very important here!


Solving Algebraic Equations
Chemists use algebraic equations to express relationships between quantities. An example of such an
equation is the relationship between the density, mass, and volume of a substance: D = mV (density is
equal to mass divided by volume). Given (or making) measurements of the mass and volume of a substance,
you can use this equation to determine the density. Suppose, for example, that you have measured the
mass and volume of a sample of liquid mercury and found that 5.00 mL of mercury has a mass of 67.5
grams. Plugging these measurements into the density formula gives you the density of mercury.


D = massvolume =
67.5 g


5.00 mL = 13.5 g/mL


You should notice both units and significant figures are carried through the mathematical operations.
Frequently, you may be asked to use the density equation to solve for a variable other than density. For
example, you may be given measurements for density and mass and be asked to determine the volume.
Example:
The density of solid lead is 11.34 g/mL. What volume will 81.0 g of lead occupy?


V = 81.0 g11.34 g/mL = 7.14 mL
Since D = mV , then V = mD .


A common equation used in chemistry is PV = nRT . Even without knowing what these variables represent,
you can manipulate this equation to solve for any of the five quantities.


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P = nRTV V =
nRT
P n =


PV
RT R =


PV
nT T =


PV
nR


Make sure you recall these skills from algebra. If necessary, you should practice them.
Example:
Use the equation AB = CD and the values A = 15.1 g, B = 3.000 mL, and C = 326.96 grams to determine
the value of D.


D = BCA =
(3.000 mL)(326.96 g)


(15.1 g) = 65.0 mL


The calculator-determined value for this arithmetic may yield 64.956954 mL but you now know not to
report such a value. Since this answer only allows three significant figures, you must round the answer to
65.0 mL.


Conversion Factors
A conversion factor is a factor used to convert one unit of measurement into another unit. A simple
conversion factor can be used to convert meters into centimeters, or a more complex one can be used to
convert miles per hour into meters per second. Since most calculations require measurements to be in
certain units, you will find many uses for conversion factors. What must always be remembered is that a
conversion factor has to represent a fact; because the conversion factor is a fact and not a measurement,
the numbers in a conversion factor are exact. This fact can either be simple or complex. For instance, you
probably already know the fact that 12 eggs equal 1 dozen. A more complex fact is that the speed of light
is 1.86 × 105 miles/second. Either one of these can be used as a conversion factor, depending on the type
of calculation you might be working with.


Dimensional Analysis
Frequently, it is necessary to convert units measuring the same quantity from one form to another. For
example, it may be necessary to convert a length measurement in meters to millimeters. This process is
quite simple if you follow a standard procedure called dimensional analysis (also known as unit analysis or
the factor-label method). Dimensional analysis is a technique that involves the study of the dimensions
(units) of physical quantities. It is a convenient way to check mathematical equations. Dimensional analysis
involves considering the units you presently have and the units you wish to end up with, as well as designing
conversion factors that will cancel units you don’t want and produce units you do want. The conversion
factors are created from the equivalency relationships between the units. For example, one unit of work
is a newton meter (abbreviated N · m). If you have measurements in newtons (a unit for force, F) and
meters (a unit for distance, d), how would you calculate work? An analysis of the units will tell you that
you should multiply force times distance to get work: W = F × d.
Suppose you want to convert 0.0856 meters into millimeters. In this case, you need only one conversion
factor that will cancel the meters unit and create the millimeters unit. The conversion factor will be created
from the relationship 1000 mL = 1 m.


(0.0856 m) · (1000 mm1 m ) = (0.0856m) · (1000 mm1


m ) = 85.6 mm


In the above expression, the meter units will cancel and only the millimeter unit will remain.
Example:


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Convert 1.53 g to cg.
The equivalency relationship is 1.00 g = 100 cg, so the conversion factor is constructed from this equivalency
in order to cancel grams and produce centigrams.


(1.53 g) · (100 cg1 g ) = 153 cg


Example:
Convert 1000. in. to ft.
The equivalency between inches and feet is 12 in. = 1 ft. The conversion factor is designed to cancel inches
and produce feet.


(1000. in.) · ( 1 ft12 in.) = 83.33 ft


Each conversion factor is designed specifically for the problem. In the case of the conversion above, we
need to cancel inches, so we know that the inches component in the conversion factor needs to be in the
denominator.
Example:
Convert 425 klums to piks given the equivalency relationship 10 klums = 1 pik.


(425 klums) · ( 1 pik10 klums) = 42.5 piks


Sometimes, it is necessary to insert a series of conversion factors. Suppose we need to convert miles to
kilometers, and the only equivalencies we know are 1 mi = 5, 280 ft, 12 in. = 1 ft, 2.54 cm = 1 in.,
100 cm = 1 m, and 1000 m = 1 km. We will set up a series of conversion factors so that each conversion
factor produces the next unit in the sequence.
Example:
Convert 12 mi to km.


(12 mi) · (5280 ft1 mi ) · (12 in.1 ft ) · (2.54 cm1 in. ) · ( 1 m100 cm) · ( 1 km1000 m) = 19 km


In each step, the previous unit is canceled and the next unit in the sequence is produced.
Conversion factors for area and volume can also be produced by this method.
Example:
Convert 1500 cm2 to m2.


(1500 cm2) · ( 1 m100 cm)2 = (1500 cm2) · ( 1 m
2


10,000 cm2 ) = 0.15 m2


Example:
Convert 12.0 in3 to cm3.


(12.0 in3) · (2.54 cm1 in )3 = (12.0 in3) · (16.4 cm
3


1 in3 ) = 197 cm
3


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Lesson Summary
• Conversion factors are used to convert one unit of measurement into another unit.
• Dimensional analysis involves considering both the units you presently have and the units you wish
to end up with, as well as designing conversion factors that will cancel units you don’t want and
produce units you do want.


Further Reading / Supplemental Links
Visit this website for a video series that reviews topics on measurement


• http://www.learner.org/resources/series184.html


This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 2-4 is
on dimensional analysis.


• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson24.htm


Review Questions
1. For the equation PV = nRT , re-write it so that it is in the form of T =.
2. The equation for density is D = mV . If D is 12.8 g/cm


3 and m is 46.1 g, solve for V, keeping significant
figures in mind.


3. The equation P1 ·V1 = P2 ·V2, known as Boyle’s law, shows that gas pressure is inversely proportional
to its volume. Re-write Boyle’s law so it is in the form of V1 =.


4. The density of a certain solid is measured and found to be 12.68 g/mL. Convert this measurement
into kg/L.


5. In a nuclear chemistry experiment, an alpha particle is found to have a velocity of 14, 285 m/s.
Convert this measurement into miles/hour (mi/h).


1.6 Scientific Notation
Lesson Objectives
The student will:


• use scientific notation to express large and small numbers.
• add, subtract, multiply, and divide using scientific notation.


Vocabulary
• scientific notation


Introduction
Work in science frequently involves very large and very small numbers. The speed of light, for example,
is 300,000,000 m/s; the mass of the earth is 6,000,000,000,000,000,000,000,000 kg; and the mass of an


www.ck12.org 24




electron is 0.0000000000000000000000000000009 kg. It is very inconvenient to write out such numbers
and even more inconvenient to attempt to carry out mathematical operations with them. Scientists and
mathematicians have designed an easier method to deal with such long numbers. This more convenient
system is called exponential notation by mathematicians and scientific notation by scientists.


What is Scientific Notation?
In scientific notation, very large and very small numbers are expressed as the product of a number
between 1 and 10 multiplied by some power of 10. For example, the number 9, 000, 000 can be written as
the product of 9 times 1, 000, 000. In turn, 1, 000, 000 can be written as 106. Therefore, 9, 000, 000 can be
written as 9 × 106. In a similar manner, 0.00000004 can be written as 4 times 1108 , or 4 × 10−8.


Table 1.7: Examples of Scientific Notation


Decimal Notation Scientific Notation
95, 672 9.5672 × 104
8, 340 8.34 × 103
100 1 × 102
7.21 7.21 × 100
0.014 1.4 × 10−2
0.0000000080 8.0 × 10−9
0.00000000000975 9.75 × 10−12


As you can see from the examples in Table 1.7, to convert a number from decimal form into scientific
notation, you count the number of spaces needed to move the decimal, and that number becomes the
exponent of 10. If you are moving the decimal to the left, the exponent is positive, and if you are moving
the decimal to the right, the exponent is negative. You should note that all significant figures are maintained
in scientific notation. You will probably realize that the greatest advantage of using scientific notation
occurs when there are many non-significant figures.


Scientific Notation in Calculations
Addition and Subtraction
When numbers in exponential form are added or subtracted, the exponents must be the same. If the
exponents are the same, the coefficients are added and the exponent remains the same.
Example:


(4.3 × 104) + (1.5 × 104) = (4.3 + 1.5) × 104 = 5.8 × 104


Note that the example above is the same as:


43, 000 + 15, 000 = 58, 000 = 5.8 × 104.


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Example:


(8.6 × 107) − (5.3 × 107) = (8.6 − 5.3) × 107 = 3.3 × 107


Example:


(8.6 × 105) + (3.0 × 104) = ?


These two exponential numbers do not have the same exponent. If the exponents of the numbers to be
added or subtracted are not the same, then one of the numbers must be changed so that the two numbers
have the same exponent. In order to add them, we can change the number 3.0 × 104 to 0.30 × 105. This
change is made by moving the decimal one place to the left and increasing the exponent by one. Now the
two numbers can be added.


(8.6 × 105) + (0.30 × 105) = (8.6 + 0.30) × 105 = 8.9 × 105


We could also have chosen to alter the other number. Instead of changing the second number to a higher
exponent, we could have changed the first number to a lower exponent.


(86 × 104) + (3.0 × 104) = (86 + 3.0) × 104 = 89 × 104
8.6 × 105 becomes 86 × 104


Even though it is not always necessary, the preferred practice is to express exponential numbers in proper
form, which has only one digit to the left of the decimal. When 89 × 104 is converted to proper form, it
becomes 8.9 × 105, which is precisely the same result as before.


Multiplication and Division
When multiplying or dividing numbers in exponential form, the numbers do not have to have the same
exponents. To multiply exponential numbers, multiply the coefficients and add the exponents. To divide
exponential numbers, divide the coefficients and subtract the exponents.
Multiplication Examples:


(4.2 × 104) · (2.2 × 102) = (4.2 · 2.2) × 104+2 = 9.2 × 106


The product of 4.2 and 2.2 is 9.24, but since we are limited to two significant figures, the coefficient is
rounded to 9.2.


(8.2 × 10−9) · (8.2 × 10−4) = (8.2 · 8.2) × 10(−9)+(−4) = 67.24 × 10−13
(2 × 10−5) · (4 × 10−4) = (2 · 4) × 10(−5)+(−4) = 8 × 10−9
(2 × 10−9) · (4 × 104) = (2 · 4) × 10−9+4 = 8 × 10−5
(2 × 109) · (4 × 1014) = (2 · 4) × 109+14 = 8 × 1023


In this last example, the product has too many significant figures and is not in proper exponential form.
We must round to two significant figures and adjust the decimal and exponent. The correct answer would
be 6.7 × 10−12.
Division Examples:


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4.6×103
2.3×10−4 = 2.0 × 10


(3)−(−4) = 2.0 × 107
8×10−7
2×10−4 = 4 × 10


(−7)−(−4) = 4 × 10−3
8×107
2×104 = 4 × 10


7−4 = 4 × 103


In the example above, since the original coefficients have two significant figures, the answer must also have
two significant figures. Therefore, the zero in the tenths place is written to indicate the answer has two
significant figures.


Lesson Summary
• Very large and very small numbers in science are expressed in scientific notation.
• All significant figures are maintained in scientific notation.
• When numbers in exponential form are added or subtracted, the exponents must be the same. If the
exponents are the same, the coefficients are added and the exponent remains the same.


• To multiply exponential numbers, multiply the coefficients and add the exponents.
• To divide exponential numbers, divide the coefficients and subtract the exponents.


Further Reading / Supplemental Links
This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 2-5 is
on scientific notation.


• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson25.htm


Review Questions
1. Write the following numbers in scientific notation.


(a) 0.0000479
(b) 251, 000, 000
(c) 4, 260
(d) 0.00206


Do the following calculations without a calculator.


2. (2.0 × 103) · (3.0 × 104)
3. (5.0 × 10−5) · (5.0 × 108)
4. (6.0 × 10−1) · (7.0 × 10−4)
5. (3.0×10−4)·(2.0×10−4)2.0×10−6


Do the following calculations.


6. (6.0 × 107) · (2.5 × 104)
7. 4.2×10−43.0×10−2


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1.7 Evaluating Measurements
Lesson Objectives
The student will:


• define accuracy and precision.
• explain the difference between accuracy and precision.
• indicate whether a given data set is precise, accurate, both, or neither.
• calculate percent error in an experiment.


Vocabulary
• accuracy
• percent error
• precision


Introduction
Accuracy and precision are two words that we hear a lot in science, math, and other everyday events. They
are also, surprisingly, two words that are often misused. For example, you may hear car advertisements
talking about the car’s ability to handle precision driving. But what do these two words mean?


Accuracy and Precision
Every measurement compares the physical quantity being measured with a fixed standard of measurement,
such as the centimeter or the gram. In describing the reliability of a measurement, scientists often use
the terms accuracy and precision. Accuracy refers to how close a measurement is to the true value of
the quantity being measured. Precision refers to how close the values in a set of measurements are to
one another. If you are using a flawed measuring instrument, you could get very precise measurements
(meaning they are very reproducible), but the measurements would be inaccurate. In many cases, the
true value of the measurement is not known, and we must take our measurement as the true value. In
such cases, instruments are checked carefully to verify that they are unflawed before a series of precise
measurements are made. It is assumed that good instruments and precise measurements imply accuracy.
Suppose a student made the same volume measurement four times and obtained the following measure-
ments: 34.25 mL, 34.45 mL, 34.33 mL, and 34.20 mL. The average of these four readings is 34.31 mL. If
the actual volume was known to be 34.30 mL, what could we say about the accuracy and precision of these
measurements, and how much confidence would we have in the answer? Since the final average is very
close to the actual value, we would say that the answer is accurate. However, the individual readings are
not close to each other, so we would conclude that the measurements were not precise. If we did not know
the correct answer, we would have very little confidence that these measurements produced an accurate
value.
Consider the values obtained by another student making the same measurements: 35.27 mL, 35.26 mL,
35.27 mL, and 35.28 mL. In this case, the average measurement is 35.27 mL, and the set of measurements
is quite precise since all readings are within 0.1 mL of the average measurement. We would normally have
confidence in this measurement since the precision is so good, but if the actual volume is 34.30 mL, the
measurements are not accurate. Generally, situations where the measurements are precise but not accurate


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are caused by a flawed measuring instrument. The ideal situation is to have quality measuring instruments
so that precision will imply accuracy.


Percent Error
Percent error is a common way of evaluating the accuracy of a measured value. Anytime an experiment
is conducted, a certain degree of uncertainty and error is expected. Scientists often express this uncertainty
and error in measurement by reporting a percent error.


percent error = (accepted value - experimental value)(accepted value) × 100%


The experimental value is what you recorded or calculated based on your own experiment in the lab. The
value that can be found in reference tables is called the accepted value. Percent error is a measure of how
far the experimental value is from the accepted value.
Example:
A student determined the density of a sample of silver to be 10.3 g/cm3. The density of silver is actually
10.5 g/cm3. What is the percent error in the experimentally determined density of silver?


percent error = 10.5 g/cm
3−10.3 g/cm3


10.5 g/cm3 × 100% = 1.90%


Lesson Summary
• Accuracy reflects how close the measured value is to the actual value.
• Precision reflects how close the values in a set of measurements are to each other.
• Accuracy is affected by the quality of the instrument or measurement.
• Percent error is a common way of evaluating the accuracy of a measured value.
• percent error = (accepted value - experimental value)(accepted value) × 100%


Further Reading / Supplemental Links
This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 2-2 is
on accuracy and precision.


• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson22.htm


The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos.
You are required to register before you can watch the videos, but there is no charge to register. The website
has a video that apply to this lesson called ‘‘Measurement: The Foundation of Chemistry” that details the
value of accuracy and precision.


• http://learner.org/resources/series61.html


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Review Questions
1. Suppose you want to hit the center of this circle with a paint ball gun. Which of the following are
considered accurate? Precise? Both? Neither?


2. Four students take measurements to determine the volume of a cube. Their results are 15.32 cm3,
15.33 cm3, 15.33 cm3, and 15.31 cm3. The actual volume of the cube is 16.12 cm3. What statement(s)
can you make about the accuracy and precision in their measurements?


3. Distinguish between accuracy and precision.
4. Nisi was asked the following question on her lab exam: When doing an experiment, what term best
describes the reproducibility in your results? What should she answer?
(a) accuracy
(b) care
(c) precision
(d) significance
(e) uncertainty


5. Karen was working in the lab doing reactions involving mass. She needed to weigh out 1.50 g of each
reactant and put them together in her flask. She recorded her data in her data table (Table 1.8).
What can you conclude by looking at Karen’s data?
(a) The data is accurate but not precise.
(b) The data is precise but not accurate.
(c) The data is neither precise nor accurate.
(d) The data is precise and accurate.
(e) You really need to see the balance Karen used.


Table 1.8: Data Table for Problem 5


Mass of Reactant A Mass of Reactant B
Trial 1 1.47 ± 0.02 g 1.48 ± 0.02 g
Trial 2 1.46 ± 0.02 g 1.46 ± 0.02 g
Trial 3 1.48 ± 0.02 g 1.50 ± 0.02 g


6. John uses his thermometer and finds the boiling point of ethanol to be 75◦C. He looks in a reference
book and finds that the actual boiling point of ethanol is 78◦C. What is his percent error?


7. The density of water at 4◦C is known to be 1.00 g/mL. Kim experimentally found the density of
water to be 1.085 g/mL. What is her percent error?


8. An object has a mass of 35.0 g. On a digital balance, Huey finds the mass of the object to be 34.92 g.
What is the percent error of his balance?


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1.8 Graphing
Lesson Objectives
The student will:


• correctly graph data with the proper scale, units, and best fit curve.
• recognize patterns in data from a graph.
• solve for the slope of given line graphs.


Vocabulary
• extrapolation
• graph
• interpolation
• slope


Introduction
Scientists search for regularities and trends in data. To make it easier to find these regularities and trends,
scientists often present data in either a table or a graph. The table below presents data about the pressure
and volume of a sample of gas. You should note that all tables have a title and include the units of the
measurements. The unit of pressure used here is atm (atmosphere).


You may note a regularity that appears in this table: as the pressure of the gas increases, its volume
decreases. This regularity or trend becomes even more apparent in a graph of this data. A graph is a
pictorial representation of the relationship between variables on a coordinate system.


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When the data from Data Table A is plotted as a graph, the trend in the relationship between the pressure
and volume of a gas sample becomes more apparent. The graph aids the scientist in the search for any
regularity that may exist in the data.


Drawing Line Graphs
Reading information from a line graph is easier and more accurate as the size of the graph increases. In the
example below, the graph on the left uses only a small fraction of the space available on the graph paper.
The graph on the right shows the same data but uses all the space available. If you were attempting to
determine the pressure at a temperature of 110 K, using the graph on the left would give a less accurate
result than using the graph on the right.


When you draw a line graph, you should arrange the numbers on the axes to use as much of the graph
paper as you can. If the lowest temperature in your data is 100 K and the highest temperature in your
data is 160 K, you should arrange for 100 K to be on the extreme left of your graph and 160 K to be on
the extreme right of your graph. The creator of the graph on the left did not take this advice and did not
produce a very good graph. You should also make sure that the axes on your graph are labeled and that
your graph has a title.


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Reading Information from a Graph


When we draw a line graph from a set of data points, we are inferring a trend and constructing new data
points between known data points. This process is called interpolation. Even though we may only have a
few data points, we are estimating the values between measured points, assuming that the line connecting
these data points is a good model of what we’re studying.
Consider the following set of data for the solubility of KClO3 in water. Data Table B shows that there are
exactly six measured data points. When the data is graphed, however, the graph maker assumes that the
relationship between the temperature and the solubility exists for all points within the data range. The
graph maker draws a line by interpolating the data points between the actual data points. Note that the
line is not drawn by just connecting the data points in a connect-the-dot manner. Instead, the line is a
smooth curve that reasonably connects the known data points.


We can now read Graph B1, shown below, for points that were not actually measured. If we wish to
determine the solubility of KClO3 at 70◦C, we follow the vertical grid line for 70◦C up to where it touches
the graphed line and then follow the horizontal grid line to the axis to read the solubility. In this case, we
would read the solubility to be 30.0 g/100 mL of H2O at 70◦C.


There are also occasions when scientists wish to know more about points that are outside the range of
measured data points. Extending the line graph beyond the ends of the original line, using the basic shape
of the curve as a guide, is called extrapolation.


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Suppose the graph for the solubility of potassium chlorate has been made from just three measured data
points. If the actual data points for the curve were the solubility at 60◦C, 80◦C, and 100◦C, the graph would
be the solid line shown in Graph B2 above. If the solubility at 30◦C was desired, we could extrapolate the
curve (the dotted line) and obtain a solubility of 5.0 g/100 mL of H2O. If we check the more complete graph
above (Graph B1), you can see that the solubility at 30◦C is closer to 10. g/100 mL of H2O. The reason the
second graph produces such a different answer is because the real behavior of potassium chlorate in water
is more complicated than the behavior suggested by the extrapolated line. For this reason, extrapolation is
only acceptable for graphs where there is evidence that the relationship shown in the graph will hold true
beyond the ends of the graph. Extrapolation is more dangerous that interpolation in terms of producing
possibly incorrect data.
In situations where it is unreasonable to interpolate or extrapolate data points from the actual measured
data points, a line graph should not be used. If it is desirable to present data in a graphic form but a line
graph is not useful, a bar graph can often be used instead. Consider the data in the following table.


For this set of data, you would not plot the data on a line graph because interpolating between years does


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not make sense; the concept of the average yearly rainfall halfway between the years 1980 and 1981 would
not make sense. Looking at the general trend exhibited by Data Table C also does not provide the slightest
amount of evidence about the rainfall in 1979 or 1990. Therefore, the interpolation and extrapolation of
the data in this table is not reasonable. If we wish to present this information in a graphic form, a bar
graph like the one seen in Graph C would be best.


From this bar graph, you could very quickly answer questions like, ‘‘Which year was most likely a drought
year for Trout Creek?” and ‘‘Which year was Trout Creek most likely to have suffered from a flood?”


Finding the Slope of a Graph


As you may recall from algebra, the slope of the line may be determined from the graph. The slope
represents the rate at which one variable is changing with respect to the other variable. For a straight-line
graph, the slope is constant for the entire line, but for a non-linear graph, the slope varies at different
points along the line. For a straight-line graph, the slope for all points along the line can be determined
from any section of the graph. Consider the following data table and the linear graph that follows.


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The relationship in this set of data is linear; in other words, the data produces a straight-line graph. The
slope of this line is constant at all points on the line. The slope of a line is defined as the rise (change
in vertical position) divided by the run (change in horizontal position). For a pair of data points, the
coordinates of the points are identified as (x1, y1) and (x2, y2). In this case, the data points selected are
(40◦C, 65 mL) and (100◦C, 80 mL). The slope can then be calculated in the following manner:


slope m = riserun =
(y2−y1)
(x2−x1) =


(80 mL−65 mL)
(100◦C−40◦C) = 0.25 mL/◦C


Therefore, the slope of the line is 0.25 mL/◦C. The fact that the slope is positive indicates that the line
is rising as it moves from left to right and that the volume increases by 0.25 mL for each 1◦C increase in
temperature. A negative slope would indicate that the line was falling as it moves from left to right.
For a non-linear graph, the slope must be calculated for each point independently. Since the line will be
a curve, the slope is calculated from the tangent to the curve at the point in question. Data Table E and
Graph E are for a reaction in which the concentration of one of the reactants, bromine, was measured
against time. The concentration is expressed in moles/liter, which is symbolized by M.


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In order to determine the slope at some point on a curved line, a tangent (approximate) is drawn in as a
line that just touches the point in question. Once the tangent has been drawn, the slope of the tangent is
determined, which is also the slope of the curve at that point. In the graph above, the tangent has been
drawn at the point where t = 2 seconds. We determine the x- and y-coordinates for two points along the
tangent line (as best we can) and use the coordinates of those two points to calculate the slope of the
tangent. The coordinates of the point at the left end of the tangent line is determined to be (1.00 s, 0.056
M). The coordinates of the point at the right end of the line is harder to determine, and we are guessing
that the coordinates are (3.25 s, 0.031 M).


slope m = riserun =
(y2−y1)
(x2−x1) =


(0.031 M−0.056 M)
(3.25 s−1.00 s) = −0.011 M/s


Since the slope is a negative number, we know the line is decreasing in height. At t = 2 seconds, the
concentration of bromine is decreasing at a rate of 0.011 moles/liter per second. At other points along
this curve, the slope would be different. From the appearance of the curve, it is apparent that the slope
is negative (the concentration of bromine is decreasing) all along the line, but it appears to be decreasing
more quickly at the beginning of the reaction and less quickly as time increases.


Lesson Summary
• Tables and graphs are two common methods of presenting data that aid in the search for regularities
and trends within the data.


• When we draw a line graph from a set of data points, we are inferring a trend and constructing new
data points between known data points. This process is called interpolation.


• Constructing data points beyond the end of a line graph, using the basic shape of the curve as a
guide, is called extrapolation.


• The slope of a graph represents the rate at which one variable is changing with respect to the other
variable.


• For a straight-line graph, the slope for all points along the line can be determined from any section
of the graph.


• For a non-linear graph, the slope must be determined for each point by drawing a tangent line to the
curve at the point in question.


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Further Reading / Supplemental Links
This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 16-5 is
on graphing. The lesson provides some tips and some sample data to practice graphing with.


• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson165.htm


Review Questions
1. What would you do to find the slope of a curved line?
2. Andrew was completing his density lab for his chemistry lab exam. He collected the following data
in his data table (shown in Table 1.9).
(a) Draw a graph to represent the data.
(b) Calculate the slope.
(c) What does the slope of the line represent?


Table 1.9: Data Table for Problem 2


Mass of Solid (g) Volume of Solution (mL)
3.4 0.3
6.8 0.6
10.2 0.9
21.55 1.9
32.89 2.9
44.23 3.9
55.57 4.9


3. Donna is completing the last step in her experiment to find the effect of the concentration of ammonia
on the reaction. She has collected the following data from her time trials and is ready for the analysis.
Her data table is Table 1.10. Help Donna by graphing the data, describing the relationship, finding
the slope, and then discussing the meaning of the slope.


Table 1.10: Data Table for Problem 3


Time (s) Concentration (mol/L)
0.20 49.92
0.40 39.80
0.60 29.67
0.81 20.43
1.08 14.39
1.30 10.84
1.53 5.86
2.00 1.95
2.21 1.07
2.40 0.71
2.60 0.71


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All images, unless otherwise stated, are created by CK-12 Foundation and are under the Creative Commons
license CC-BY-NC-SA.


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Chapter 2


Independent and Dependent
Events


Introduction
Probability is present in many parts of our everyday lives. When you say it is raining and your numbers
came up in the lottery, you are talking about two independent events. The fact that it is raining is not
dependent on the fact that your numbers came up in the lottery, and vice versa. If you say that you have
the flu and you are taking medicine, you are talking about dependent events. The terms dependent and
independent in mathematics are the same as those found in the English language. In order to determine
probability mathematically, we need to understand the differences in these definitions. Independent events
are those where the outcome of one event is not affected by the other, and dependent events are events
where the outcome of one is affected by the other. Other terms, such as mutually inclusive and mutually
exclusive, are also important. With mutually inclusive events, the concept of double counting using the
addition principle is used in the calculation of the probability.


2.1 Independent Events
Learning Objectives


• Know the definition of the notion of independent events.
• Use the rules for addition, multiplication, and complementation to solve for probabilities of particular
events in finite sample spaces.


What’s in a word?
The words dependent and independent are used by students and teachers on a daily basis. In fact, they
are probably used frequently! You may tell your parent or guardian that you are independent enough to
go to the movies on your own with your friends. You could say that when you bake a cake or make a cup
of hot chocolate, the taste of these are dependent on what ingredients you use. In the English language,
the term dependent means to be unable to do without, whereas independent means to be free from any
outside influence.
What about in mathematics? What do the terms dependent and independent actually mean? This lesson
will explore the mathematics of independence and dependence.
What are Venn Diagrams and Why are They Used?


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In probability, a Venn diagram is a graphic organizer that shows a visual representation for all possible
outcomes of an experiment and the events of the experiment in ovals. Normally, in probability, the Venn
diagram will be a box with overlapping ovals inside. Look at the diagram below.


The S represents all of the possible outcomes of a set. It is called the sample space. The ovals A and B
represent the outcomes of the events that occur in the sample space. Let’s look at an example.
Let’s say our sample space is the numbers from 1 to 10. Event A will be the odd numbers from 1 to 10,
and event B will be all the prime numbers. Remember that a prime number is a number where the only
factors are 1 and itself. Now let’s draw the Venn diagram to represent this example.
We know that:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 3, 5, 7, 9}
B = {1, 3, 5, 7}


Notice that the prime numbers are part of both sets and are, therefore, in the overlapping part of the Venn
diagram. The numbers 2, 4, 6, 8, and 10 are the numbers not part of A or B, but they are still members
of the sample space. Now you try.
Example 1
Two coins are tossed. Event A consists of the outcomes when tossing heads on the first toss. Event B
consists of the outcomes when tossing heads on the second toss. Draw a Venn diagram to represent this
example.
Solution:
We know that:
S = {HH,HT,TH,TT }
A = {HH,HT }
B = {HH,TH}


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Notice that Event A and Event B share the Heads + Heads outcome and that the sample space contains
Tails + Tails, which is neither in Event A nor Event B.
Example 2
In ABC high school, 30 percent of the students have a part-time job. 25 percent of the students from
the high school are on the honor roll. Event A represents the students holding a part-time job. Event B
represents the students on the honor roll. Draw a Venn diagram to represent this example.
Solution:
We know that:
S = {students in ABC high school}
A = {students holding a part-time job}
B = {students on the honor roll}


Notice that the overlapping oval for A and B represents the students who have a part-time job and are on
the honor roll. The sample space S outside the ovals represents students neither holding a part-time job
nor on the honor roll.
In a Venn diagram, when events A and B occur, the symbol used is ∩. Therefore, A ∩ B is the intersection
of events A and B and can be used to find the probability of both events occurring. If, in a Venn diagram,
either A or B occurs, the symbol is ∪. This symbol would represent the union of events A and B, where
the outcome would be in either A or B.
Example 3
You are asked to roll a die. Event A is the event of rolling a 1, 2, or a 3. Event B is the event of rolling a
3, 4, or a 5. Draw a Venn diagram to represent this example. What is A ∩ B? What is A ∪ B?
Solution
We know that:
S = {1, 2, 3, 4, 5, 6}
A = {1, 2, 3}
B = {3, 4, 5}


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A ∩ B = {3}
A ∪ B = {1, 2, 3, 4, 5}
Independent Events
In mathematics, the term independent means to have one event not dependent on the other. It is similar
to the English definition. Suppose you are trying to convince your parent/guardian to let you go to the
movies on your own. Your parent/guardian is thinking that if you go, you will not have time to finish
your homework. For this reason, you have to convince him/her that you are independent enough to go to
the movies and finish your homework. Therefore, you are trying to convince your parent/guardian that
the two events, going to the movies and finishing your homework, are independent. This is similar to the
mathematical definition. Say you were asked to pick a card from a deck of cards and roll a six on a die. It
does not matter if you choose the card first and roll a six second, or vice versa. The probability of rolling
the six would remain the same, as would the probability of choosing the card.
Going back to our Venn diagrams, independent events are represented as those events that occur in
both sets. If we look just at Example 2, Event A is a student holding a part-time job. Event B is the
student being on the honor roll. These two events are independent of each other. In other words, whether
you hold a part-time job is not dependent on you being on the honor roll, or vice versa. The outcome of
one event is not dependent on the outcome of the second event. To calculate the probabilities, you would
look at the overlapping part of the diagram. A and B represent the probability of both events occurring.
Let’s look at the probability calculation.


P(A) = 30% or 0.30
P(B) = 25% or 0.25


P(A and B) = P(A) × P(B)
P(A and B) = 0.30 × 0.25
P(A and B) = 0.075


In other words, 7.5% of the students of ABC high school are both on the honor roll and have a part-time
job.
In Example 1, two coins are tossed. Remember that Event A consists of the outcomes when getting heads
on the first toss, and Event B consists of the outcomes when getting heads on the second toss. What would
be the probability of tossing the coins and getting a head on both the first coin and the second coin? We


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know that the probability of getting a head on a coin toss is 12 , or 50%. In other words, we have a 50%
chance of getting a head on a toss of a fair coin and a 50% chance of getting a tail.


P(A) = 50% or 0.50
P(B) = 50% or 0.50


P(A and B) = P(A) × P(B)
P(A and B) = 0.50 × 0.50
P(A and B) = 0.25


Therefore, there is a 25% chance of getting two heads when tossing two fair coins.
Example 4
Two cards are chosen from a deck of cards. The first card is replaced before choosing the second card.
What is the probability that they both will be sevens?


Solution
Let A = 1st seven chosen.
Let B = 2nd seven chosen.
A little note about a deck of cards


A deck of cards consists of 52 cards.
Each deck has four parts (suits) with 13 cards in them.
Each suit has 3 face cards.


4 suits 1 seven per suit
↘ ↙


The total number of sevens in the deck = 4 × 1 = 4.


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Since the card was replaced, these events are independent:


P(A) = 4
52


Note: the total number of cards is


P(B) = 4
52


↙ 52 after choosing the first card,
because the first card is replaced.


P(A and B) = 4
52


× 4
52
or P(A ∩ B) = 4


52
× 4


52


P(A ∩ B) = 16
2704


P(A ∩ B) = 1
169


Example 5
The following table represents data collected from a grade 12 class in DEF High School.


Table 2.1: Plans after High School


Gender University Community College Total
Males 28 56 84
Females 43 37 80
Total 71 93 164


Suppose one student was chosen at random from the grade 12 class.
(a) What is the probability that the student is female?
(b) What is the probability that the student is going to university?
Now suppose two people both randomly chose one student from the grade 12 class. Assume that it’s
possible for them to choose the same student.
(c) What is the probability that the first person chooses a student who is female and the second person
chooses a student who is going to university?
Solution:


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Probabilities: P(female) = 80
164


↙ 164 total students


P(female) = 20
41


P(going to university) = 71
164


P(female) × P(going to university) = 20
41


× 71
164


= 1420
6724


= 355
1681


= 0.211


Therefore, there is a 21.1% probability that the first person chooses a student who is female and the second
person chooses a student who is going to university.


2.2 Dependent Events
For two events to be dependent, the probability of the second event is dependent on the probability of the
first event. In English, remember, the term dependent means to be unable to do without. This is similar
to the mathematical definition of dependent events, where the second event is unable to happen without
the first event occurring.
Example 6
Remember in Example 4 when you were asked to determine the probability of drawing 2 sevens from a
standard deck of cards? What would happen if one card is chosen and not replaced? In this case, the
probability of drawing a seven on the second draw is dependent on drawing a seven on the first draw. Now
let’s calculate the probability of the two cards being drawn without replacement. This can be done with
the Multiplication Rule.
Solution
Let A = 1st seven chosen.
Let B = 2nd seven chosen.


4 suits 1 seven per suit
↘ ↙


The total number of sevens in the deck = 4 × 1 = 4.


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P(A) = 4
52


Note: the total number of cards is


P(B) = 3
51


↙ 51 after choosing the first card if
it is not replaced.


P(A and B) = 4
52


× 3
51
or P(A ∩ B) = 4


52
× 3


51


P(A ∩ B) = 12
2652


P(A ∩ B) = 1
221


Notice in this example that the numerator and denominator decreased from P(A) to P(B). Once we picked
the first card, the number of cards available from the deck dropped from 52 to 51. The number of sevens
also decreased from 4 to 3. Again, the explanation given in the example was that the first card chosen was
kept in your hand and not replaced into the deck before the second card was chosen.
Example 7
A box contains 5 red marbles and 5 purple marbles. What is the probability of drawing 2 purple marbles
and 1 red marble in succession without replacement?
Solution
On the first draw, the probability of drawing a purple marble is:


On the third draw the probability of drawing a red marble is:


P(red) = 5
8


Therefore, the probability of drawing 2 purple marbles and 1 red marble is:


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P(1 purple and 1 purple and 1 red) = P(1P ∩ 1P ∩ 1R)
= P1(purple) × P2(purple) × P(red)


= 5
10


× 4
9
× 5


8
= 100


720
= 5


36


2.3 Mutually Inclusive and Mutually Exclusive
Events


When determining the probabilities of events, we must also look at two additional terms. These terms are
mutually inclusive and mutually exclusive. When we add probability calculations of events described by
these terms, we can apply the words and (∩) and or (∪) using the addition rule. Let’s take another look
at the Venn diagrams when defining these terms.
Two events A and B that cannot occur at the same time are mutually exclusive events. They have no
common outcomes. See in the diagram below that P(A and B) = 0. Notice that there is no intersection
between the possible outcomes in Event A and the possible outcomes in event B. For example, if you were
asked to pick a number between one and ten, you cannot pick a number that is both even and odd. These
events are mutually exclusive.


To calculate the probability of picking a number from 1 to 10 that is even or picking a number from 1 to
10 that is odd, you would follow the steps below.
A = {2, 4, 6, 8, 10}


P(A) = 510


B = {1, 3, 5, 7, 9}


P(B) = 510


P(A or B) = 510 + 510


P(A or B) = 1010


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P(A or B) = 1
The probability of picking a number from 1 to 10 that is even and picking a number from 1 to 10 that is
odd would just be 0, since these are mutually exclusive events. In other words, P(A and B) = 0.
If events A and B share some overlap in the Venn diagram, they may be considered not mutually exclusive
events, but mutually inclusive events. Look at the diagrams below to see how these events can occur.
Mutually inclusive events can occur at the same time. Say, for example, in the problem above, you wanted
to pick a number from 1 to 10 that is less than 4 and pick an even number.
A = {1, 2, 3}


P(A) = 310


B = {2, 4, 6, 8, 10}


P(B) = 510


P(A and B) = 110


The reason why P(A and B) = 110 is because there is only one number from 1 to 10 that is both less than
4 and even, and that number is 2.
When representing this on the Venn diagram, we would see something like the following:


Mutually exclusive events, remember, cannot occur at the same time. Mutually inclusive events can. Look
at the Venn diagram below. What do you think we need to do in order to calculate the probability of A∪B
just from looking at this diagram?


If you look at the diagram, you see that the calculation involves not only P(A) and P(B), but also P(A∩B).
However, the items in A ∩ B are also part of Event A and Event B. To represent the probability of A or B,
we need to subtract the P(A ∩ B), otherwise we are double counting. In other words:


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P(A or B) = P(A) + P(B) − P(A and B)
or


P(A ∪ B) = P(A) + P(B) − P(A ∩ B)


where ∩ represents and and ∪ represents or.
This is known as the Addition Principle (Rule).
Addition Principle
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A ∩ B) = 0 for mutually exclusive events.
Think about the idea of rolling a die. Event A is the sample set where the number rolled on the die is odd.
Event B is the sample set where the number rolled is greater than 2.
Event A = {1, 3, 5}
Event B = {3, 4, 5, 6}
Notice that the events have two elements in common. Therefore, they are not mutually exclusive but
mutually inclusive.
What if we said that we are choosing a card from a deck of cards? Event A is the sample set where the
card chosen is an 8. Event B is the sample set where the card chosen is an Ace (A).


Notice that the events have no elements in common. Therefore, they are mutually exclusive.
Look at the example below to understand the concept of double counting.
Example 8
What is the probability of choosing a card from a deck of cards that is a club or a ten?
Solution


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P(A) = probability of selecting a club


P(A) = 13
52


P(B) = probability of selecting a ten


P(B) = 4
52


P(A ∩ B) = 1
52


P(A ∪ B) = P(A) + P(B) − P(A ∩ B)


P(A ∪ B) = 13
52


+ 4
52


− 1
52


P(A ∪ B) = 16
52


P(A ∪ B) = 4
13


Example 9
What is the probability of choosing a number from 1 to 10 that is less than 5 or odd?
Solution


A = {1, 2, 3, 4}
P(A) = probability of selecting a number less than 5


P(A) = 4
10


P(A) = 2
5


B = {1, 3, 5, 7, 9}
P(B) = probability of selecting a number that is odd


P(B) = 5
10


P(B) = 1
2


P(A ∩ B) = 2
10


P(A ∩ B) = 1
5


P(A ∪ B) = P(A) + P(B) − P(A ∩ B)


P(A ∪ B) = 2
5


+ 1
2
− 1


5
P(A ∪ B) = 4


10
+ 5


10
− 2


10
P(A ∪ B) = 7


10


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Notice in the previous two examples how the concept of double counting was incorporated into the calcu-
lation by subtracting the P(A ∪ B). Let’s try a different example where you have two events happening.
Example 10: Two fair die are rolled. What is the probability of getting a sum less than seven or a sum
equal to 10?
Solution:
P(A) = probability of obtaining a sum less than seven


P(A) = 1536


P(B) = probability of obtaining a sum equal to 10


P(B) = 136


There are no elements that are common, so the events are mutually exclusive.


P(A and B) = 0
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)


P(A ∪ B) = 15
36


+ 1
36


− 0


P(A ∪ B) = 16
36


P(A ∪ B) = 4
9


Example 11: Two fair die are rolled. What is the probability of getting a sum less than seven or a sum
less than 4?
Solution
P(A) = probability of obtaining a sum less than seven


P(A) = 1536


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P(B) = probability of obtaining a sum less than four


P(B) = 336


Notice that there are three elements in common. Therefore, the events are not mutually exclusive, and we
must account for the double counting.


P(A and B) = 3
36


P(A ∩ B) = 3
36


P(A ∩ B) = 1
12


P(A ∪ B) = P(A) + P(B) − P(A ∩ B)


P(A ∪ B) = 15
36


+ 3
36


− 1
12


P(A ∪ B) = 15
36


+ 3
36


− 3
36


P(A ∪ B) = 15
36


P(A ∪ B) = 5
12


Points to Consider


• What is the difference between the probabilities calculated with the multiplication rule versus the
addition rule?


• Can mutually exclusive events be independent? Can they be dependent?


Vocabulary


Addition Principle (Rule) With two events, the probability of one event occurring or another is given
by: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).


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Dependent events Two or more events whose outcomes affect each other. The probability of occurrence
of one event depends on the occurrence of the other.


Independent events Two or more events whose outcomes do not affect each other.


Multiplication Rule For two independent events (A and B) where the outcome of A does not change
the probability of B, the probability of A and B is given by: P(A and B) = P(A) × P(B).


Mutually exclusive events Two outcomes or events are mutually exclusive when they cannot both
occur simultaneously.


Mutually inclusive events Two outcomes or events are mutually inclusive when they can both occur
simultaneously.


Outcomes The possible results of one trial of a probability experiment.


Probability The chance that something will happen.


Random When everyone in a population has an equal chance of being selected; not only is each person
or thing equally likely, but all groups of persons or things are also equally likely.


Sample space The set of all possible outcomes of an event or group of events.


Venn diagram A diagram of overlapping circles that shows the relationships among members of different
sets.


∪ The union of two events, where the sample space contains two events, A and B, and the set belongs to
A or B.


∩ The intersection of two events, where the sample space contains two events, A and B, and the set belongs
to A and B.


2.4 Review Questions
Answer the following questions and show all work (including diagrams) to create a complete answer.


1. Determine which of the following are examples of independent events.
(a) Rolling a 5 on one die and rolling a 5 on a second die.
(b) Choosing a cookie from the cookie jar and choosing a jack from a deck of cards.
(c) Winning a hockey game and scoring a goal.


2. Determine which of the following are examples of independent events.
(a) Choosing an 8 from a deck of cards, replacing it, and choosing a face card.
(b) Going to the beach and bringing an umbrella.
(c) Getting gasoline for your car and getting diesel fuel for your car.


3. Determine which of the following are examples of dependent events.


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(a) Selecting a marble from a container and selecting a jack from a deck of cards.
(b) Rolling a number less than 4 on a die and rolling a number that is even.
(c) Choosing a jack from a deck of cards and choosing a heart.


4. Determine which of the following are examples of dependent events.
(a) Selecting a book from the library and selecting a book that is a mystery novel.
(b) Rolling a 2 on a die and flipping a coin to get tails.
(c) Being lunchtime and eating a sandwich.


5. Two die are tossed. What is the probability of obtaining a sum equal to 6?
6. Two die are tossed. What is the probability of obtaining a sum less than 6?
7. Two die are tossed. What is the probability of obtaining a sum greater than 6?
8. Two die are tossed. What is the probability of obtaining a sum of at least 6?
9. A coin and a die are tossed. Calculate the probability of getting tails and a 5.
10. ABC school is debating whether or not to write a policy where all students must have uniforms and


wear these during school hours. In an survey, 45% of the students wanted uniforms, 35% did not,
and 10% said they did not mind a uniform and did not care if there was no uniform. Represent this
information in a Venn diagram.


11. ABC school is debating whether or not to write a policy where all students must have uniforms and
wear them during school hours. In an survey, 45% of the students wanted uniforms, and 55% did
not. Represent this information in a Venn diagram.


12. For question 11, calculate the probability that a person selected at random from ABC school will
want the school to have uniforms.


13. Consider a sample set as S = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}. Event A is the multiples of 4, while Event
B is the multiples of 5. What is the probability that a number chosen at random will be from both
A and B?


14. For question 13, what is the probability that a number chosen at random will be from either A or B?
15. Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12


green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue
or green jelly bean?


16. Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12
green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue
or green jelly bean and then reaching in again and pulling out a red jelly bean?


17. Jack is a student in Bluenose High School. He noticed that a lot of the students in his math class
were also in his chemistry class. In fact, of the 60 students in his grade, 32 students were in his math
class and 28 students were in his chemistry class. He decided to do a survey to find out what the
probability was of selecting a student at random from his grade who is taking either chemistry or
math, but not both. Draw a Venn diagram and help Jack with his calculation.


18. Brenda did a survey of the students in her class about whether they liked to get a candy bar or a
new math pencil as their reward for positive behavior. She asked all 75 students she taught, and 32
said they would like a candy bar, 25 said they wanted a new pencil, and 4 said they wanted both. If
Brenda were to select a student at random from her classes, what is the probability that the student
chosen would want:
(a) a candy bar or a pencil?
(b) neither a candy bar nor a pencil?


19. A card is chosen at random from a standard deck of cards. What is the probability that the card
chosen is a heart and spade? Are these events mutually exclusive?


20. A card is chosen at random from a standard deck of cards. What is the probability that the card
chosen is a heart and a face card? Are these events mutually exclusive?


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Chapter 3


The Next Step...Conditional
Probability


Introduction


This chapter builds on the concepts learned in the previous chapter on probability. Starting with tree
diagrams as a means of displaying outcomes for various trials, we will learn how to read the diagrams
and find probabilities. We will also find that order does not matter unless working with permutations.
Permutations, such as the combination of your lock at the gym, require their own special formula. When
outcomes for permutations have repetitions, these repetitions must also be included in the calculations in
order to account for the multiple entries. Combinations, like permutations, also have their own special
formula. Combinations, such as the number of teams of four that can be arranged in a class of 15 students,
are different from permutations, because the order in combinations is insignificant. In this chapter, we will
also learn about conditional probability. Conditional probability works when the probability of the second
event occurring is dependent on the probability of the first event.


3.1 What are Tree Diagrams?
Learning Objectives


• Know the definition of conditional probability.
• Use conditional probability to solve for probabilities in finite sample spaces.


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In the last chapter, we studied independent and dependent events, as well as mutually inclusive and
mutually exclusive events. We used the addition rule for dependent events, as well as mutually inclusive and
mutually exclusive events. The addition rule or principle is used to find P(A or B), while the multiplication
rule is used for independent events.
Addition Rule – For two events, A and B, the probability of selecting one event or another is given by:
P(A or B) = P(A) + P(B) − P(A and B).
Multiplication Rule – For two independent events, A and B, where the outcome of A does not change
the probability of B, the probability of A and B is given by: P(A and B) = P(A) × P(B).


Tree diagrams are another way to show the outcomes of simple probability events. In a tree diagram,
each outcome is represented as a branch on the tree.
Let’s say you were going to toss a coin two times and wanted to find the probability of getting two heads.
This is an example of independent events, because the outcome of one event does not affect the outcome
of the second event. What does this mean? Well, when you flip the coin once, you have an equal chance
of getting a head (H) or a tail (T). On the second flip, you also have an equal chance of getting a a head
or a tail. In other words, whether the first flip was heads or tails, the second flip could just as likely be
heads as tails. You can represent the outcomes of these events on a tree diagram.


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From the tree diagram, you can see that the probability of getting a head on the first flip is 12 . Starting with
heads, the probability of getting a second head will be, again, 12 . But how do we calculate the probability
of getting two heads? These are independent events, since the outcome of tossing the first coin in no way
affects the outcome of tossing the second coin. Therefore, we can calculate the probability as follows.


P(A and B) = 1
2
× 1


2
P(A and B) = 1


4


Therefore, we can conclude that the probability of getting 2 heads when tossing a coin twice is 25%, or 14 .
Let’s try an example that is a little more challenging.
Example 1
Irvin opens up his sock drawer to get a pair socks to wear to school. He looks in the sock drawer and sees
4 red socks, 8 white socks, and 6 brown socks. Irvin reaches in the drawer and pulls out a red sock. He is
wearing blue shorts, so he replaces it. He then draws out a white sock. What is the probability that Irvin
pulls out a red sock, replaces it, and then pulls out a white sock?
Solution
First let’s draw a tree diagram.


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There are 18 socks in Irvin’s sock drawer. The probability of getting a red sock when he pulls out the first
sock is:


P(red) = 4
18


P(red) = 2
9


Irvin puts the sock back in the drawer and pulls out the second sock. The probability of getting a white
sock on the second draw is:


P(white) = 6
18


P(white) = 1
3


Therefore, the probability of getting a red sock and then a white sock when the first sock is replaced is:


P(red and white) = 2
9
× 1


3
P(red and white) = 2


27


One important part of these types of problems is that order is not important.
Let’s say Irvin picked out a white sock, replaced it, and then picked out a red sock. Calculate this
probability.


P(white and red) = 1
3
× 2


9
P(white and red) = 2


27


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So no matter the order in which he takes the socks out, the probability is the same. In other words,
P(red and white) = P(white and red).
Example 2
In Example 1, what happens if the first sock is not replaced?
Solution
The probability that the first sock is red is:


P(red) = 4
18


P(red) = 2
9


The probability of picking a white sock on the second pick is now:


So now, the probability of selecting a red sock and then a white sock, without replacement, is:


P(red and white) = 2
9
× 6


17
P(red and white) = 12


153
P(red and white) = 4


51


If the first sock is white, will P(red and white) = P(white and red) as we found in Example 1?


P(white) = 6
18


P(white) = 1
3


The probability of picking a red sock on the second pick is now:


As with the last example, P(red and white) = P(white and red). So when does order really matter?


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3.2 Order and Probability
Permutations and combinations are the next step in the learning of probability. It is by using permutations
and combinations that we can find the probabilities of various events occurring at the same time, such as
choosing 3 boys and 3 girls from a class of grade 12 math students.
In mathematics, we use more precise language:
If the order doesn’t matter, it is a combination.
If the order does matter, it is a permutation.
Say, for example, you are making a salad. You throw in some lettuce, carrots, cucumbers, and green
peppers. The order in which you throw in these vegetables doesn’t really matter. Here we are talking
about a combination. For combinations, you are merely selecting. Say, though, that Jack went to the
ATM to get out some money and that he has to put in his PIN number. Here the order of the digits in
the PIN number is quite important. In this case, we are talking about a permutation. For permutations,
you are ordering objects in a specific manner.
Permutations
The Fundamental Counting Principle states that if an event can be chosen in p different ways and
another independent event can be chosen in q different ways, the number of different ways the two events
can occur is p × q. In other words, the Fundamental Counting Principle tells you how many ways you can
arrange items. Permutations are the number of possible arrangements in an ordered set of objects.
Example 3
How many ways can you arrange the letters in the word MATH?
Solution
You have 4 letters in the word, and you are going to choose one letter at a time. When you choose the
first letter, you have 4 possibilities (’M’, ’A’, ’T’, or ’H’). Your second choice will have 3 possibilities, your
third choice will have 2 possibilities, and your last choice will have only 1 possibility.
Therefore, the number of arrangements is: 4 × 3 × 2 × 1 = 24 possible arrangements.
The notation for a permutation is: nPr,
where:
n is the total number of objects.
r is the number of objects chosen.
For simplifying calculations, when n = r, then nPr = n!.
The factorial function (!) requires us to multiply a series of descending natural numbers.
Examples:


5! = 5 × 4 × 3 × 2 × 1 = 120
4! = 4 × 3 × 2 × 1 = 24
1! = 1


Note: It is a general rule that 0! = 1.
With TI calculators, you can find the factorial function using:
MATH I I I (PRB) H H H ( 4 )


Example 4:


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Solve for 4P4.
Solution
4P4 = 4 · 3 · 2 · 1 = 24
This represents the number of ways to arrange 4 objects that are chosen from a set of 4 different objects.
Example 5:
Solve for 6P3.
Solution
6P3 = 6 · 5 · 4 = 120
↑ Starting with 6, multiply the first 3 numbers of the factorial.
This represents the number of ways to arrange 3 objects that are chosen from a set of 6 different objects.
The formula to solve permutations like these is:


nPr =
n!


(n − r)!


Look at Example 5 above. In this example, the total number of objects (n) is 6, while the number of
objects chosen (r) is 3. We can use these two numbers to calculate the number of possible permutations
(or the number of arrangements) of six objects chosen three at a time.


nPr =
n!


(n − r)!


6P3 =
6!


(6 − 3)!


6P3 =
6!
3!


= 6 × 5 × 4(
(


(


(


(×3 × 2 × 1
(


(


(


(3 × 2 × 1


6P3 =
120
1


6P3 = 120


Example 6
What is the total number of possible 4-letter arrangements of the letters ’s’, ’n’, ’o’, and ’w’ if each letter
is used only once in each arrangement?
Solution
In this problem, there are 4 letters to choose from, so n = 4. We want 4-letter arrangements; therefore, we
are choosing 4 objects at a time. In this example, r = 4.


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Example 7
A committee is formed with a president, a vice president, and a treasurer. If there are 10 people to select
from, how many committees are possible?
Solution
In this problem, there are 10 committee members to choose from, so n = 10. We want to choose 3
members to be president, vice-president, and treasurer; therefore, we are choosing 3 objects at a time. In
this example, r = 3.


nPr =
n!


(n − r)!


10P3 =
10!


(10 − 3)!


10P3 =
10!
7!


= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
7 × 6 × 5 × 4 × 3 × 2 × 1


10P3 =
720
1


10P3 = 720


Permutations with Repetition
There is a subset of permutations that takes into account that there are double objects or repetitions in a
permutation problem. In general, repetitions are taken care of by dividing the permutation by the factorial
of the number of objects that are identical.
If you look at the word TOOTH, there are two O’s in the word. Both O’s are identical, and it does not
matter in which order we write these two O’s, since they are the same. In other words, if we exchange ’O’
for ’O’, we still spell TOOTH. The same is true for the T’s, since there are two T’s in the word TOOTH
as well.
If we were to ask the question, ”In how many ways can we arrange the letters in the word TOOTH?” we
must account for the fact that these two O’s are identical and that the two T’s are identical. We do this
using the formula:


nPr
x1!x2! , where x is the number of times a letter is repeated.


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nPr
x1!x2!


= 5P5
2!2!


5P5
2!2!


= 5 × 4 × 3 × 2 × 1
2 × 1 × 2 × 1


5P5
2!2!


= 120
4


5P5
2!2!


= 30


Tech Tip: Calculating Permutations on the Calculator
Permutations (nPr)
Enter the n value. Press MATH . You should see modes across the top of the screen. You want the fourth
mode: PRB (arrow right three times). You will see several options: nPr is the second. Press 2 . Enter
the r value. Press ENTER .
Example 8
Compute 9P5.
Solution
9 MATH I I I (PRB) 2 (nPr) 5 ENTER


9P5 = 15120
Example 9
How many different 5-letter arrangements can be formed from the word APPLE?
Solution
There are 5 letters in the word APPLE, so n = 5. We want 5-letter arrangements; therefore, we are
choosing 5 objects at a time. In this example, r = 5. In this problem, we are using a word with letters
that repeat. In the word APPLE, there are two P’s, so x1 = 2.


Example 10
How many different six-digit numerals can be written using all of the following seven digits?
3, 3, 4, 4, 4, 5, 6.
Solution
There are 7 digits, so n = 7. We want 6-digit arrangements; therefore, we are choosing 6 objects at a time.


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In this example, r = 6, and we are using a group of digits with numbers that repeat. In the group of seven
digits (3, 3, 4, 4, 4, 5, 6), there are two 3’s and three 4’s, so x1 = 2 and x2 = 3.


When there are no repetitions, remember that we use the standard permutation formula:


nPr =
n!


(n − r)!


Example 11
How many ways can first and second place be awarded to 10 people?
Solution
There are 10 people (n = 10), and there are two prize winners (r = 2).


Example 12
In how many ways can 3 favorite desserts be listed in order from a menu of 10? (i.e., permutations without
repetition)


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Solution
There are 10 menu items (n = 10), and you are choosing three (r = 3) in order.


Combinations
If you think about the lottery, you choose a group of lucky numbers in hopes of winning millions of dollars.
When the numbers are drawn, the order they are drawn does not have to be the same order as on your
lottery ticket. The numbers drawn simply have to be on your lottery ticket in order for you to win. You
can imagine how many possible combinations of numbers exist, which is why your odds of winning are so
small!


Combinations are arrangements of objects without regard to order and without repetition, selected from
a distinct number of objects. A combination of n objects taken r at a time (nCr) can be calculated using
the formula:


nCr =
n!


r!(n − r)!


Example 13
Evaluate: 7C2.
Solution


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7C2 =
7!


2!(7 − 2)!


7C2 =
7!


2!(5)!


7C2 =
7 × 6 × 5 × 4 × 3 × 2 × 1
(2 × 1)(5 × 4 × 3 × 2 × 1)


7C2 =
5040


(2)(120)


7C2 =
5040
240


7C2 = 21


Example 14
In how many ways can 3 desserts be chosen in any order from a menu of 10?
Solution
There are 10 menu items (n = 10), and you are choosing 3 desserts in any order (r = 3).


10C3 =
10!


3!(10 − 3!)


10C3 =
10!


3!(7!)
10C3 = 120


Example 15
There are 12 boys and 14 girls in Mrs. Cameron’s math class. Find the number of ways Mrs. Cameron
can select a team of 3 students from the class to work on a group project. The team must consist of 2 girls
and 1 boy.
Solution
There are groups of both boys and girls to consider. From the 14 girls (n = 14) in the class, we are choosing
2 (r = 2).


Girls


14C2 =
14!


2!(14 − 2)!


14C2 =
14!


2!(12)!


14C2 =
87178291200
2(479001600)


14C2 =
87178291200
958003200


14C2 = 91


From the 12 boys (n = 12) in the class, we are choosing 1(r = 1).


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Boys


12C1 =
12!


1!(12 − 1)!


12C1 =
12!


1!(11)!


12C1 =
479001600
1(39916800)


12C1 =
479001600
39916800


12C1 = 12


Therefore, the number of ways Mrs. Cameron can select a team of 3 students (2 girls and 1 boy) from the
class of 26 students to work on a group project is:
Total combinations = 14C2 × 12C1 = 91 × 12 = 1092
Example 16
If there are 20 rock songs and 20 rap songs to choose from, in how many different ways can you select 12
rock songs and 7 rap songs for a mix CD?
Solution
As in Example 13, we have multiple groups from which we are required to select, so we have to calculate
the possible combinations for each group (rock songs and rap songs in this example) separately and then
multiply together.
Using TI technology: for nCr, type the n value (the total number of items), and then MATH I I I
PRB H H (to number 3) ENTER . Then type the r value (the number of items your want to choose),
and finally, ENTER .


Rock


20C12 =
20!


12!(20 − 12)!


20C12 =
20!


12!(8)!
20C12 = 125970


Rap


20C7 =
20!


7!(20 − 7)!


20C7 =
20!


7!(13)!
20C7 = 77520


Therefore, the number of possible combinations is:


Rock Rap
20C12 ×20 C7 = 125970 × 77520 = 9.765 × 109 possible combinations


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Tech Tip: Calculating Combinations on the Calculator
Combinations (nCr)
Enter the n value. Press MATH . You should see modes across the top of the screen. You want the fourth
mode: PRB (arrow right three times). You will see several options: nCr is the third. Press 3 . Enter the
r value. Press ENTER .
Example 17
Compute 10C6.
Solution


10C6 1 0 MATH I I I (PRB) 3 (nCr) 6 ENTER
10C6 = 210


3.3 Conditional Probability
What if the probability of a second event is affected by the probability of the first event? This type of
probability calculation is known as conditional probability.
When working with events that are conditionally probable, you are working with two events, where the
second event is conditional on the first event occurring. Say, for example, that you want to know the
probability of drawing two kings from a deck of cards.


P(first king) = 1
13


P(second king) = 3
51


P(two kings) = 1
13


× 3
51


P(two kings) = 3
663


P(two kings) = 1
221


Now let’s assume you are playing a game where you need to draw two kings to win. You draw the first card
and get a king. What is the probability of getting a king on the second card? The probability of getting
two kings can now be thought of as a conditional probability. The formula for calculating conditional
probability is given as:
Conditional Probability


P(B|A) = P(A ∩ B)
P(A)


P(A ∩ B) = P(A) × P(B|A)


Another way to look at the conditional probability formula is as follows. Assuming the first event has
occurred, the probability of the second event occurring is:


P(second choice|first choice) = P(first choice and second choice)
P(first choice)


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Let’s work through a few problems using the formula for conditional probability.
Example 18
You are playing a game of cards where the winner is determined when a player gets two cards of the same
suit. You draw a card and get a club (♣). What is the probability that the second card will be a club?
Solution
Step 1: List what you know.
First event = drawing the first club


P(first club) = 13
52


P(club and club) = 13
52


× 12
51


P(club and club) = 156
2652


P(club and club) = 1
17


Step 2: Calculate the probability of choosing a club as a second card when a club is chosen as a first card.


Probability of drawing the second club = P(club and club)
P(first club)


P(club|club) =
1
17
13
52


P(club|club) = 1
17


× 52
13


P(club|club) = 52
221


P(club|club) = 4
17


Step 3: Write your conclusion: Therefore, the probability of selecting a club as a second card when a club
is chosen as the first card is 24%.
Example 19
In the next round of the game, the first person to be dealt a black ace wins the game. You get your first
card and it is a queen. What is the probability of obtaining a black ace?
Solution
Step 1: List what you know.
First event = being dealt the queen


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P(queen) = 4
52


P(black ace and queen) = 4
52


× 2
51


P(black ace and queen) = 8
2652


P(black ace and queen) = 4
663


Step 2: Calculate the probability of choosing black ace as a second card when a queen is chosen as a first
card.


P(black ace|queen) = P(black ace and queen)
P(queen)


P(black ace|queen) =
4


663
4
52


P(black ace|queen) = 4
663


× 52
4


P(black ace|queen) = 52
663


P(black ace|queen) = 4
51


Step 3: Write your conclusion: Therefore, the probability of selecting a black ace as the second card when
a queen is chosen as the first card is 7.7%.
Example 20
At Bluenose High School, 90% of the students take physics and 35% of the students take both physics and
statistics. What is the probability that a student from Bluenose High School who is taking physics is also
taking statistics?


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Solution:
Step 1: List what you know.


P(physics) = 0.90
P(physics and statistics) = 0.35


Step 2: Calculate the probability of choosing statistics as a second course when physics is chosen as a
first course.


P(statistics|physics) = P(physics and statistics)
P(physics)


P(statistics|physics) = 0.35
0.90


P(statistics|physics) = 0.388
P(statistics|physics) = 39%


Step 3: Write your conclusion: Therefore, the probability that a student from Bluenose High School who
is taking physics is also taking statistics is 39%.
Example 21
Sandra went out for her daily run. She goes on a path that has alternate routes to give her a variety of
choices to make her run more enjoyable. The path has three turns where she can go left or right at each
turn. The probability of turning right the first time is 12 . Based on past runs, the probability of turning
right the second time is 23 . Draw a tree diagram to represent the path. What is the probability that she
will turn left the second time after turning right the first time?
Solution


Step 1: List what you know.


P(right) = 1
2


P(right and left) = 1
2
× 1


3
P(right and left) = 1


6


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Step 2: Calculate the probability of choosing left as the second turn when right is chosen as the first turn.


P(left|right) = P(right and left)
P(right)


P(left|right) =
1
6
1
2


P(left|right) = 1
6
× 2


1
P(left|right) = 2


6
P(left|right) = 1


3
P(left|right) = 0.333
P(left|right) = 33%


Step 3: Write your conclusion: Therefore, the probability of choosing left as the second turn when right
is chosen as the first turn is 33%.
Points to Consider


• How does a permutation differ from a combination?
• How are tree diagrams helpful for determining probability?


Vocabulary


Combinations The number of possible arrangements (nCr) of objects (r) without regard to order and
without repetition selected from a distinct number of objects (n).


Conditional probability The probability of a particular dependent event, given the outcome of the
event on which it depends.


Factorial function (!) To multiply a series of consecutive descending natural numbers.


Fundamental Counting Principle If an event can be chosen in p different ways and another indepen-
dent event can be chosen in q different ways, the probability of the two events occurring in p × q.


Permutations The number of possible arrangements (nPr) in an ordered set of objects, where n = the
number of objects and r = the number of objects selected.


Tree diagrams A way to show the outcomes of simple probability events, where each outcome is repre-
sented as a branch on the tree.


3.4 Review Questions
Answer the following questions and show all work (including diagrams) to create a complete answer.


1. A bag contains 3 red balls and 4 blue balls. Thomas reaches in the bag and picks a ball out at
random from the bag. He places it back into the bag. Thomas then reaches in the bag and picks
another ball at random.


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(a) Draw a tree diagram to represent this problem.
(b) What is the probability that Thomas picks:


i. two red balls
ii. a red ball in his second draw


2. A teacher has a prize box on her front desk for when students do exceptional work in math class.
Inside the box there are 20 math pencils and 10 very cool erasers. Janet completed a challenge
problem for Ms. Cameron, and Ms. Cameron rewarded Janet’s innovative problem-solving approach
with a trip to the prize box. Janet reaches into the box and picks out a prize and then drops it back
in. Then she reaches in again and picks out a prize a second time.
(a) Draw a tree diagram to represent this problem.
(b) What is the probability that Janet reaches into the box and picks out an eraser on the second


pick?


3. Determine whether the following situations would require calculating a permutation or a combination:
(a) Selecting three students to attend a conference in Washington, D.C.
(b) Selecting a lead and an understudy for a school play.
(c) Assigning students to their seats on the first day of school.


4. Solve for 7P5.
5. Evaluate 4P2 × 5P3.
6. How many different 4-digit numerals can be made from the digits of 56987 if a digit can appear just
once in a numeral?


7. How many ways can the letters in the word REFERENCE be arranged?
8. How many ways can the letters in the word MISSISSIPPI be arranged?
9. How many ways can the letters in the word MATHEMATICS be arranged?
10. If there are 4 chocolate chip, 2 oatmeal, and 2 double chocolate cookies in a box, in how many


different orders is it possible to eat all of these cookies?
11. A math test is made up of 15 multiple choice questions. 5 questions have the answer A, 4 have the


answer B, 3 have the answer C, 2 have the answer D, and 1 has the answer E. How many answer
sheets are possible?


12. In how many ways can you select 17 songs from a mix CD of a possible 38 songs?
13. If an ice cream dessert can have 2 toppings, and there are nine available, how many different selections


can you make?
14. If there are 17 randomly placed dots on a circle, how many lines can be formed using any two dots?
15. A committee of 4 is to be formed from a group of 13 people. How many different committees can be


formed?
16. There are 4 kinds of meat and 10 veggies available to make wraps at the school cafeteria. How many


possible wraps have 1 kind of meat and 3 veggies?
17. There are fifteen freshmen and thirty seniors in the Senior Math Club. The club is to send four


representatives to the State Math Championships.
(a) How many different ways are there to select a group of four students to attend the Champi-


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onships?
(b) If the members of the club decide to send two freshmen and two seniors, how many different


groupings are possible?


18. Students in BDF High School were asked about their preference regarding the new school colors.
They were given a choice between green and blue as the primary color and red and yellow as the
secondary color. The results of the survey are shown in the tree diagram below. You can see that 75%
of the students choose green as the primary color. Of this 75%, 45% chose yellow as the secondary
color. What is the probability that a student in BDF High School selected red as the secondary color
if he or she chose blue as the primary color?


19. Two fair die are rolled. What is the probability that the sum is even given that the first die that is
rolled is a 2?


20. Two fair die are rolled. What is the probability that the sum is even given that the first die rolled is
a 5?


21. Two fair die are rolled. What is the probability that the sum is odd given that the first die rolled is
a 5?


22. Steve and Scott are playing a game of cards with a standard deck of playing cards. Steve deals Scott
a king. What is the probability that Scott’s second card will be a red card?


23. Sandra and Karen are playing a game of cards with a standard deck of playing cards. Sandra deals
Karen a seven. What is the probability that Karen’s second card will be a black card?


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24. Donna discusses with her parents the idea that she should get an allowance. She says that in her
class, 55% of her classmates receive an allowance for doing chores, and 25% get an allowance for doing
chores and are good to their parents. Her mom asks Donna what the probability is that a classmate
will be good to his or her parents given that he or she receives an allowance for doing chores. What
should Donna’s answer be?


25. At the local high school, the probability that a student speaks English and French is 15%. The
probability that a student speaks French is 45%. What is the probability that a student speaks
English, given that the student speaks French?


26. At the local high school, the probability that a student takes statistics and art is 10%. The probability
that a student takes art is 60%. What is the probability that a student takes statistics, given that
the student takes art?


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Chapter 4


Introduction to Discrete
Random Variables


Introduction
In this chapter, you will learn about discrete random variables. Random variables are simply quantities
that take on different values depending on chance (or probability). Discrete random variables can take on
a finite number of values in an interval, or as many values as there are positive integers. In other words,
a discrete random variable can take on an infinite number of values, but not all the values in an interval.
When you find the probabilities of these values, you are able to show the probability distribution. A
probability distribution consists of all the values of the random variable, along with the probability of the
variable taking on each of these values. Each probability must be between 0 and 1, and the probabilities
must sum to 1.
You will also be introduced the concept of a binomial distribution. This will be discussed in depth in the
next chapter, but in this chapter, you will use a binomial distribution when talking about the number
of successful events or the value of a random variable. A binomial distribution is only used when there
are two possible outcomes. For example, you will use the binomial distribution formula for coin tosses
(heads or tails). Other examples include yes/no responses, true or false questions, and voting Democrat
or Republican. When the number of outcomes goes beyond two (2), you use a multinomial distribution.
Rolling a die is a common example of a multinomial distribution problem.


In addition, you will use factorials again for solving these problems. Factorials were introduced in Chapter
2 for permutations and combinations, but they are also used in many other probability problems. Finally,
you will use a graphing calculator to show the difference between theoretical and experimental probability.
The calculator is an effective and efficient tool for illustrating the difference between these two probabilities,
and also for determining the experimental probability when the number of trials is large.


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4.1 What are Variables?
Learning Objectives


• Demonstrate an understanding of the notion of discrete random variables by using them to solve for
the probabilities of outcomes, such as the probability of the occurrence of five heads in 14 coin tosses.


In the previous two chapters, we looked at the mathematics involved in probability events. We looked at
examples of event A occurring if event B had occurred (conditional events), of event B being affected by
the outcome of event A (dependent events) and of event A and event B not being affected by each other
(independent events). We also looked at examples where events cannot occur at the same time (mutually
exclusive events), or when events were not mutually exclusive and there was some overlap, so that we
had to account for the double counting (mutually inclusive events). If you recall, we used Venn Diagrams
(below), tree diagrams, and even tables to help organize information in order to simplify the mathematics
for the probability calculations.


Our examination of probability, however, began with a look at the English language. Although there are
a number of differences in what terms mean in mathematics and English, there are a lot of similarities as
well. We saw this with the terms independent and dependent. In this chapter, we are going to learn about
variables. In particular, we are going to look at discrete random variables. When you see the sequence
of words “discrete random variables,” it may, at first, send a shiver down your spine, but let’s look at the
words individually and see if we can ”simplify” the sequence!
The term discrete, in English, means to constitute a separate thing or to be related to unconnected parts.
In mathematics, we use the term discrete when we are talking about pieces of data that are not connected.
Random, in English, means to lack any plan or to be without any prearranged order. In mathematics, the
definition is the same. Random events are fair, meaning that there is no way to tell what outcome will
occur. In the English language, the term variable means to be likely to change or subject to variation. In
mathematics, the term variable means to have no fixed quantitative value.
Now that we have seen the three terms separately, let’s combine them and see if we can come up with a
definition of a discrete random variable. We can say that discrete variables have values that are unconnected
to each other and have variations within the values. Think about the last time you went to the mall.


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Suppose you were walking through the parking lot and were recording how many cars were made by Ford.
The variable is the number of Ford cars you see. Therefore, since each car is either a Ford or it is not, the
variable is discrete. If you randomly selected 20 cars from the parking lot and determined whether or not
each was manufactured by Ford, you would then have a discrete random variable.


Now let’s define discrete random variables. Discrete random variables represent the number of distinct
values that can be counted of an event. For example, when Robert was randomly chosen from all the
students in his classroom and asked how many siblings there are in his family, he said that he has 6 sisters.
Joanne picked a random bag of jelly beans at the store, and only 15 of 250 jelly beans were green. When
randomly selecting from the most popular movies, Jillian found that Iron Man 2 grossed 3.5 million dollars
in sales on its opening weekend. Jack, walking with his mom through the parking lot, randomly selected
10 cars on his way up to the mall entrance and found that only 2 were Ford vehicles.


4.2 The Probability Distribution
When we talk about the probability of discrete random variables, we normally talk about a probability
distribution. In a probability distribution, you may have a table, a graph, or a chart that would show
you all the possible values of X (your variable), and the probability associated with each of these values
(P(X)).
It is important to remember that the values of a discrete random variable are mutually exclusive. Think
back to our car example with Jack and his mom. Jack could not, realistically, find a car that is both a Ford
and a Mercedes (assuming he did not see a home-built car). He would either see a Ford or not see a Ford
as he went from his car to the mall doors. Therefore, the values for the variable are mutually exclusive.
Now let’s look at an example.
Example 1
Say you are going to toss two coins. Show the probability distribution for this toss.
Solution
Let the variable be the number of times your coin lands on tails. The table below lists all of the possible
events that can occur from the tosses.


Table 4.1:


Toss First Coin Second Coin X
1 H H 0
2 H T 1
3 T T 2
4 T H 1


We can add a fifth column to the table above to show the probability of each of these events (the tossing


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of the two coins).


Table 4.2:


Toss First Coin Second Coin X P(X)
1 H H 0 14
2 H T 1 14
3 T T 2 14
4 T H 1 14


As you can see in the table, each event has an equally likely chance of occurring. You can see this by
looking at the column P(X). From here, we can find the probability distributions. In the X column, we
have three possible discrete values for this variable: 0, 1, and 2.


p(0) = toss 1 = 1
4


p(1) = toss 2 + toss 4


= 1
4


+ 1
4


= 1
2


p(2) = toss 3 = 1
4


Now we can represent the probability distribution with a graph, called a histogram. A histogram is a
graph that uses bars vertically arranged to display data. Using the TI-84 PLUS calculator, we can draw
the histogram to represent the data above. Let’s start by first adding the data into our lists. Below you
will find the key sequence to perform this task. We will use this sequence frequently throughout the rest
of this book, so make sure you follow along with your calculator.


This key sequence allows you to erase any data that may be entered into the lists already. Now let’s enter
our data.


Now we can draw our histogram from the data we just entered.


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The result:


We can see the values of P(X) if we press TRACE . Look at the screenshot below. You can see the value
of P(X) = 0.25 for X = 0.


It’s clear that the histogram shows the probability distribution for the discrete random variable. In other
words, P(0) gives the probability that the discrete random variable is equal to 0, P(1) gives the probability
that the discrete random variable is equal to 1, and P(2) gives the probability that the discrete random
variable is equal to 2. Notice that the probabilities add up to 1. One of the rules for probability is that
the sum of the probabilities of all the possible values of a discrete random variable must be equal to 1.
Example 2
Does the following table represent the probability distribution for a discrete random variable?


X 0 1 2 3
P(X) 0.1 0.2 0.3 0.4


Solution
Yes, it does, since ∑ P(X) = 0.1 + 0.2 + 0.3 + 0.4, or ∑ P(X) = 1.0.


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4.3 A Glimpse at Binomial and Multinomial Dis-
tributions


In Chapter 4, we will learn more about binomial and multinomial distributions. However, we are now
talking about probability distributions, and as such, we should at least see how the problems change for
these distributions. We will briefly introduce the concepts and their formulas here, and then we will get
into more detail in Chapter 4. Let’s start with a problem involving a binomial distribution.
Example 3
The probability of scoring above 75% on a math test is 40%. What is the probability of scoring below
75%?
Solution
P(scoring above 75%) = 0.40
Therefore, P(scoring below 75%) = 1 − 0.40 = 0.60.
The randomness of an individual outcome occurs when we take one event and repeat it over and over again.
One example is if you were to flip a coin multiple times. In order to calculate the probability of this type
of event, we need to look at one more formula.
The probability of getting x successes in n trials is given by:


P(x = a) = nCa × pa × q(n−a)


where:
a = the number of successes from the trials
p = the probability of the event occurring
q = the probability of the event not occurring
Now, remember in Chapter 2, you learned about the formula nCr. The formula is shown below.


nCr =
n!


r!(n − r)!


Also, remember the symbol ! means factorial. As a review, recall the explanation of factorial from Chapter
2.
Note: The factorial function (!) just means to multiply a series of consecutive descending natural
numbers.
Examples:


4! = 4 × 3 × 2 × 1 = 24
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
1! = 1


Note: it is generally agreed that 0! = 1.
Technology Tip: You can find the factorial function using:
MATH I I I (PRB) H H H ( 4 )
Now let’s try a few problems with the new formula.


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Example 4
A fair die is rolled 10 times. Let X be the number of rolls in which we see a two.
(a) What is the probability of seeing a two in any one of the rolls?
(b) What is the probability that we see a two exactly once in the 10 rolls?
Solution
(a) P(X) = 16 = 0.167
(b) p = 0.167
q = 1 − 0.167 = 0.833
n = 10
a = 1


P(X = a) = nCa × pa × q(n−a)


P(X = 1) = 10C1 × p1 × q(10−1)


P(X = 1) = 10C1 × (0.167)1 × (0.833)(10−1)


P(X = 1) = 10 × 0.167 × 0.193
P(X = 1) = 0.322


Therefore, the probability of seeing a 2 exactly once when a die is rolled 10 times is 32.2%.
Interestingly, it was Blaise Pascal (pictured below) with Pierre de Fermat who provided the world with
the basics of probability. These two mathematicians studied many different theories in mathematics, one
of which was odds and probability. To learn more about Pascal, go to http://en.wikipedia.org/wiki/
Blaise_Pascal. To learn more about Fermat, go to http://en.wikipedia.org/wiki/Fermat. These two
mathematicians have contributed greatly to the world of mathematics.


Example 5
A fair die is rolled 15 times. What is the probability of rolling two 2’s?
(a) What is the probability of seeing a two in any one of the rolls?
(b) What is the probability that we see a two exactly twice in the 15 rolls?
Solution
(c) P(X) = 16 = 0.167
(d) p = 0.167
q = 1 − 0.167 = 0.833
n = 15
a = 2


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P(X = a) = nCa × pa × q(n−a)


P(X = 2) = 15C2 × p2 × q(15−2)


P(X = 2) = 15C2 × (0.167)2 × (0.833)(15−2)


P(X = 2) = 105 × 0.0279 × 0.0930
P(X = 2) = 0.272


Therefore, the probability of seeing a two exactly twice when a die is rolled 15 times is 27.2%.
Example 6
A pair of fair dice is rolled 10 times. Let X be the number of rolls in which we see at least one two.
(a) What is the probability of seeing at least one two in any one roll of the pair of dice?
(b) What is the probability that in exactly half of the 10 rolls, we see at least one two?
Solution
If we look at the chart below, we can see the number of times a 2 shows up when rolling two dice.


(a) The probability of seeing at least one two in any one roll of the pair of dice is:


P(X) = 11
36


= 0.306


(b) The probability of seeing at least one two in exactly 5 of the 10 rolls is:


p = 0.306
q = 1 − 0.306 = 0.694
n = 10
a = 5


P(X = a) = nCa × pa × q(n−a)


P(X = 5) = 10C5 × p5 × q(10−5)


P(X = 5) = 10C5 × (0.306)5 × (0.694)(10−5)


P(X = 5) = 252 × 0.00268 × 0.161
P(X = 5) = 0.109


Therefore, the probability of rolling at least one two five times when two dice are rolled 10 times is 10.9%.
It should be noted here that the previous two examples are examples of binomial experiments. We will
be learning more about binomial experiments and distributions in Chapter 4. For now, we can visualize


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a binomial distribution experiment as one that has a fixed number of trials, with each trial being
independent of the others. In other words, rolling a die twice to see if a two appears is a binomial
experiment, because there is a fixed number of trials (2), and each roll is independent of the others. Also,
for binomial experiments, there are only two outcomes (a successful event and a non-successful event). For
our rolling of the die, a successful event is seeing a 2, and a non-successful event is not seeing a 2.
Example 7
You are given a bag of marbles. Inside the bag are 5 red marbles, 4 white marbles, and 3 blue marbles.
Calculate the probability that with 6 trials, you choose 3 that are red, 1 that is white, and 2 that are blue,
replacing each marble after it is chosen.
Solution
Notice that this is not a binomial experiment, since there are more than 2 outcomes. For binomial
experiments, k = 2 (2 outcomes). Therefore, we use the binomial experiment formula for problems involving
heads or tails, yes or no, or success or failure. In this problem, there are three possible outcomes: red,
white, or blue. This type of experiment produces what we call a multinomial distribution. In order to
solve this problem, we need to use one more formula:


P = n!
n1!n2!n3! . . . nk!


× (p1n1 × p2n2 × p3n3 . . . pknk)


where:
n = number of trials
p = probability for each possible outcome
k = number of possible outcomes
Notice that in this example, k equals 3. If we had only red marbles and white marbles, k would be equal
to 2, and we would have a binomial distribution.
The probability of choosing 3 red marbles, 1 white marble, and 2 blue marbles in exactly 6 picks is:


n = 6(6 picks)


p1 =
5
12


= 0.416 (probability of choosing a red marble)


p2 =
4
12


= 0.333 (probability of choosing a white marble)


p3 =
3
12


= 0.25 (probability of choosing a blue marble)
n1 = 3(3 red marbles chosen)
n2 = 1(1 white marble chosen)
n3 = 2(2 blue marbles chosen)
k = 3 (3 possibilities)


P(x = 6) = n!
n1!n2!n3! . . . nk!


× (p1n1 × p2n2 × p3n3 . . . pknk)


P(x = 6) = 6!
3!1!2!


× (0.4163 × 0.3331 × 0.252)


P(x = 6) = 60 × 0.0720 × 0.333 × 0.0625
P(x = 6) = 0.0899


Therefore, the probability of choosing 3 red marbles, 1 white marble, and two blue marbles is 8.99%.


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Example 8
You are randomly drawing cards from an ordinary deck of cards. Every time you pick one, you place it
back in the deck. You do this five times. What is the probability of drawing 1 heart, 1 spade, 1 club, and
2 diamonds?
Solution


n = 5(5 trials)


p1 =
13
52


= 0.25 (probability of drawing a heart)


p2 =
13
52


= 0.25 (probability of drawing a spade)


p3 =
13
52


= 0.25 (probability of drawing a club)


p4 =
13
52


= 0.25 (probability of drawing a diamond)
n1 = 1(1 heart)
n2 = 1(1 spade)
n3 = 1(1 club)
n4 = 2(2 diamonds)
k = 1 (for each trial, there is only 1 possible outcome)


P(x = 5) = n!
n1!n2!n3! . . . nk!


× (p1n1 × p2n2 × p3n3 . . . pknk)


P(x = 5) = 5!
1!1!1!2!


× (0.251 × 0.251 × 0.251 × 0.252)


P(x = 5) = 60 × 0.25 × 0.25 × 0.25 × 0.25
P(x = 5) = 0.0586


Therefore, the probability of choosing one heart, one spade, one club, and two diamonds is 5.86%.


4.4 Using Technology to Find Probability Distri-
butions


If we look back at Example 1, we were tossing two coins. If you were to repeat this experiment 100 times,
or if you were going to toss 10 coins 50 times, these experiments would be very tiring and take a great deal
of time. On the TI-84, there are applications built in to determine the probability of such experiments. In
this section, we will look at how you can use your graphing calculator to calculate probabilities for larger
trials and draw the corresponding histograms.
On the TI-84 calculator, there are a number of possible simulations you can do. You can do a coin toss,
spin a spinner, roll dice, pick marbles from a bag, or even draw cards from a deck.


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After pressing ENTER , you will have the following screen appear.


Let’s try a spinner problem. Choose Spin Spinner.


Example 9
You are spinning a spinner like the one shown to the right 20 times. How many times does it land on blue?


Solution
In Spin Spinners, a wheel with four possible outcomes is shown. You can adjust the number of spins, graph
the frequency of each number, and use a table to see the number of spins for each number. Let’s try this
problem. We want to set this spinner to spin 20 times. Look at the keystrokes below and see how this is
done.


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In order to match our color spinner with the one found in the calculator, you will see that we have added
numbers to our spinner. This is not necessary, but it may help in the beginning to remember that 1 =
blue (for this example).


Now that the spinner is set up for 20 trials, choose SPIN by pressing WINDOW .


We can see the result of each trial by choosing TABL, or pressing GRAPH .


And we see the graph of the resulting table (go back to the first screen) simply by choosing GRPH, or
pressing GRAPH .
Now, the question asks how many times we landed on blue (outcome 1). We can actually see how many
times we landed on blue for these 20 spins. If you press the right arrow


(


I


)


, the frequency label will show
you how many of the times the spinner landed on blue (number 1).


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To go back to the question, how many times will the spinner land on blue if it is spun 20 times? The
answer is 3. To calculate the probability, we have to divide by the total number of spins.


P(blue) = 3
20


= 0.15


Therefore, for this experiment, the probability of landing on blue with 20 spins is 15%.
The above example introduces us to a new concept. We know that the spinner has 4 equal parts (blue,
purple, green, and red). In a single trial, we can assume that:


P(blue) = 1
4


= 0.25


However, we know that we did the experiment and found that the probability of landing on blue, if the
spinner is spun 20 times, is 0.15. Why the difference?
The difference between these two numbers has to do with the difference between theoretical and experi-
mental probability. Theoretical probability is defined as the number of desired outcomes divided by
the total number of outcomes.
Theoretical Probability


P(desired) = number of desired outcomestotal number of outcomes


For our spinner example, the theoretical probability of landing on blue is 0.25. Finding the theoretical
probability requires no collection of data.
In the case of the experiment of spinning the spinner 20 times, the probability of 0.15, found by counting
the number of times the spinner landed on blue, is called the experimental probability. Experimental
probability is, just as the name suggests, dependent on some form of data collection. To calculate the
experimental probability, divide the number of times the desired outcome has occurred by the total number
of trials.
Experimental Probability


P(desired) = number of times desired outcome occurstotal number of trials


You can try a lot of examples and trials yourself using the NCTM Illuminations page found at http:
//illuminations.nctm.org/activitydetail.aspx?ID=79.
What is interesting about theoretical and experimental probabilities is that the more trials you do, the
closer the experimental probability gets to the theoretical probability. To show this, try spinning the
spinner for the next example.
Example 10
You are spinning a spinner like the one shown to the right 50 times. How many times does it land on blue?


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Solution
Set the spinner to spin 50 times and choose SPIN by pressing WINDOW .


You can see the result of each trial by choosing TABL, or pressing GRAPH .


Again, we can see the graph of the resulting table (go back to the first screen) simply by choosing GRPH,
or pressing GRAPH .
The question asks how many times we landed on blue (outcome 1) for the 50 spins. Press the right arrow
(


I


)


, and the frequency label will show you how many of the times the spinner landed on blue (number
1).


Now go back to the question. How many times will the spinner land on blue if it is spun 50 times? The
answer is 11. To calculate the probability, we have to divide by the total number of spins.


P(blue) = 11
50


= 0.22


Therefore, for this experiment, the probability of landing on blue with 50 spins is 22%.
If we tried 100 trials, we would see the following.


We see that the frequency of 1 is 23.


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So how many times will the spinner land on blue if it is spun 100 times? The answer is 23. To calculate
the probability, we have to divide by the total number of spins.


P(blue) = 23
100


= 0.23


Therefore, for this experiment, the probability of landing on blue with 100 spins is 23%. You see that as
we get more trials, we get closer to 25%, which is the theoretical probability.
Example 11
How many times do you predict we would have to spin the spinner in Example 10 to have the theoretical
probability equal the experimental probability?
Solution
With 170 spins, we get a frequency of 42.


P(blue) = 42
170


= 0.247


So the experimental probability is 24.7%, which is even closer to the theoretical probability of 25%. While
we’re getting closer to the theoretical probability, there is no number of trials that will guarantee that the
experimental probability will exactly equal the theoretical probability.
Let’s try an example using the coin toss simulation.
Example 12
A fair coin is tossed 50 times. What is the theoretical and experimental probability for tossing heads on
the fair coin?
Solution
To calculate the theoretical probability, we need to remember that the probability of getting tails is 12 , or:


P(tails) = 1
2


= 0.50


To find the experimental probability, we need to run the coin toss in the probability simulator. We could
also actually take a coin and flip it 50 times, each time recording if we get heads or tails.
If we follow the same keystrokes to get into the app probability simulation, we get to the main screen.


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Choose 1. Toss Coins and then press ZOOM .


Choose OK by pressing GRAPH and go back to the main screen. Then choose TOSS by pressing
WINDOW .


To find the frequency, we need to press the I arrow to view the frequency for the tossing experiment.


We see the frequency is 30. Now we can calculate the experimental probability.


P(tails) = 30
50


= 0.60


Example 13
What if the fair coin is tossed 100 times? What is the experimental probability? Is the experimental
probability getting closer to the theoretical probability?
Solution
To find the experimental probability for this example, we need to run the coin toss in the probability
simulator again. You could also, like in Example 12, actually take a coin and flip it 100 times, each time


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recording if you get heads or tails. You can see how the technology is going to make this experiment take
a lot less time.
Choose 1. Toss Coins and then press ZOOM .


Choose OK by pressing GRAPH and go back to the main screen. Then choose TOSS by pressing
WINDOW .


To find the frequency, we need to press the I arrow to view the frequency for the tossing experiment.


Notice that the frequency is 59. Now you can calculate the experimental probability.


P(tails) = 59
100


= 0.59


With 50 tosses, the experimental probability was 60%, and with 100 tosses, the experimental probability
is 59%. This means that the experimental probability is getting closer to the theoretical probability.
You can also use this same program to toss 2 coins or 5 coins. Actually, you can use this simulation to
toss any number of coins any number of times.
Example 14
Two fair coins are tossed 10 times. What is the theoretical probability of both coins landing on heads?
What is the experimental probability of both coins landing on heads?
Solution
The theoretical probability of getting heads on the first coin is 12 . Flipping the second coin, the theoretical
probability of getting heads is again 12 . The overall theoretical probability is


(


1
2


)2 for 2 coins, or:


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P(2H) = 1
2
× 1


2


P(2H) =
(1
2


)2


P(2H) = 1
4


To determine the experimental probability, let’s go to the probability simulator. Again, you may also do
this experiment manually by taking 2 coins, tossing them 10 times, and recording your observations.


Now let’s toss the coins.


Find the frequency of getting two heads (2H).


The frequency is equal to 4. Therefore, for 2 coins tossed 10 times, there were 4 times that both coins
landed on heads. You can now calculate the experimental probability.


P(2H) = 4
10


P(2H) = 0.40 or 40%


To try another type of probability simulation, you can use the Texas Instruments Activities Exchange. Look
up simple probability simulations on http://education.ti.com/educationportal/activityexchange/
Activity.do?cid=US&aId=9327.
You can also use randBin to simulate the tossing of a coin. The randBin function on your calculator is
used to produce experimental values for discrete random variables. If you wanted to toss 2 coins 10 times
as in Example 14 above, you would follow the keystrokes below.


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This list contains the count of heads resulting from each set of 4 coin tosses. If you use the right arrow
(


I


)


, you can see how many times from the 10 trials you actually had 4 heads.


Example 15
You are in math class. Your teacher asks what the probability is of obtaining five heads if you were to toss
15 coins.
(a) Determine the theoretical probability for the teacher.
(b) Use the TI calculator to determine the actual probability for a trial experiment of 10 trials.
Solution
(a) Let’s calculate the theoretical probability of getting 5 heads in the 15 tosses. In order to do this type
of calculation, let’s bring back the concept of factorial from an earlier lesson.
Numerator (Top)
In the example, you want to have 5 H’s and 10 T‘s. Our favorable outcomes would be HHHHHTTTTTTTTTT,
with the H’s and T’s coming in any order. The number of favorable outcomes would be:


number of favorable outcomes = number of tosses!number of heads! × number of tails!
number of favorable outcomes = 15!


5! × 10!
number of favorable outcomes = 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1


(5 × 4 × 3 × 2 × 1) × (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)


number of favorable outcomes = 1.31 × 10
12


120 × 3628800
number of favorable outcomes = 3003


Denominator (Bottom)


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The number of possible outcomes = 215


The number of possible outcomes = 32,768
Now you just divide the numerator by the denominator:


P(5 heads) = 3003
32768


P(5 heads) = 0.0916


Therefore, the theoretical probability would be 9.16% of getting 5 heads when tossing 15 coins.
b) To calculate the experimental probability, let’s use the randBin function on the TI-84 calculator.


From the list, you can see that you only have 5 heads 1 time in the 10 trials.


Therefore, the experimental probability can be calculated as follows:


P(5 heads) = 1
10


= 10%


Points to Consider


• How is the calculator a useful tool for calculating probabilities in discrete random variable experi-
ments?


• How are these experimental probabilities different from what you would expect the theoretical prob-
abilities to be? When can the two types of probability possibly be equal?


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Vocabulary


Binomial distribution A distribution produced by an experiment where there is a fixed number of
successes in X (random variable) trials, and each trial is independent of the others.


Discrete random variables Only have a specific (or finite) number of numerical values within a certain
range.


Experimental probability The actual probability of an event resulting from an experiment.


Factorial function (!) The function of multiplying a series of consecutive descending natural numbers.


Histogram A graph that uses vertically arranged bars to display data.


Multinomial distribution A distribution produced by an experiment where the number of possible
outcomes is greater than 2 and where each outcome has a specific probability.


Probability distribution A table, a graph, or a chart that shows you all the possible values of X (your
variable), and the probability associated with each of these values (P(X)).


Random variables Variables that takes on numerical values governed by a chance experiment.


Theoretical probability A probability that is the ratio of the number of different ways an event can
occur to the total number of equally likely possible outcomes. The numerical measure of the likelihood
that an event, E, will happen.


P(E) = number of favorable outcomestotal number of possible outcomes


4.5 Review Questions
Answer the following questions and show all work (including diagrams) to create a complete answer.


1. Match the following statements from the first column with the probability values in the second
column.


Table 4.3:


Probability Statement P(X)
a. The probability of this event will never occur. P(X) = 1.0
b. The probability of this event is highly likely. P(X) = 0.33
c. The probability of this event is very likely. P(X) = 0.67
d. The probability of this event is somewhat likely. P(X) = 0.00
e. The probability of this event is certain. P(X) = 0.95


2. Match the following statements from the first column with the probability values in the second


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column.


Table 4.4:


Probability Statement P(X)
a. I bought a ticket for the State Lottery. The
probability of a successful event (winning) is likely
to be:


P(X) = 0.80


b. I have a bag of equal numbers of red and green
jelly beans. The probability of reaching into the
bag and picking out a red jelly bean is likely to be:


P(X) = 0.50


c. My dad teaches math and my mom chemistry.
The probability that I will be expected to study
science or math is likely to be:


P(X) = 0.67


d. Our class has the highest test scores in the State
Math Exams. The probability that I have scored a
great mark is likely to be:


P(X) = 1.0


e. The Chicago baseball team has won every game
in the season. The probability that the team will
make it to the play offs is likely to be:


P(X) = 0.01


3. Read each of the following statements and match the following words to each statement. You can
put your answers directly into the table.
Here is the list of terms you can add:
• certain or sure
• impossible
• likely or probable
• unlikely or improbable
• maybe
• uncertain or unsure


Table 4.5:


Statement Probability Term
Tomorrow is Friday.
I will be in New York on Friday.
It will be dark tonight.
It is snowing in August!
China is cold in January.


4. Read each of the following statements and match the following words to each statement. You can
put your answers directly into the table.
Here is the list of terms you can add:
• certain or sure
• impossible
• likely or probable


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• unlikely or improbable
• maybe
• uncertain or unsure


Table 4.6:


Statement Probability Term
I am having a sandwich for lunch.
I have school tomorrow.
I will go to the movies tonight.
January is warm in New York.
My dog will bark.


5. The probability of scoring above 80% on a math test is 20%. What is the probability of scoring below
80%?


6. The probability of getting a job after university is 85%. What is the probability of not getting a job
after university?


7. Does the following table represent the probability distribution for a discrete random variable?
X 2 4 6 8
P(X) 0.2 0.4 0.6 0.8


8. Does the following table represent the probability distribution for a discrete random variable?
X 1 2 3 4 5
P(X) 0.202 0.174 0.096 0.078 0.055


9. Does the following table represent the probability distribution for a discrete random variable?
X 1 2 3 4 5 6
P(X) 0.302 0.251 0.174 0.109 0.097 0.067


10. A fair die is rolled 10 times. Let X be the number of rolls in which we see a six.
(a) What is the probability of seeing a six in any one of the rolls?
(b) What is the probability that we will see a six exactly once in the 10 rolls?


11. A fair die is rolled 15 times. Let X be the number of rolls in which we see a six.
(a) What is the probability of seeing a six in any one of the rolls?
(b) What is the probability that we will see a six exactly once in the 15 rolls?


12. A fair die is rolled 15 times. Let X be the number of rolls in which we see a five.
(a) What is the probability of seeing a five in any one of the rolls?
(b) What is the probability that we will see a five exactly 7 times in the 15 rolls?


13. A pair of fair dice is rolled 10 times. Let X be the number of rolls in which we see at least one five.
(a) What is the probability of seeing at least one five in any one roll of the pair of dice?
(b) What is the probability that in exactly half of the 10 rolls, we see at least one five?


14. A pair of fair dice is rolled 15 times. Let X be the number of rolls in which we see at least one five.
(a) What is the probability of seeing at least one five in any one roll of the pair of dice?
(b) What is the probability that in exactly 8 of the 15 rolls, we see at least one five?


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15. You are randomly drawing cards from an ordinary deck of cards. Every time you pick one, you place
it back in the deck. You do this seven times. What is the probability of drawing 2 hearts, 2 spades,
2 clubs, and 3 diamonds?


16. A telephone survey measured the percentage of students in ABC town who watch channels NBX,
FIX, MMA, and TSA. After the survey, analysis showed that 35 percent watch channel NBX, 40
percent watch channel FIX, 10 percent watch channel MMA, and 15 percent watch channel TSA.
What is the probability that from seven randomly selected students, one will be watching channel
NBX, two will be watching channel FIX, three will be watching channel MMA, and two will be
watching channel TSA?


17. Based on what you know about probabilities, write definitions for theoretical and experimental prob-
ability.


18. (a) What is the difference between theoretical and experimental probability?
(b) As you add more data, do your experimental probabilities get closer to or further away from


your theoretical probabilities?
(c) Is tossing one coin 100 times the same as tossing 100 coins one time? Why or why not?


19. Use randBin to simulate 5 tosses of a coin 25 times to determine the probability of getting two tails.
20. Use randBin to simulate 10 tosses of a coin 50 times to determine the probability of getting four


heads.
21. Calculate the theoretical probability of getting 3 heads in 10 tosses.
22. Find the experimental probability using technology of getting 3 heads in 10 tosses.
23. Calculate the theoretical probability of getting 8 heads in 12 tosses.
24. Calculate the theoretical probability of getting 7 heads in 14 tosses.


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Chapter 5


Probability Distributions


Introduction
For a standard normal distribution, the data presented is continuous. In addition, the data is centered
at the mean and is symmetrically distributed on either side of that mean. This means that the resulting
data forms a shape similar to a bell and is, therefore, called a bell curve. Binomial experiments are
discrete probability experiments that involve a fixed number of independent trials, where there are only
two outcomes. As a rule of thumb, these trials result in successes and failures, and the probability of
success for one trial is the same as for the next trial (i.e., independent events). As the sample size increases
for a binomial distribution, the resulting histogram approaches the appearance of a normal distribution
curve. With this increase in sample size, the accuracy of the distribution also increases. An exponential
distribution is a distribution of continuous data, and the general equation is in the form y = abx. The
closer the correlation coefficient is to 1, the more likely the equation for the exponential distribution is
accurate.


5.1 Normal Distributions
Learning Objectives


• Be familiar with the standard distributions (normal, binomial, and exponential).
• Use standard distributions to solve for events in problems in which the distribution belongs to one
of those families.


In Chapter 3, you spent some time learning about probability distributions. A distribution, itself, is
simply a description of the possible values of a random variable and the possible occurrences of these
values. Remember that probability distributions show you all the possible values of X (your variable), and
the probability associated with each of these values (P(X)). You were also introduced to the concept of
binomial distributions, or distributions of experiments where there are a fixed number of successes in X
(random variable) trials, and each trial is independent of the other. In addition, you were introduced to
binomial distributions in order to compare them with multinomial distributions. Remember that multino-
mial distributions involve experiments where the number of possible outcomes is greater than 2 and the
probability is calculated for each outcome for each trial.


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In this first lesson on probability distributions, you are going to begin by learning about normal distri-
butions. A normal distribution curve can be easily recognized by its shape. The first two diagrams
above show examples of normal distributions. What shape do they look like? Do they look like a bell
to you? Compare the first two diagrams above to the third diagram. A normal distribution is called a
bell curve because its shape is comparable to a bell. It has this shape because the majority of the data is
concentrated at the middle and slowly decreases symmetrically on either side. This gives it a shape similar
to a bell.


Actually, the normal distribution curve was first called a Gaussian Curve after a very famous mathe-
matician, Carl Friedrich Gauss. He lived between 1777 and 1855 in Germany. Gauss studied many
aspects of mathematics. One of these was probability distributions, and in particular the bell curve. It
is interesting to note that Gauss also spoke about global warming and postulated the eventual finding
of Ceres, the planet residing between Mars and Jupiter. A neat fact about Gauss is that he was also
known to have beautiful handwriting. If you want to read more about Carl Friedrich Gauss, look at
http://en.wikipedia.org/wiki/Carl_Friedrich_Gauss.


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In Chapter 3, you also learned about discrete random variables. Remember that discrete random variables
are ones that have a finite number of values within a certain range. In other words, a discrete random
variable cannot take on all values within an interval. For example, say you were counting from 1 to 10;
you would count 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. These are discrete values. 3.5 would not count as
a discrete value within the limits of 1 to 10. For a normal distribution, however, you are working with
continuous variables. Continuous variables, unlike discrete variables, can take on any value within the
limits of the variable. So, for example, if you were counting from 1 to 10, 3.5 would count as a value for
the continuous variable. Lengths, temperatures, ages, and heights are all examples of continuous variables.
Compare these to discrete variables, such as the number of people attending your class, the number of
correct answers on a test, or the number of tails on a coin flip. You can see how a continuous variable
would take on an infinite number of values, whereas a discrete variable would take on a finite number of
values. As you may know, you can actually see this when you graph discrete and continuous data. Look
at the two graphs below. The first graph is a graph of the height of a child as he or she ages. The second
graph is the cost of a a gallon of gasoline as the years progress.


If you look at the first graph, the data points are joined, because as the child grows from birth to age 1,
for example, his height also increases. As he continues to grow, he continues to age. The data is said to
be continuous and, therefore, you can connect the points on the graph. For the second graph, the price
of a gallon of gas at the end of each year is recorded. In 1930, a gallon of gas cost 10




c. You would not
have gone in and paid 10.2




c or 9.75


c. The data is, therefore, called discrete, and the data points cannot
be connected.
Let’s look at a few problems to show how histograms approximate normal distribution curves.


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Example 1
Jillian takes a survey of the heights of all of the students in her high school. There are 50 students in her
school. She prepares a histogram of her results. Is the data normally distributed?


Solution
If you take a normal distribution curve and place it over Jillian’s histogram, you can see that her data does
not represent a normal distribution.


If the histogram were actually shaped like a normal distribution, it would have a shape like the curve
below.


Example 2
Thomas did a survey similar to Jillian’s in his school. His high school had 100 students. Is his data
normally distributed?


Solution
If you take a normal distribution curve and place it over Thomas’s histogram, you can see that his data
also does not represent a normal distribution.


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Example 3
Joanne posted a problem to her friends on FaceBook. She told her friends that her grade 12 math project
was to measure the lifetime of the batteries used in different toys. She surveyed people in her neighborhood
and asked them, on average, how many hours their typical battery lasts. Her results are shown below.


98 108 107 79 100 112 97 79 41 127
135 100 92 80 66 62 119 118 56 112
99 83 86 62 127 155 107 140 144 122
110 116 134 102 133 157 100 96 55 132
126 171 169 146 122 74 70 82 84 93


Is her data normally distributed?
Solution
If you took a normal distribution curve and placed it over Joanne’s histogram, you can see that her data
appears to come from a normal distribution.


This means that the data fits a normal distribution with a mean around 105. Using the TI-84 calculator,
you can actually find the mean of this data to be 105.7.


What Joanne’s data does tell us is that the mean score (105.7) is at the center of the distribution, and
the data from all of the other scores (times) are spread from that mean. You will be learning much more
about standard normal distributions in a later chapter. But for now, remember the two key points about
a standard normal distribution. The first key point is that the data represented is continuous. The second
key point is that the data is centered at the mean and is symmetrically distributed on either side of that
mean.


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Standard normal distributions are special kinds of distributions and differ from the binomial distributions
you learned about in the last chapter. Let’s now take a more detailed look at binomial distributions and
see how they differ dramatically from the standard normal distribution.


5.2 Binomial Distributions
In the last chapter, you found that binomial experiments are ones that involve only two choices. Each
observation from the experiment, therefore, falls into the category of a success or a failure. For example,
if you tossed a coin to see if a six appears, it would be a binomial experiment. A successful event is the 6
appearing. Every other roll (1, 2, 3, 4, or 5) would be a failure. Asking your classmates if they watched
American Idol last evening is an example of a binomial experiment. There are only two choices (yes and
no), and you can deem a yes answer to be a success and a no answer to be a failure. Coin tossing is another
example of a binomial experiment, as you saw in Chapter 3. There are only two possible outcomes (heads
and tails), and you can say that heads are successes if you are looking to count how many heads can be
obtained. Tails would then be failures. You should also note that the observations are independent of each
other. In other words, whether or not Alisha watched American Idol does not affect whether or not Jack
watched the show. In fact, knowing that Alisha watched American Idol does not tell you anything about
any of your other classmates. Notice, then, that the probability for success for each trial is the same.
The distribution of the observations in a binomial experiment is known as a binomial distribution.
For binomial experiments, there is also a fixed number of trials. As the number of trials increases, the
binomial distribution becomes closer to a normal distribution. You should also remember that for normal
distributions, the random variable is continuous, and for binomial distributions, the random variable is
discrete. This is because binomial experiments have two outcomes (successes and failures), and the counts
of both are discrete.
Of course, as the sample size increases, the accuracy of a binomial distribution also increases. Let’s look
at an example.
Example 4
Keith took a poll of the students in his school to see if they agreed with the new “no cell phones” policy.
He found the following results.


Table 5.1:


Age Number Responding No
14 56
15 65
16 90
17 95
18 60


Solution
Keith plotted the data and found that it was distributed as follows:


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Clearly, Keith’s data is not normally distributed.
Example 5
Jake sent out the same survey as Keith, except he sent it to every junior high school and high school in
the district. These schools had students from grade 6 (age 12) to grade 12 (up to age 20). He found the
following data.


Table 5.2:


Age Number Responding No
12 274
13 261
14 259
15 289
16 233
17 225
18 253
19 245
20 216


Solution
Jake plotted his data and found that the data was distributed as follows:


Because Jake expanded the survey to include more people (and a wider age range), his data turned out to
fit more closely to a normal distribution.
Notice that in Jake’s sample, the number of people responding no on the survey was 2,225. This number
has a huge advantage over the number of students who responded no on Keith’s survey, which was 366.
You can make much more accurate conclusions with Jake’s data. For example, Jake could say that the
biggest opposition to the new “no cell phone” policy came from the middle or junior high schools, where
students were 12–15 years old. This is not the same conclusion Keith would have made based on his small
sample.
In Chapter 3, you did a little work with the formula used to calculate probability for binomial experiments.
Here is the general formula for finding the probability of a binomial experiment from Chapter 3.


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The probability of getting x successes in n trials is given by:


P(X = a) = nCa × pa × q(n−a)


where:
a is the number of successes from the trials.
p is the probability of success.
q is the probability of failure.
Let’s look at a simple example of a binomial probability distribution just to recap what you learned in
Chapter 3.
Example 6
A coin is tossed 5 times. Find the probability of getting exactly 2 tails.
Solution
There are 5 trials, so n = 5.
A success is getting a tail. We are interested in exactly 2 successes. Therefore, a = 2.
The probability of a success is 12 , and, thus, p = 12 .
Therefore, the probability of a failure is 1 − 12 , or 12 . From this, you know that q = 12 .


P(X = a) = nCa × pa × q(n−a)


P(2 tails) = 3C2 × p2 × q1


P(2 tails) = 3C2 ×
(1
2


)2
×
(1
2


)1


P(2 tails) = 3 × 1
4
× 1


2
P(2 tails) = 3


8


Therefore, the probability of seeing exactly 2 tails in 5 tosses is 38 , or 37.5%.
Example 7
A local food chain has determined that 40% of the people who shop in the store use an incentive card,
such as air miles. If 10 people walked into the store, what is the probability that half of them will be using
an incentive card?
Solution
There are 10 trials, so n = 10.
A success is a person using a card. You are interested in 5 successes. Therefore, a = 5.
The probability of a success is 40%, or 0.40, and, thus, p = 0.40.
Therefore, the probability of a failure is 1 − 0.40, or 0.60. From this, you know that q = 0.60.


P(X = a) = nCa × pa × q(n−a)


P(5 people using a card) = 10C5 × p5 × q5


P(5 people using a card) = 10C5 × (0.40)5 × (0.60)5
P(5 people using a card) = 252 × 0.01024 × 0.07776
P(5 people using a card) = 0.201


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Therefore, the probability of seeing 5 people using a card in a random set of 10 people is 20.1%.
Technology Note
You could have used technology to solve this problem, rather than pencil and paper calculations. However,
with technology, it is often very helpful to check our answers using pencil and paper as well. With Example
7, you could have used binompdf on the TI-84 calculator to solve this problem.
The key sequence to find binompdf is as follows:


If you used the data from Example 7, you would find the following:


Notice that you typed in binompdf(n, p, a). What if you changed the problem to include the phrase at
most? In reality, probability problems are normally ones that use at least, more than, less than, or at
most. You do not always have a probability problem using the word exactly. Take a look at Example 8, a
problem similar to Example 7.
Example 8
A local food chain has determined that 40% of the people who shop in the store use an incentive card,
such as air miles. If 10 people walked into the store, what is the probability that at most half of these will
be using an incentive card?
Solution
There are 10 trials, so n = 10.
A success is using a card. We are interested in at most 5 people using a card. That is, we are interested
in 0, 1, 2, 3, 4, or 5 people using a card. Therefore, a = 5, 4, 3, 2, 1, and 0.
The probability of a success is 40%, or 0.40, and, thus, p = 0.40.
Therefore, the probability of a failure is 1 − 0.40, or 0.60. From this, you know that q = 0.60.


P(X = a) = nCa × pa × q(n−a)


P(5 people using a card) = 10C5 × p5 × q5


P(5 people using a card) = 10C5 × (0.40)5 × (0.60)5
P(5 people using a card) = 252 × 0.01024 × 0.07776
P(5 people using a card) = 0.201


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P(4 people using a card) = 10C4 × p4 × q6


P(4 people using a card) = 10C4 × (0.40)4 × (0.60)6
P(4 people using a card) = 210 × 0.0256 × 0.04666
P(4 people using a card) = 0.251


P(3 people using a card) = 10C3 × p3 × q7


P(3 people using a card) = 10C3 × (0.40)3 × (0.60)7
P(3 people using a card) = 120 × 0.064 × 0.02799
P(3 people using a card) = 0.215


P(2 people using a card) = 10C2 × p2 × q8


P(2 people using a card) = 10C2 × (0.40)2 × (0.60)8
P(2 people using a card) = 45 × 0.16 × 0.01680
P(2 people using a card) = 0.121


P(1 person using a card) = 10C1 × p1 × q9


P(1 person using a card) = 10C1 × (0.40)1 × (0.60)9
P(1 person using a card) = 10 × 0.40 × 0.01008
P(1 person using a card) = 0.0403


P(0 people using a card) = 10C0 × p0 × q10


P(0 people using a card) = 10C0 × (0.40)0 × (0.60)10
P(0 people using a card) = 1 × 1 × 0.00605
P(0 people using a card) = 0.00605


The total probability for this example is:


P(X ≤ 5) = 0.201 + 0.251 + 0.215 + 0.121 + 0.0403 + 0.0605
P(X ≤ 5) = 0.834


Therefore, the probability of seeing at most 5 people using a card in a random set of 10 people is 82.8%.
Technology Note
You can see now that the use of the TI-84 calculator can save a great deal of time when solving problems
involving the phrases at least, more than, less than, or at most. This is due to the fact that the calculations
become much more cumbersome. You could have used binomcdf on the TI-84 calculator to solve Example
8. Binomcdf stands for binomial cumulative probability. Binompdf is simply binomial probability.
The key sequence to find binompdf is as follows:


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If you used the data from Example 8, you would find the following:


You can see how using binomcdf is a lot easier than actually calculating 6 probabilities and adding them
up. If you were to round 0.8337613824 to three decimal places, you would get 0.834, which is our calculated
value found in Example 8.
Example 9
Karen and Danny want to have 5 children after they get married. What is the probability that they will
have exactly 3 girls?
Solution
There are 5 trials, so n = 5.
A success is when a girl is born, and we are interested in 3 girls. Therefore, a = 3.
The probability of a success is 50%, or 0.50, and thus, p = 0.50.
Therefore, the probability of a failure is 1 − 0.50, or 0.50. From this, you know that q = 0.50.


P(X = a) = nCa × pa × q(n−a)


P(3 girls) = 5C3 × p3 × q2


P(3 girls) = 5C3 × (0.50)3 × (0.50)2
P(3 girls) = 10 × 0.125 × 0.25
P(3 girls) = 0.3125


Therefore, the probability of having exactly 3 girls from the 5 children is 31.3%.
When using technology, you will select binompdf, because you are looking for the probability of exactly 3
girls from the 5 children.


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Using the TI-84 calculator gave us the same result as our calculation (and was a great deal quicker).
Example 10
Karen and Danny want to have 5 children after they get married. What is the probability that they will
have less than 3 girls?
Solution
There are 5 trials, so n = 5.
A success is when a girl is born, and we are interested in less than 3 girls. Therefore, a = 2, 1, and 0.
The probability of a success is 50%, or 0.50, and, thus, p = 0.50.
Therefore, the probability of a failure is 1 − 0.50, or 0.50. From this, you know that q = 0.50.


P(X = a) = nCa × pa × q(n−a)


P(2 girls) = 5C2 × p2 × q3


P(2 girls) = 5C2 × (0.50)2 × (0.50)3
P(2 girls) = 10 × 0.25 × 0.125
P(2 girls) = 0.3125


P(1 girl) = 5C2 × p1 × q4


P(1 girl) = 5C1 × (0.50)1 × (0.50)4
P(1 girl) = 5 × 0.50 × 0.0625
P(1 girl) = 0.1563


P(0 girls) = 5C0 × p0 × q5


P(0 girls) = 5C0 × (0.50)0 × (0.50)5
P(0 girls) = 1 × 1 × 0.03125
P(0 girls) = 0.03125


The total probability for this example is:


P(X < 3) = 0.3125 + 0.1563 + 0.3125
P(X < 3) = 0.500


Therefore, the probability of having less than 3 girls in 5 children is 50.0%.
When using technology, you will select binomcdf, because you are looking for the probability of less than
3 girls from the 5 children.


Example 11


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A fair coin is tossed 50 times. What is the probability that you will get heads in 30 of these tosses?
Solution
There are 50 trials, therefore n = 50.
A success is getting a head, and we are interested in exactly 30 heads. Therefore, a = 30.
The probability of a success is 50%, or 0.50, and, thus, p = 0.50.
Therefore, the probability of a failure is 1 − 0.50, or 0.50. From this, you know that q = 0.50.


P(X = a) = nCa × pa × q(n−a)


P(3 girls) = 50C30 × p30 × q20


P(3 girls) = 50C30 × (0.50)30 × (0.50)20


P(3 girls) = (4.713 × 1013) × (9.313 × 10−10) × (9.537 × 10−7)
P(3 girls) = 0.0419


Therefore, the probability of getting exactly 30 heads from 50 tosses of a fair coin is 4.2%.
Using technology to check, you get:


Example 12
A fair coin is tossed 50 times. What is the probability that you will get heads in at most 30 of these tosses?
Solution
There are 50 trials, so n = 50.
A success is getting a head, and we are interested in at most 30 heads. Therefore, a = 30, 29, 28, 27, 26, 25,
24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and, 0.
The probability of a success is 50%, or 0.50, and, thus, p = 0.50.
Therefore, the probability of a failure is 1 − 0.50, or 0.50. From this, you know that q = 0.50.
Obviously, you will be using technology here to solve this problem, as it would take us a long time to
calculate all of the individual probabilities.


Therefore, the probability of having at most 30 heads from 50 tosses of a fair coin is 94.1%.
Example 13


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A fair coin is tossed 50 times. What is the probability that you will get heads in at least 30 of these tosses?
Solution
There are 50 trials, so n = 50.
A success is getting a head, and we are interested in at least 30 heads. Therefore, a = 50, 49, 48, 47, 46, 45,
44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, and 30.
The probability of a success is 50%, or 0.50, and, thus, p = 0.50.
Therefore, the probability of a failure is 1 − 0.50, or 0.50. From this, you know that q = 0.50.
Again, you will obviously be using technology here to solve this problem, as it would take us a long time
to calculate all of the individual probabilities.


Notice that when you use the phrase at least, you used the numbers 50, 0.5, 29. In other words, you would
type in 1 - binomcdf (n, p, a − 1). Since a = 30, at least a would be anything greater than 30. Therefore,
the probability of having at least 30 heads from 50 tosses of a fair coin is 10.1%.
Example 14
You have a summer job at a jelly bean factory as a quality control clerk. Your job is to ensure that the
jelly beans coming through the line are the right size and shape. If 90% of the jelly beans you see are the
right size and shape, you give your thumbs up, and the shipment goes through to processing and on to the
next phase to shipment. A normal day at the jelly bean factory means 15 shipments are produced. What
is the probability that exactly 10 will pass inspection?
Solution
There are 15 shipments, so n = 15.
A success is a shipment passing inspection, and we are interested in exactly 10 passing inspection.
Therefore, a = 10.
The probability of a success is 90%, or 0.90, and, thus, p = 0.90.
Therefore, the probability of a failure is 1 − 0.90, or 0.10. From this, you know that q = 0.10.


P(X = a) = nCa × pa × q(n−a)


P(10 shipments passing) = 15C10 × p10 × q5


P(10 shipments passing) = 15C10 × (0.90)10 × (0.10)5


P(10 shipments passing) = 3003 × 0.3487 × (1.00 × 10−5)
P(10 shipments passing) = 0.0105


Therefore, the probability that exactly 10 of the 15 shipments will pass inspection is 1.05%.


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5.3 Exponential Distributions
A third type of probability distribution is an exponential distribution. When we discussed normal dis-
tributions, or standard distributions, we talked about the fact that these distributions used continuous
data, so you could use standard distributions when talking about heights, ages, lengths, temperatures,
and the like. The same types of data are used when discussing exponential distributions. Exponential
distributions, contrary to standard distributions, deal more with rates or changes over time. For example,
the length of time the battery in your car will last will be an exponential distribution. The length of time is
a continuous random variable. A continuous random variable is one that can form an infinite number
of groupings. So time, for example, can be broken down into hours, minutes, seconds, milliseconds, and so
on. Another example is the lifetime of computer parts. Different computer parts have different life spans,
depending on their use (and abuse). The rate of decay of a computer part is an example of an exponential
distribution.


Let’s look at the differences between our normal distribution curve, the binomial distribution histogram,
and now our exponential distribution. List some of the similarities and differences that you see in the
figures below.


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Notice with the standard distribution and the exponential distribution curves, the data represents contin-
uous variables. The data in the binomial distribution histogram is discrete. The curve for the standard
distribution is symmetrical about the mean. In other words, if you draw a horizontal line through the cen-
ter of the curve, the two halves of the standard distribution curve would be mirror images of each other.
This symmetry is not the case for the exponential distribution curve (nor for the binomial distribution).
Did you notice anything else?
Let’s look at some examples where the resulting graphs would show you an exponential distribution.
Example 15
ABC Computer Company is doing a quality control check on their newest core chip. They randomly chose
25 chips from a batch of 200 to test and ran them to see how long they would continuously run before
failing. The following results were obtained.


Table 5.3:


Number of chips Hours to failure
8 1000
6 2000
4 3000
3 4000
2 5000
2 6000


What kind of data is represented in the table?
Solution
In order to solve this problem, you need to graph it to see what it looks like. You can use graph paper or
your calculator. Entering the data into the TI-84 involves the following keystrokes. There are a number
of them, because you have to enter the data into lists 1 and 2, and then plot the lists using STAT PLOT.


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After you press GRAPH , you get the following curve.


This curve looks somewhat like an exponential distribution curve, but let’s test it out. You can do this on
the TI-84 using the CALC feature under STAT.


Notice that the r2 value is close to 1. This value indicates that the exponential is a good fit for this data
and that the data, therefore, represents an exponential distribution.
You use regression to determine a rule that best explains the data you are observing. There is a standard
quantitative measure of this best fit, known as the coefficient of determination (r2). The value of r2
can be from 0 to 1, and the closer the value is to 1, the better the fit. In our data above, the r2 value is
0.9856 for the exponential regression. If we had gone and done a quadratic regression, our r2 value would
have been 0.9622. The data is not linear, but if we thought it might be, the r2 value would have been
0.9161. Remember, the higher the value, the better the fit.
You can even go one step further and graph this line on top of our plotted points. Follow the keystrokes
below and test it out.


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Note: It was not indicated that the data was in L1 and L2 when finding the exponential regression. This
is because it is the default of the calculator. If you had used L2 and L3, you would have had to add this
to your keystrokes.


Take a look at the formula that you used with the exponential regression calculation (ExpReg) above.
The general formula was y = abx. This is the characteristic formula for the exponential distribution
curve. Siméon Poisson was one of the first to study exponential distributions with his work in applied
mathematics. The Poisson distribution, as it is known, is a form of an exponential distribution. He
received little credit for his discovery during his lifetime, as it only found application in the early part
of the 20th century, almost 70 years after Poisson had died. To read more about Siméon Poisson, go to
http://en.wikipedia.org/wiki/Sim%C3%A9on_Denis_Poisson.
Example 16
Radioactive substances are measured using a Geiger-Müller counter (or a Geiger counter for short). Robert
was working in his lab measuring the count rate of a radioactive particle. He obtained the following data.


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Table 5.4:


Time (hr) Count (atoms)
15 544
12 272
9 136
6 68
3 34
1 17


Is this data representative of an exponential function? If so, find the equation. What would be the count
at 7.5 hours?
Solution
Remember, we can plot this data using pencil and paper, or we can use a graphing calculator. We will use
a graphing calculator here.


The resulting graph appears as follows:


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At a glance, it does look like an exponential curve, but we really have to take a closer look by doing the
exponential regression.


In the analysis of the exponential regression, we see that the r2 value is close to 1, and, therefore, the curve
is indeed an exponential curve. We should go one step further and graph this exponential equation onto
our stat plot and see how close a match it is.


The equation, therefore, is y = 15.06(1.274x).
The last part of our question asked us to determine what the count was after 7.5 hours. In other words,
what is y when x = 7.5?


y = 15.06 (1.274x)


y = 15.06
(


1.2747.5
)


y = 15.06 (6.149)
y = 92.6 atoms


We can check this on our calculator as follows:


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Our calculation is a bit over, because we rounded the values for a and b in the equation y = abx, whereas
the calculator did not.
Example 17
Jack believes that the concentration of gold decreases exponentially as you move further and further away
from the main body of ore. He collects the following data to test out his theory.


Table 5.5:


Distance (m) Concentration (g/t)
0 320
400 80
800 20
1200 5
1600 1.25
2000 0.32


Is this data representative of an exponential function? If so, find the equation. What is the concentration
at 1000 m?
Solution
Again, we can plot this data using pencil and paper, or we can use a graphing calculator. We will, as with
Example 16, use a graphing calculator here.


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The resulting graph appears as follows:


At a glance, it does look like an exponential curve, but we really have to take a closer look by doing the
exponential regression.


In the analysis of the exponential regression, we see that the r2 value is close to 1, and, therefore, the curve
is indeed an exponential curve. We will go one step further and graph this exponential equation onto our
stat plot and see how close a match it is.


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The equation, therefore, is y = 318.56 (0.9965x).
The question asks, “What is the concentration at 1000 m?”


y = 318.56 (0.9965x)


y = 318.56
(


0.99651000
)


y = 318.56(0.03001)
y = 9.56 g/t


Therefore, the concentration of gold is 9.56 grams of gold per ton of rock.
We can check this on our calculator as follows:


Our calculation is a bit under, because we rounded the values for a and b in the equation y = abx, whereas
the calculator did not.
Points to Consider


• Why is a normal distribution considered to be a continuous probability distribution, whereas a
binomial distribution is considered to be a discrete probability distribution?


• How can you tell that a curve is truly an exponential distribution curve?


Vocabulary


Binomial experiments Experiments that include only two choices, with distributions that involve a
discrete number of trials of these two possible outcomes.


Binomial distribution A probability distribution of the successful trials of a binomial experiment.


Continuous random variable A variable that can form an infinite number of groupings.


Continuous variables Variables that take on any value within the limits of the variable.


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Continuous data Data where an infinite number of values exist between any two other values. Data
points are joined on a graph.


Coefficient of determination (r2) A standard quantitative measure of best fit. Has values from 0 to 1,
and the closer the value is to 1, the better the fit.


Discrete values Data where a finite number of values exist between any two other values. Data points
are not joined on a graph.


Distribution The description of the possible values of a random variable and the possible occurrences
of these values.


Exponential distribution A probability distribution showing a relation in the form y = abx.


Normal distribution curve A symmetrical curve that shows the highest frequency in the center (i.e.,
at the mean of the values in the distribution) with an equal curve on either side of that center.


Standard distributions Normal distributions, which are often referred to as bell curves.


5.4 Review Questions
Answer the following questions and show all work (including diagrams) to create a complete answer.


1. Look at the following graphs and indicate whether they are binomial distributions, normal (standard)
distributions, or exponential distributions. Explain how you know.
(a)


(b)


(c)


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(d)


2. Look at the following graphs and indicate whether they are binomial distributions, normal (standard)
distributions, or exponential distributions. Explain how you know.


(a)


(b)


(c)


(d)


3. It is determined that for a particular genetic trend in a family, the probability of having a boy is
60%. Janet and David decide to have 4 children. What is the probability that they will have exactly
2 boys?


4. For question 3, what is the probability that Janet and David will have at least 2 boys?


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5. For question 3, what is the probability that Janet and David will have at most 2 boys?
6. The following data was collected on a recent 25-point math quiz. Does the data represent a normal
distribution? Can you determine anything from the data?


20 17 22 23 25
14 15 14 17 9
18 2 11 18 19
14 21 19 20 18
16 13 14 10 12


7. A recent blockbuster movie was rated PG, with an additional violence warning. The manager of a
movie theater did a survey of movie goers to see what ages were attending the movie in an attempt
to see if people were adhering to the warnings. Is his data normally distributed? Do movie goers at
the theater regularly adhere to warnings?


17 9 20 27 16
15 14 24 19 14
19 7 21 18 12
5 10 15 23 14
17 13 13 12 14


8. The heights of coniferous trees were measured in a local park in a regular inspection. Is the data
normally distributed? Are there areas of the park that seem to be in danger? The measurements are
all in feet.


22.8 9.7 23.2 21.2 23.5
18.2 7.0 8.8 25.7 19.4
25.0 8.8 23.0 23.2 20.1
23.1 18.5 21.7 21.7 9.1
4.3 7.8 3.4 20.0 8.5


9. Thomas is studying for his AP Biology final. In order to complete his course, he must do a self-
directed project. He decides to swab a tabletop in the student lounge and test for bacteria growing on
the surface. Every hour, he looks in his Petri dish and makes an estimate of the number of bacteria


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present. The following results were recorded.


Table 5.6:


Time (hr) Bacteria count
0 1
1 6
2 40
3 215
4 1300
5 7800


Is this data representative of an exponential function? If so, find the equation. What is the count after 1
day?


10. If you watch a grasshopper jump, you will notice the following trend.


Table 5.7:


Jump number Distance (m)
1 4
2 2
3 1.1
4 0.51
5 0.25
6 0.13


Is this data representative of an exponential function? If so, find the equation. Why do you think the
grasshopper’s distance decreased with each jump?


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Chapter 6


Measures of Central Tendency


Introduction
Here’s an activity that will involve all the students in your class and will also serve as a learning tool to
enhance your understanding of the measures of central tendency, which are mean, median and mode.
Prior to the beginning of class, fill a pail with single, plastic interlocking blocks similar to those shown
below. You and your classmates will each use only one hand to gather a handful of blocks from the pail.


Before you and your classmates begin to pick your handfuls of blocks, have a brain-storming discussion to
reveal your knowledge of the measures of central tendency. Record the various responses and refer to these
as the lessons progress.
You and your classmates can now each proceed to the pail to collect a handful of blocks. Once you have
had some time to compare your handful with those of your classmates, record each of your numbers of
blocks on post-it notes. The post-it notes for you and your classmates can now be placed in order on a
large sheet of grid paper. The grid paper allows for repeated numbers to be posted in the same column.
What do you think you would be finding if you were to determine the mean number of blocks that had
been picked from the pail? Now share your blocks with your classmates, and have your classmates do the
same. so that you each have a similar number of blocks. From this sharing process, it is very likely that
two groups of students will be created. One group will have stacks of one number of blocks, and another
group will have stacks of another number of blocks. You and your classmates may come to realize that
further sharing will not create stacks of the same size for each of you. Is it clearer to you now what we are
talking about when we use the term mean?
Place your stacks of cubes in a safe place, for they will be used again in the discovery of the mode and the
median. The numbers that were placed on the grid paper can also be used for mathematical calculations
of the mean, median, and mode of your data.


6.1 The Mean
Learning Objectives


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• Understand the mean of a set of numerical data.
• Compute the mean of a given set of data.
• Understand the affect of an outlier on the mean of a set of data.
• Understand the mean of a set of data as it applies to real world situations.


Now that you have had some fun discovering what you are finding when you are looking for the mean of
a set of data, it is time to actually calculate the mean of your handfuls of blocks.
The term “central tendency” refers to the middle, or typical, value of a set of data, which is most commonly
measured by using the three m’s−mean, median and mode. In this lesson, we will explore the mean, and
then we will move on to the median and the mode in the following lessons.
The mean, often called the ‘average’ of a numerical set of data, is simply the sum of the data values
divided by the number of values. This is also referred to as the arithmetic mean. The mean is the balance
point of a distribution.
To calculate the actual mean of your handfuls of blocks, you can use the numbers that were posted on
your grid paper. These posted numbers represent the number of blocks that were picked by each student
in your class. Therefore, you are calculating the mean of a population. A population is a collection of all
elements whose characteristics are being studied. You are not calculating the mean number of some of the
blocks, but you are calculating the mean number of all of the blocks. We will use the example below for
our calculations:


From the grid paper, you can see that there were 30 students who posted their numbers of blocks. The
total number of blocks picked by all the students was:


1 × 2 + 2 × 3 + 3 × 4 + 5 × 5 + 3 × 6 + 4 × 7 + 3 × 8 + 2 × 9 + 3 × 10 + 3 × 11 + 1 × 12
2 + 6 + 12 + 25 + 18 + 28 + 24 + 18 + 30 + 33 + 12 = 208


The sum of all the blocks is 208, and the mean is the number you get when you divide the sum by the
number of students who placed a post-it-note on the grid paper. The mean number of blocks is, therefore,
208
30 ≈ 6.93. This means that, on average, each student picked 7 blocks from the pail.
When calculations are done in mathematics, formulas are often used to represent the steps that are being
applied. The symbol ∑ means “the sum of“ and is used to represent the addition of numbers. The numbers
in every question are different, so the variable x is used to represent the numbers. To make sure that all the
numbers are included, a subscript is often used to name the numbers. Therefore, the first number in the
example can be represented as x1. The number of data values for a population is written as N. The mean
of the population is denoted by the symbol µ, which is pronounced ”mu.” The following formula represents
the steps that are involved in calculating the mean of a set of data:


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Mean = sum of the valuesthe number of values


This formula can also be written using symbols:


µ =


x1 + x2 + x3 + . . . + xn
N


You can now use the formula to calculate the mean of the blocks:


µ =


x1 + x2 + x3 + . . . + xn
N


µ = 2 + 6 + 12 + 25 + 18 + 28 + 24 + 18 + 30 + 33 + 12
30


µ = 208
30


µ ≈ 6.93


This means that, on average, each student picked 7 blocks from the pail.
Example 1
Stephen has been working at Wendy’s for 15 months. The following numbers are the number of hours that
Stephen worked at Wendy’s for each of the past seven months:


24, 24, 31, 50, 53, 66, 78


What is the mean number of hours that Stephen worked each month?
Solution
Step 1: Add the numbers to determine the total number of hours he worked.


24 + 25 + 33 + 50 + 53 + 66 + 78 = 329


Step 2: Divide the total by the number of months.
329
7 = 47
The mean number of hours that Stephen worked each month was 47.
Stephen has worked at Wendy’s for 15 months, but the numbers given above are for seven months. There-
fore, this set of data represents a sample, which is a portion of the population. The formula that was used
to calculate the mean of the blocks must be changed slightly to represent a sample. The mean of a sample
is denoted by x̄, which is called “x bar.”
The number of data values for a sample is written as n. The following formula represents the steps that
are involved in calculating the mean of a sample:


Mean = sum of the valuesthe number of values


This formula can now be written using symbols:


x =


x1 + x2 + x3 + . . . + xn
n


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You can now use the formula to calculate the mean of the hours that Stephen worked:


x =


x1 + x2 + x3 + . . . + xn
n


x = 24 + 25 + 33 + 50 + 53 + 66 + 78
7


x = 329
7


x = 47


The mean number of hours that Stephen worked each month was 47.
The formulas only differ in the symbol used for the mean and the case of the variable used for the number of
data values (N or n). The calculations are done the same way for both a population and a sample. However,
the mean of a population is constant, while the mean of a sample changes from sample to sample.
Example 2
Mark operates a shuttle service that employs eight people. Find the mean age of these workers if the ages
of the eight employees are:


55 63 34 59 29 46 51 41


Solution
Since the data set includes the ages of all eight employees, it represents a population.


µ =


x1 + x2 + x3 + . . . + xn
N


µ = 55 + 63 + 34 + 59 + 29 + 46 + 51 + 41
8


µ = 378
8


µ = 47.25


The mean age of all eight employees is 47.25 years, or 47 years and 3 months.
If you were to take a sample of three employees from the group of eight and calculate the mean age for
those three workers, would the result change? Let’s use the ages 55, 29, and 46 for one sample of three,
and the ages 34, 41, and 59 for another sample of three:


x =


x1 + x2 + x3 + . . . + xn
n


x =


x1 + x2 + x3 + . . . + xn
n


x = 55 + 29 + 46
3


x = 34 + 41 + 59
3


x = 130
3


x = 134
3


x = 43.33 x = 44.66


The mean age of the first three employees is 43.33 years.
The mean age of the second group of three employees is 44.66 years.
The mean age for a sample of a population depends upon what values of the population are included in
the sample. From this example, you can see that the mean of a population and that of a sample from the
population are not necessarily the same.


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Example 3
The selling prices of the last 10 houses sold in a small town are listed below:


$125, 000 $142, 000 $129, 500 $89, 500 $105, 000
$144, 000 $168, 300 $96, 000 $182, 300 $212, 000


Calculate the mean selling price of the last 10 homes that were sold.
Solution
The prices are those of a sample.


x =


x1 + x2 + x3 + . . . + xn
n


x = 125, 000 + 142, 000 + 129, 500 + 89, 500 + 105, 000 + $144, 000 + 168, 300 + 96, 000 + 182, 300 + 212, 000
10


x = $1, 393, 600
10


x = $139, 360


The mean selling price of the last 10 homes that were sold was $139,360.
The mean value is one of the three m’s and is a measure of central tendency. It is a summary statistic that
gives you a description of the entire data set and is especially useful with large data sets, where you might
not have the time to examine every single value. You can also use the mean to calculate further descriptive
statistics, such as the variance and standard deviation. These topics will be explored in a future lesson.
The mean assists you in understanding and making sense of your data, since it uses all of the values in the
data set in its calculation.
When a data set is large, a frequency distribution table is often used to display the data in an organized
way. A frequency table lists the data values, as well as the number of times each value appears in the data
set. A frequency table is easy to both read and interpret.
The numbers in a frequency table do not have to be put in order. To make it easier to enter the values in
the table, a tally column is often inserted. Inserting a tally column allows you to account for every value
in the data set, without having to continually scan the numbers to find them in the list. A slash (/) is
used to represent the presence of a value in the list, and the total number of slashes will be the frequency.
If a tally column is inserted, the table will consist of three columns, and if no tally column is inserted, the
table will consist of two columns. Let’s examine this concept with an actual problem and data.
Example 4
Sixty students were asked how many books they had read over the past twelve months. The results are
listed in the frequency table below. Calculate the mean number of books read by each student.


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Table 6.1:


Number of Books Number of Students (Frequency)
0 1
1 6
2 8
3 10
4 13
5 8
6 5
7 6
8 3


Solution
To determine the total number of books that were read by the students, each number of books must be
multiplied by the number of students who read that particular number of books. Then, all the products
must be added to determine the total number of books read. This total number divided by 60 will tell you
the mean number of books read by each student. The formula that was written to determine the mean,
x =




x1+x2+x3+...+xn
n , does not show any multiplication of the numbers by their frequencies. However, this


can be easily inserted into this formula as shown below:


x =


x1 f1 + x2 f2 + x3 f3 + . . . + xn fn
f1 + f2 + f3 + . . . + fn


This formula will be used to calculate the mean number of books read by each student.


x =


x1 f1 + x2 f2 + x3 f3 + . . . + xn fn
f1 + f2 + f3 + . . . + fn


x =


(0)(1) + (1)(6) + (2)(8) + (3)(10) + (4)(13) + (5)(8) + (6)(5) + (7)(6) + (8)(3)
1 + 6 + 8 + 10 + 13 + 8 + 5 + 6 + 3


x =


0 + 6 + 16 + 30 + 52 + 40 + 30 + 42 + 24
60


x = 240
60


x = 4


The mean number of books read by each student was 4 books.


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Suppose the numbers of books read by each student were randomly listed, and it was your job to determine
the mean of the numbers.


0 5 1 4 4 6 7 2 4 3 7 2 6 4 2
8 5 8 3 4 3 6 4 5 6 1 1 3 5 4
1 5 4 1 7 3 5 4 3 8 7 2 4 7 2
1 4 6 3 2 3 5 3 2 4 7 2 5 4 3


An alternative to entering all the numbers into a calculator would be to create a frequency distribution
table like the one shown below.


Table 6.2:


Number of Books Tally Number of Students (Fre-
quency)


0 | 1
1




|||| | 6
2




|||| ||| 8
3




||||


|||| 10
4




||||


|||| ||| 13
5




|||| ||| 8
6




|||| 5
7




|||| | 6
8 ||| 3


Now that the data has been organized, the numbers of books read and the numbers of students who read
the books are evident. The mean can be calculated as it was above.
Example 5
The following data shows the height in centimeters of a group of grade 10 students:


183 171 158 171 182 158 164 183
179 170 182 183 170 171 167 176
176 164 176 179 183 176 170 183
183 167 167 176 171 182 179 170


Organize the data in a frequency table and calculate the mean height of the students.


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Solution
Table 6.3:


Height of Students(cm) Tally Number of Students (Fre-
quency)


171 |||| 4
158 || 2
176




|||| 5
182 ||| 3
164 || 2
179 ||| 3
170 |||| 4
183




|||| | 6
167 ||| 3


x =


x1 f1 + x2 f2 + x3 f3 + . . . + xn fn
f1 + f2 + f3 + . . . + fn


x =


(171)(4) + (158)(2) + (176)(5) + (182)(3) + (164)(2) + (179)(3) + (170)(4) + (183)(6) + (167)(3)
4 + 2 + 5 + 3 + 2 + 3 + 4 + 6 + 3


x =


684 + 316 + 880 + 546 + 328 + 537 + 680 + 1098 + 501
32


x = 5570
32


≈ 174.1 cm


The mean height of the students is approximately 174.1 cm.
The mean is often used as a summary statistic. However, it is affected by extreme values, or outliers. This
means that when there are extreme values at one end of a data set, the mean is not a very good summary
statistic. For example, if you were employed by a company that paid all of its employees a salary between
$60,000 and $70,000, you could probably estimate the mean salary to be about $65,000. However, if you
had to add in the $150,000 salary of the CEO when calculating the mean, then the value of the mean
would increase greatly. It would, in fact, be the mean of the employees’ salaries, but it probably would not
be a good measure of the central tendency of the salaries.
Technology is a major tool that is available for you to use when doing mathematical calculations, and
its use goes beyond entering numbers to perform simple arithmetic operations. For example, the TI-83
calculator can be used to determine the mean of a set of given data values. You will first learn to calculate
the mean by simply entering the data values in a list and determining the mean. The second method that
you will learn about utilizes the frequency table feature of the TI-83.
Example 6
Using technology, determine the mean of the following set of numbers:


24, 25, 25, 25, 26, 26, 27, 27, 28, 28, 31, 32


Solution
Step One:


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Step Two:


Notice that the sum of the data values is 324 (∑ x = 324).
Notice that the number of data values is 12 (n = 12).
Notice the mean of the data values is 27 (x = 27).
Now we will use the same data values and use the TI-83 to create a frequency table.
Step One:


Step Two:


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Step Three:


Step Four:
Press 2ND 0 to obtain the CATALOGUE function of the calculator. Scroll down to sum( and enter
L3 → .


You can repeat this step to determine the sum of L2 → .


x = 324
12


= 27


Note that not all the data values and frequencies are visible in the screenshots, but rest assured that they
were all entered into the calculator.
After entering the data into L1 and the frequencies into L2, another way to solve this problem with the
calculator would have been to press 2ND STAT , go to the MATH menu, choose option 3, and enter L1
and L2 so that you have mean(L1, L2). Then press ENTER to get the answer. This way, the calculator
will do all the calculations for you.
In addition to calculating the mean for a given set of data values, you can also apply your understanding
of the mean to determine other information that may be asked for in everyday problems.
Example 7
During his final season with the Cadillac Selects, Joe Sure Shot played 14 regular season basketball games
and had an average of 24.5 points per game. In the first two playoff games, Joe scored 18 and 26 points,
respectively. Determine his new average for the season.
Solution
Step One: Multiply the given average by 14 to determine the total number of points he had
scored before the playoff games.


24.5 × 14 = 343


Step Two: Add the points from the two playoff games to this total.


343 + 18 + 26 = 387


Step Three: Divide this new total by 16 to determine the new average.


x = 387
16


≈ 24.19


All of the values for the means that you have calculated so far have been for ungrouped, or listed, data.
A mean can also be determined for data that is grouped, or placed in intervals. Unlike listed data, the


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individual values are not available, and you are not able to calculate their sum. To calculate the mean of
grouped data, the first step is to determine the midpoint of each interval, or class. These midpoints must
then be multiplied by the frequencies of the corresponding classes. The sum of the products divided by
the total number of values will be the value of the mean. The following example will show how the mean
value for grouped data can be calculated.
Example 8
In Tim’s school, there are twenty-five teachers. Each teacher travels to school every morning in his or her
own car. The distribution of the driving times (in minutes) from home to school for the teachers is shown
in the table below:


Table 6.4:


Driving Times (minutes) Number of Teachers
0 to less than 10 3
10 to less than 20 10
20 to less than 30 6
30 to less than 40 4
40 to less than 50 2


The driving times are given for all twenty-five teachers, so the data is for a population. Calculate the mean
of the driving times.
Solution
Step One: Determine the midpoint for each interval.
For 0 to less than 10, the midpoint is 5.
For 10 to less than 20, the midpoint is 15.
For 20 to less than 30, the midpoint is 25.
For 30 to less than 40, the midpoint is 35.
For 40 to less than 50, the midpoint is 45.
Step Two: Multiply each midpoint by the frequency for the class.
For 0 to less than 10, (5)(3) = 15
For 10 to less than 20, (15)(10) = 150
For 20 to less than 30, (25)(6) = 150
For 30 to less than 40, (35)(4) = 140
For 40 to less than 50, (45)(2) = 90
Step Three: Add the results from step two and divide the sum by 25.


15 + 150 + 150 + 140 + 90 = 545


µ = 545
25


= 21.8


Each teacher spends a mean time of 21.8 minutes driving from home to school each morning.
To better represent the problem and its solution, a table can be drawn:


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Table 6.5:


Driving Times (min-
utes)


Number of Teachers
f


Midpoint Of Class m Product mf


0 to less than 10 3 5 15
10 to less than 20 10 15 150
20 to less than 30 6 25 150
30 to less than 40 4 35 140
40 to less than 50 2 45 90


For the population, N = 25 and ∑mf = 545, where m is the midpoint of the class and f is the frequency.
The mean for the population was found by dividing ∑mf by N. As a result, the formula µ =




mf
N can


be written to summarize the steps used to determine the value of the mean for a set of grouped data. If
the set of data represented a sample instead of a population, the process would remain the same, and the
formula would be written as x =




mf
n .


Example 9
The following table shows the frequency distribution of the number of hours spent per week texting messages
on a cell phone by 60 grade 10 students at a local high school.


Table 6.6:


Time Per Week (Hours) Number of Students
0 to less than 5 8
5 to less than 10 11
10 to less than 15 15
15 to less than 20 12
20 to less than 25 9
25 to less than 30 5


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Calculate the mean number of hours per week spent by each student texting messages on a cell phone.
Hint: A table may be useful.
Solution


Table 6.7:


Time Per Week
(Hours)


Number of Students
f


Midpoint of Class m Product mf


0 to less than 5 8 2.5 20.0
5 to less than 10 11 7.5 82.5
10 to less than 15 15 12.5 187.5
15 to less than 20 12 17.5 210.0
20 to less than 25 9 22.5 202.5
25 to less than 30 5 27.5 137.5


x =


mf
n


x = 20.0 + 82.5 + 187.5 + 210.0 + 202.5 + 137.5
60


x = 840
60


x = 14


The mean time spent per week by each student texting messages on a cell phone is 14 hours.
Now that you have created several distribution tables for grouped data, it’s time to point out that the
first column of the table can be represented in another way. As an alternative to writing the interval, or
class, in words, the words can be expressed as [# - #), where the front square bracket closes the class,
so the first number is included in the designated interval, but the open bracket at the end does not close
the class, so the last number is not included in the designated interval. Keeping this in mind, the table in
Example 9 can be presented as follows:


Table 6.8:


Time Per Week
(Hours)


Number of Students
f


Midpoint of Class m Product mf


[0-5) 8 2.5 20.0
[5-10) 11 7.5 82.5
[10-15) 15 12.5 187.5
[15-20) 12 17.5 210.0
[20-25) 9 22.5 202.5
[25-30) 5 27.5 137.5


Lesson Summary
You have learned the significance of the mean as it applies to a set of numerical data. You have also
learned how to calculate the mean using appropriate formulas for the given data for both a population and
a sample. When the data was presented as a list of numbers, you learned how to represent the values in a
frequency table, and when the data was grouped, you learned how to represent the data in a distribution


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table, as well as how to calculate the mean of this data. The use of technology in calculating the mean
was also demonstrated in this chapter.
Points to Consider


• Is the mean only used as a measure of central tendency, or is it applied to other representations of
data?


• If the mean is applied to other representations of data, can its value be calculated or estimated from
this representation?


• What other measures of central tendency can be used as a statistical summary when the mean is not
the best measure to use?


6.2 The Median
Learning Objectives


• Understand the median of a set of data as being an important measure of central tendency.
• Determine the median of a set of numerical data when there is an odd number of values and an even
number of values.


• Understand the application of the median to real-world problems.


Before class begins, bring out the blocks that you and your classmates chose from the pail for the lesson
on mean. In addition, have the grid paper on display where each student in your class posted his or her
number of blocks.
To begin the class, refer to the comments on the measures of central tendency that were recorded from the
previous lesson, when the brainstorming session occurred. Highlight the comments that were made with
regard to the median of a set of data and discuss this measure of central tendency with your classmates.
Once the discussion has been completed, choose a handful of blocks like you did before, with your classmates
doing the same.
Next, form a line with your classmates as a representation of the data from smallest to largest. In other
words, those with the largest number of blocks should be at one end of the line, and those with the fewest
number of blocks should be at the other end of the line. Have those at the ends of the line move out from
the line two at a time, with one from each end leaving the line at the same time. This movement from the
ends should enforce the concept of what is meant by “center”, and as you and your classmates move away,
you will see that the last one or two people remaining in the line actually represent the center of the data.
A similar activity can be done by using the grid paper chart. Instead of you and your classmates moving
from a line, you could simply remove your post-it notes the same way.


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Another simple activity to enforce the concept of center is to place five desks in a single row and have one
of your classmates sit in the middle desk.


Your classmate should be sitting in desk number 3. The other students in your class will quickly notice
that there are two desks in front of your classmate and two desks behind your classmate. Therefore, your
classmate is sitting in the desk in the middle position, which is the median of the desks.
From the discussion, the activity with the blocks, and the activity with the desks, you and your classmates
should have an understanding of the meaning of the median with respect to a set of data. Here is another
example. The test scores for seven students were 25, 55, 58, 64, 66, 68 and 70. The mean mark is 48.6,
which is lower than all but one of the student’s marks. The one very low mark of 25 has caused the mean to
be skewed. A better measure of the average performance of the seven students would be the middle mark
of 64. The median is the number in the middle position once the data has been organized. Organized
data is simply the numbers arranged from smallest to largest or from largest to smallest. 64 is the only
number for which there are as many values above it as below it in the set of organized data. The median
for an odd number of data values is the value that divides the data into two halves. If n represents the
number of data values and n is an odd number, then the median will be found in the n+12 position.
Example 1
Find the median of the following data:
a) 12, 2, 16, 8, 14, 10, 6
b) 7, 9, 3, 4, 11, 1, 8, 6, 1, 4
Solution
a) The first step is to organize the data, or arrange the numbers from smallest to largest.


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12, 2, 16, 8, 14, 10, 6 → 2, 6, 8, 10, 12, 14, 16


The number of data values is seven, which is an odd number. Therefore, the median will be found in the
n+1
2 position.


n + 1
2


= 7 + 1
2


= 8
2


= 4


In this case, the median is the value that is found in the 4th position of the organized data.


2, 6, 8, 10 , 12, 14, 16


This means that the median is 10.
b) The first step is to organize the data, or arrange the numbers from smallest to largest.
7, 9, 3, 4, 11, 1, 8, 6, 1, 4 → 1, 1, 3, 4, 4, 6, 7, 8, 9, 11
The number of data values is ten, which is an even number. Therefore, the median will be the mean of the
number found before the n+12 position and the number found after the n+12 position.


n + 1
2


= 10 + 1
2


= 11
2


= 5.5


The number found before the 5.5 position is 4, and the number found after the 5.5 position is 6.


1, 1, 3, 4, 4, 6 , 7, 8, 9, 11


This means that the median is 4+62 = 102 = 5.
Example 2
The amount of money spent by fifteen high school girls for a prom dress is shown below:


$250 $175 $325 $195 $450 $300 $275 $350 $425
$150 $375 $300 $400 $225 $360


What is the median price spent on a prom dress?
Solution


$150 $175 $195 $225 $250 $275 $300 $300 $325 $350 $360 $375
$400 $425 $450


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The prices have been organized from least to greatest, and the number of prices is an odd number. There-
fore, the median will be in the n+12 position: n+12 = 15+12 = 162 = 8.
The median price is $300, which is the 8th position.
Example 3
The students from a local high school volunteered to clean up the playground as an act of community
service. The number of pop cans collected by 20 of the students is shown in the following table:


16 22 10 8 14
12 36 18 12 10
34 26 44 6 20
31 25 15 9 13


What is the median number of pop cans collected by the students?
Solution


6 8 9 10 10
12 12 13 14 15
16 18 20 22 25
26 31 34 36 44


There is an even number of data values in the table, so the median will be the mean of the number before
the n+12 position and the number after the n+12 position: n+12 = 20+12 = 212 = 10.5.
The number before the 10.5 position is 15, and the number after the 10.5 position is 16. Therefore, the
median is 15+162 = 312 = 15.5.
The median number of pop cans collected by the students is 15.5.
Often, the number of data values is quite large, and the task of organizing the data can take a great deal
of time. To help organize data, the TI-83 calculator can be used. The following example will show you
how to use the calculator to organize data and find the median for the data values.
Example 4
The local police department spent the holiday weekend ticketing drivers who were speeding. Fifty locations
within the state were targeted as being ideal spots for drivers to exceed the posted speed limit. The number
of tickets issued during the weekend in each of the locations is shown in the following table. What is the
median number of speeding tickets issued?


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32 12 15 8 16 42 9 18 11 10
24 18 6 17 21 41 3 5 35 27
13 26 16 28 31 3 7 37 10 19
23 33 7 25 36 40 15 21 38 46
17 37 9 2 33 41 23 29 19 40


Solution
Using the TI-83 calculator


Step One:


Step Two:


The numbers that you entered into list one are now sorted from smallest to largest.
Step Three:


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You can now scroll down the list to reveal the ordered numbers.
There are fifty data values in the table. The median will be the mean of the number before the n+12 position
and the number after the n+12 position: n+12 = 50+12 = 512 = 25.5. The number before the 25.5 position is
19, and the number after the 25.5 position is 21. This means that the median is 19+212 = 402 = 20.


2 3 3 5 6 7 7 8 9 9
10 10 11 12 13 15 15 16 16 17
17 18 18 19 19 21 21 23 23 24
25 26 27 28 29 31 32 33 33 35
36 37 37 38 40 40 41 41 42 46


The median number of speeding tickets is 20.
The median of the data can also be determined by using technology. The following are two ways that you
can use technology to determine the median of the values by using the TI-83.
Method One:
All of the data values have been entered into L1. Using a feature of the STAT function will reveal the
median of the data.


Med = 20 indicates that the median of the data values in list one is 20.
Method Two:
Above the zero key is the word CATALOG, and this function acts like the yellow pages of a telephone
book. When you press 2ND 0 to access CATALOG, an alphabetical list of terms appears. You can
either scroll down to the word median (this will take a long time) or press the blue ÷ to access all terms
beginning with the letter ‘m’.


Again, the median of the values in L1 is 20.


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Whichever method you use, the result will be same. Using technology will save you time when you are
determining the median of a set of data values. When a set of data values is given in the form of a frequency
table, technology is often used to determine the median.
Example 5
The following numbers represent the number of chocolate bars sold by students during a recent fundraising
campaign. What was the median number of chocolate bars sold?


Table 6.9:


Number of Bars Number of Students
11 6
12 8
13 5
14 13
15 17
16 15


Solution
Using the TI-83 calculator:


When data is entered into a frequency table, a column that displays the cumulative frequency is often
included. This column is simply the sum of the frequencies up to and including that frequency. The median
can be determined by using the information that is presented in the cumulative frequency column.
Example 6
The following table shows the number of goals by Ashton during 26 hockey games.


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Solution
There are 26 scores, which is an even number. The middle position is n+12 = 26+12 = 272 = 13.5, and the
median is the sum of the numbers above and below position 13.5 divided by 2. According to the table, the
numbers in the 13th and 14th positions are 2’s. Therefore, the median is 2+22 = 42 = 2.
Lesson Summary
You have learned the significance of the median as it applies to a set of numerical data. You have also
learned how to calculate the median of a set of data values, whether the number of values is an odd number
or an even number. You also learned that if an outlier affects the mean of a set of data values, then the
median is the better measure of central tendency to use. The use of technology in calculating the median
was also demonstrated in this chapter.
Points to Consider


• The median of a set of data values cuts the data in half. Is only the median of an entire set of data
a useful value?


• Is the median of a set of data useful in any other aspect of statistics?
• Other than as a numerical value, can the median be used to represent data in any other way?


6.3 The Mode
Learning Objectives


• Understand the concept of mode.
• Identify the mode or modes of a data set for both quantitative and qualitative data.
• Describe the distribution of data as being modal, bimodal or multimodal.


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• Identify the mode of a set of data given in different representations.


Before class begins, bring out the blocks that you and your classmates chose from the pail for the lesson
on mean. In addition, have the grid paper on display where each student in your class posted his or her
number of blocks.
To begin the class, refer to the comments on the measures of central tendency that were recorded from the
lesson on mean, when the brainstorming session occurred. Highlight the comments that were made with
regard to the mode of a set of data and discuss this measure of central tendency with your classmates. Once
the discussion has been completed, choose a handful of blocks like you did before, with your classmates
doing the same.
The mode of the set of blocks can be given as a quantitative value or as a qualitative value. You and your
classmates can tell from the grid paper which number of blocks was picked most. The chart below shows
that 5 students each picked 5 blocks from the pail. This is a mode for quantitative data, since the answer
is in the form of a number.


To extend the mode to include qualitative data, you and your classmates should each now determine the
color or colors of block(s) that appear most often in each of your handfuls. To do this, group your blocks
according to color and count them, and have your classmates do the same. There may be more than one
color that occurs with the same highest frequency. The color(s) that appear most often for each handful
of blocks is the mode for that particular handful.
The mode of a set of data is simply the value that appears most frequently in the set. If two or more
values appear with the same frequency, each is a mode. The downside to using the mode as a measure
of central tendency is that a set of data may have no mode or may have more than one mode. However,
the same set of data will have only one mean and only one median. The word ‘modal’ is often used when
referring to the mode of a data set. If a data set has only one value that occurs the most often, the set
is called unimodal. Likewise, a data set that has two values that occur with the greatest frequency is
referred to as bimodal. Finally, when a set of data has more than two values that occur with the same
greatest frequency, the set is called multimodal. When determining the mode of a data set, calculations
are not required, but keen observation is a must. The mode is a measure of central tendency that is simple
to locate, but it is not used much in practical applications.
Example 1
The posted speed limit along a busy highway is 65 miles per hour. The following values represent the
speeds (in miles per hour) of ten cars that were stopped for violating the speed limit.


76 81 79 80 78 83 77 79 82 75


What is the mode?


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Solution
There is no need to organize the data, unless you think that it would be easier to locate the mode if the
numbers were arranged from least to greatest. In the above data set, the number 79 appears twice, but all
the other numbers appear only once. Since 79 appears with the greatest frequency, it is the mode of the
data values.
Mode = 79 miles per hour
Example 2
The weekly wages of seven randomly selected employees of Wendy’s were $98.00, $125.00, $75.00, $120.00,
$86.00, $92.00, and $110.00. What is the mode of these wages?
Solution
Each value in the above data set occurs only once. Therefore, this data has no mode.
Example 3
The ages of twelve randomly selected customers at a local coffee shop are listed below:


23, 21, 29, 24, 31, 21, 27, 23, 24, 32, 33, 19


What is the mode of the above ages?
Solution
The above data set has three values that each occur with a frequency of 2. These values are 21, 23, and
24. All other values occur only once. Therefore, this set of data has 3 modes.
Modes = 21, 23, and 24
Remember that the mode can be determined for qualitative data as well as quantitative data, but the
mean and the median can only be determined for quantitative data.
Example 4
Six students attending a local swimming competition were asked what color bathing suit they were wearing.
The responses were red, blue, black, pink, green, and blue.
What is the mode of these responses?
Solution
The color blue was the only response that occurred more than once and is, therefore, the mode of this data
set.
Mode = blue
When data is arranged in a frequency table, the mode is simply the value that has the highest frequency.
Example 5
The following table represents the number of times that one hundred randomly selected students ate at
the school cafeteria during the first month of school.


Number of Times Eating in the Cafeteria 2 3 4 5 6 7 8
Number of Students 3 8 22 29 20 8 10


What is the mode of the numbers of times that a student ate at the cafeteria?
Solution
The table shows that 29 students ate five times in the cafeteria. Therefore, 5 is the mode of the data set.


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Mode = 5 times
Lesson Summary
You have learned that the mode of a data set is simply the value that occurs with the highest frequency.
You have also learned that it is possible for a set of data to have no mode, one mode, two modes, or more
than two modes. Observation is required to determine the mode of a data set, and this mode can be for
either quantitative or qualitative data.
Points to Consider


• Is reference made to the mode in any other branch of statistics?
• Can the mode be useful when presenting graphical representations of data?


Vocabulary


Bimodal The term used to describe the distribution of a data set that has two modes.


Cumulative frequency The sum of the frequencies up to and including that frequency.


Measures of central tendency Values that describe the center of a distribution. The mean, median,
and mode are three measures of central tendency.


Mean A measure of central tendency that is determined by dividing the sum of all values in a data set
by the number of values.


Median The value of the middle term in a set of organized data. For a set of data with an odd number
of values, it is the value that has an equal number of data values before and after it, or the middle
value. For a set of data with an even number of values, the median is the average of the two values
in the middle positions.


Mode The value or values that occur with the greatest frequency in a data set.


Multimodal The term used to describe the distribution of a data set that has more than two modes.


Unimodal The term used to describe the distribution of a data set that has only one mode.


6.4 Review Questions
Show all work necessary to answer each question. Be sure to include any formulas that are needed.
The Mean
Section A – All the review questions in this section are selected response.


1. What is the mean of the following numbers?
10, 39, 71, 42, 39, 76, 38, 25


(a) 42
(b) 39


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(c) 42.5
(d) 35.5


2. What name is given to a value in a data set that is much lower or much higher than the other values?
(a) A sample
(b) An outlier
(c) A population
(d) A tally


3. What symbol is used to denote the mean of a population?
(a) ∑
(b) x
(c) xn
(d) µ


4. What measure of central tendency is calculated by adding all the values and dividing the sum by the
number of values?
(a) median
(b) mean
(c) mode
(d) typical value


5. The mean of four numbers is 71.5. If three of the numbers are 58, 76, and 88, what is the value of
the fourth number?
(a) 64
(b) 60
(c) 76
(d) 82


Section B – All questions in this section require you to show all the work necessary to arrive at a correct
solution.


1. Determine the means of the following sets of numbers:
(a) 20, 14, 54, 16, 38, 64
(b) 22, 51, 64, 76, 29, 22, 48
(c) 40, 61, 95, 79, 9, 50, 80, 63, 109, 42


2. The mean weight of five men is 167.2 pounds. The weights of four of the men are 158.4 pounds,
162.8 pounds, 165 pounds, and 178.2 pounds. What is the weight of the fifth man?


3. The mean height of 12 boys is 5.1 feet. The mean height of 8 girls is 4.8 feet.
(a) What is the total height of the boys?
(b) What is the total height of the girls?
(c) What is the mean height of the 20 boys and girls?


4. The following data represents the number of advertisements received by ten families during the past
month. Calculate the mean number of advertisements received by each family during the month.


43 37 35 30 41 23 33 31 16 21


5. The following table of grouped data represents the weight (in pounds) of all 100 babies born at a
local hospital last year. Calculate the mean weight for a baby.


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Table 6.10:


Weight (pounds) Number of Babies
[3 - 5) 8
[5 - 7) 25
[7 - 9) 45
[9 - 11) 18
[11 - 13) 4


6. A group of grade 6 students earned the following marks on an in-class assignment. The marks for
the boys were 90, 50, 70, 80, and 70. The marks for the girls were 60, 20, 30, 80, 90, and 20.
(a) Find the mean mark for the boys.
(b) Find the mean mark for the girls.
(c) Find the mean mark for all the students.


7. The mean of four numbers is 31.
(a) What is the sum of the four numbers?
The mean of six other numbers is 28.
(b) Calculate the mean of all ten numbers.


8. The following numbers represent the weights (in pounds) of nine dogs.
22 19 26 18 29 33 20 16 30


(a) What is the mean weight of the dogs?
(b) If the heaviest and the lightest dogs are removed from the group, find the mean weight of the


remaining dogs.
9. To demonstrate her understanding of the concept of mean, Melanie recorded the daily temperature
in degrees Celsius for her hometown at the same time each day for a period of one week. She then
calculated the mean daily temperature.


Table 6.11:


Day Sun Mon Tues Wed Thurs Fri Sat
Temperature
(◦C)


−7◦C 0◦C −1◦C 1◦C −4◦C −6◦C 3◦C


x = −7 + −1 + 1 + −4 + −6 + 3
6


x = −14
6


x = −2.3◦C


Melanie then reported the mean daily temperature to be −2.3◦C.
(a) Is Melanie correct? Justify your answer.
(b) If you do not agree with Melanie’s answer, can you tell Melanie what mistake she made?


10. Below are the points scored by two basketball teams during the regular season for their first twelve
games:


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Honest Hoopers 93 78 84 106 116 93 90 75 104 100 123 57
Bouncy Baskets 110 89 91 121 84 79 114 66 50 101 106 114


Which team had the higher mean score?
Section C – Match the words in the left column with the correct symbol from the right column.


The Median
Section A – All the review questions in this section are selected response.


1. What is the median of the following numbers?
10, 39, 71, 42, 39, 76, 38, 25


(a) 42.5
(b) 39
(c) 42
(d) 35.5


2. The front row in a movie theatre has 23 seats. If you were asked to sit in the seat that occupied the
median position, in what number seat would you have to sit?
(a) 1
(b) 11
(c) 23
(d) 12


3. What was the median mark achieved by a student who recorded the following marks on 10 math
quizzes?


68, 55, 70, 62, 71, 58, 81, 82, 63, 79


(a) 68
(b) 71
(c) 69
(d) 79


4. A set of four numbers that begins with the number 32 is arranged from smallest to largest. If the
median is 35, which of these could possibly be the set of numbers?


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(a) 32, 34, 36, 38
(b) 32, 35, 38, 41
(c) 32, 33, 34, 35
(d) 32, 36, 40, 44


5. The number of chocolate bars sold by each of 30 students is as follows:
32, 6, 21, 10, 8, 11, 12, 36, 17, 16, 15, 18, 40, 24, 21, 23, 24, 24, 29, 16, 32, 31, 10, 30, 35, 32, 18, 39, 12, 20


What is the median number of chocolate bars sold by the 30 students?
(a) 18
(b) 21
(c) 24
(d) 32


Section B – All questions in this section require you to show all the work necessary to arrive at a correct
solution.


1. The following table lists the retail price and the dealer’s costs for 10 cars at a local car lot this past
year:
Deals on Wheels


Table 6.12:


Car Model Retail Price Dealer’s Cost
Nissan Sentra $24,500 $18,750
Ford Fusion $26,450 $21,300
Hyundai Elantra $22,660 $19,900
Chevrolet Malibu $25,200 $22,100
Pontiac Sunfire $16,725 $14,225
Mazda 5 $27,600 $22,150
Toyota Corolla $14,280 $13,000
Honda Accord $28,500 $25,370
Volkswagen Jetta $29,700 $27,350
Subaru Outback $32,450 $28,775


(a) Calculate the median for the data on the retail prices for the above cars.
(b) Calculate the median for the data on the dealer’s costs for the above cars.


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2. Due to high winds, a small island in the Atlantic suffers frequent power outages. The following
numbers represent the number of outages each month during the past year.


4 5 3 4 2 1 0 3 2 7 2 3


What is the median for these power outages?
3. The Canadian Coast Guard has provided all of its auxiliary members with a list of 14 safety items
(flares, fire extinguishers, life jackets, fire buckets, etc.) that must be aboard each boat at all times.
During a recent check of fifteen boats, the number of safety items that were aboard each boat was
recorded as follows:


7 14 10 5 11 2 8 6 9 7 13 4 12 8 3


What is the median number of safety items aboard the boats that were checked?
4. A teacher’s assistant who has been substituting has been recording her biweekly wages for the past
thirteen pay periods. Her biweekly wages during this time were the following:


$700 $550 $760 $670 $500 $925 $600
$480 $390 $800 $850 $365 $525


What was her median biweekly wage?
5. A minor hockey league has fifty active players who range in age from 10 years to 15 years. The
following table shows the ages of the players:


Table 6.13:


Age of Players(yrs) Number of Players
10 6
11 8
12 7
13 11
14 10
15 8


Use the TI-83 calculator to determine the median age of the players.


6. A die was thrown fourteen times, and the results of each throw are shown:


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What is the median score for the fourteen rolls of the die?
7. At a local golf club, 100 players competed in a one-day tournament. The fifth hole of the course is a
par 6. The scores of each player on this hole were recorded, and the results are shown below:


Score 2 3 4 5 6 7 8
Number of Players 3 8 22 29 20 8 10


What is the median score for the players?
8. A math teacher at a local high school begins every class with a warm-up quiz based on the work
presented the previous day. Each test consists of six questions valued at one point for each correct
answer. The results of Monday’s quiz for the forty students in the class are shown below:


Mark 0 1 2 3 4 5 6
Number of Students 2 6 7 10 5 3 7


What is the median score for Monday’s quiz?
9. In Canada, with the loonie and the toonie, you could have a lot of coins in your pocket. A number of
high school students were asked how many coins they had in their pocket, and the results are shown
in the following table:


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Number of Coins 0 1 2 3 4 5 6 7
Number of Students 2 5 8 9 6 4 9 7


What is the median number of coins that a student had in his or her pocket?
10. A group of twelve students participated in a local dirt bike race that required them to cover a one-


mile course in the fastest time possible. The times, in minutes, of the twelve participants are shown
below:


3.5 min 4.2 min 3.1 min 5.3 min 6.2 min 4.6 min
5.1 min 6.7 min 5.4 min 4.4 min 3.9 min 5.0 min


What is the median time of the participants in the race?


The Mode
Section A – All the review questions in this section are selected response.


1. Which of the following measures can be determined for quantitative data only?
(a) mean
(b) median
(c) mode
(d) none of these


2. Which of the following measures can be calculated for qualitative data only?
(a) mean
(b) median
(c) mode
(d) all of these


3. What is the term used to describe the distribution of a data set that has one mode?
(a) multimodal
(b) unimodal
(c) nonmodal
(d) bimodal


4. What is the mode of the following numbers, which represent the ages of eight hockey players? 12,
11, 14, 10, 8, 13, 11, 9
(a) 11
(b) 10
(c) 14
(d) 8


5. Which of the following measures can have more than one value for a set of data?


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(a) median
(b) mode
(c) mean
(d) none of these


Section B – All questions in this section require your answer to be a complete sentence.


1. What are the modes of the following sets of numbers?
(a) 3, 13, 6, 8, 10, 5 6
(b) 12, 0, 15, 15, 13, 19, 16, 13, 16, 16


2. A student recorded her scores on weekly English quizzes that were marked out of a possible ten
points. Her scores were as follows:


8, 5, 8, 5, 7, 6, 7, 7, 5, 7, 5, 5, 6, 6, 9, 8, 9, 7, 9, 9, 6, 8, 6, 6, 7


What is the mode of her scores on the weekly English quizzes?
3. The following table represents the number of minutes that students spent studying for a math test.


Table 6.14:


Studying Time (minutes) Number of Students
[0 - 10) 2
[10 - 20) 10
[20 - 30) 6
[30 - 40) 4
[40 - 50) 3


What is the mode of the amounts of time spent studying for the math test?


4. A pre-test for students entering high school mathematics was given to 48 students. The following
table shows the number of questions attempted out of 50 by each of the students taking the test:43 39 40 40 41 44


42 41 41 41 42 44
43 41 40 41 42 41
42 41 42 42 41 44
41 42 43 40 42 39
42 40 39 42 43 42
42 39 41 41 42 40
43 44 40 42 44 39


What number of questions was attempted the most by the students?
5. A die was tossed fourteen times. What is the mode of the numbers that were rolled?


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6. The following table represents the number of times that twenty-four students attended school bas-
ketball games during the year:


Number of Games 5 6 7 8 9
Number of Students 1 5 3 9 6


What is the mode of the numbers of games that students attended?


7. The newly-formed high school soccer team is playing its first season. The following table shows the
number of goals it scored during each of its matches:


Number of Goals 1 2 3
Number of Matches 8 8 m


If the mean number of goals scored is 2.04, what is the smallest possible value of m if the mode of the
numbers of goals scored is 3?


8. What is the mode of the following numbers, and what word can be used to describe the distribution
of the data set?


5, 4, 10, 3, 3, 4, 7, 4, 6, 5, 11, 9, 5, 7


9. List three examples of how mode could be useful in everyday life?
10. The temperature in ◦F on 20 days during the month of June was:


70◦F, 76◦F, 76◦F, 74◦F, 70◦F, 70◦F, 72◦F, 74◦F, 78◦F, 80◦F
74◦F, 74◦F, 78◦F, 76◦F, 78◦F, 76◦F, 74◦F, 78◦F, 80◦F, 76◦F


What is the mode of the temperatures for the month of June?


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Chapter 7


The Shape, Center and Spread
of a Normal Distribution


Introduction


What a way to start math class! Imagine your math teacher allowing you to bounce a basketball around
the classroom.


Prior to the beginning of class, go to the physical education department of your school and borrow some
basketballs. These will be used in this chapter to introduce the concept of a normal distribution. You
may find your class to be somewhat noisy, but you’re sure to enjoy the activity. Begin the class by
reviewing the concept of a circle and the terms associated with the measurements of a circle. You and your
classmates should be able to provide these facts based on your previous learning. To ensure that everyone
understands the concepts of center, radius, diameter, and circumference, work in small groups to create
posters to demonstrate your understanding. These posters can then be displayed around the classroom.
The following is a sample of the type of poster you may create:


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Once you and your classmates have completed this and you are confident that you all understand the
measurements associated with a circle, distribute the basketballs to everyone in the class. While your
classmates are playing with the balls, set up a table with tape, rulers, string, scissors, markers, and any
other materials that you think you may find useful to answer the following questions:


• What is the circumference of the basketball?
• What is the diameter of the basketball?
• What is the radius of the basketball?
• How did you determine these measurements?
• What tools did you use to find your answers?


When playtime is over, use the tools provided to answer all of the above questions. It is the job of you and
your classmates to determine a method of calculating these measurements. For the questions that require
numerical measurements as answers, take two measurements for each. You must plot your results for the
diameter of the basketball, so remember to record your data.


Oops! Don’t forget that the ruler cannot go through the basketball!
When you have answered the above questions, plot your two results for the diameter of the basketball on
a large sheet of grid paper, and have your classmates do the same.


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7.1 Estimating the Mean and Standard Devia-
tion of a Normal Distribution


Learning Objectives


• Understand the meaning of normal distribution and bell-shape.
• Estimate the mean and the standard deviation of a normal distribution.


Now that you have created your plot on a large sheet of grid paper, can you describe the shape of the plot?
Do the dots seem to be clustered around one spot (value) on the chart? Do some dots seem to be far away
from the clustered dots? After you have made all the necessary observations to answer these questions,
pick two numbers from the chart to complete this statement:
“The typical measurement of the diameter is approximately______inches, give or take______inches.”
We will complete this statement later in the lesson.
Normal Distribution
The shape below should be similar to the shape that has been created with the dot plot.


When you made the observations regarding the measurements of the diameter of the basketball, you must
have noticed that they were not all the same. In spite of the different measurements, you should have seen
that the majority of the measurements clustered around the value of 9.4 inches. This value represents the
approximate diameter of a basketball. Also, you should have noticed that a few measurements were to the
right of this value, and a few measurements were to the left of this value. The resulting shape looks like a
bell, and this is the shape that represents a normal distribution of data.
In the real world, no examples match this smooth curve perfectly. However, many data plots, like the one
you made, will approximate this smooth curve. For this reason, you will notice that the term assume is
often used when referring to data that deals with normal distributions. When a normal distribution is
assumed, the resulting bell-shaped curve is symmetric. That is, the right side is a mirror image of the left
side. In the figure below, if the blue line is the mirror (the line of symmetry), you can see that the pink
section to the left of the line of symmetry is the mirror image of the yellow section to the right of the line
of symmetry. The line of symmetry also goes through the x-axis.


If you knew all of the measurements that were plotted for the diameter of the basketball, you could calculate
the mean (average) diameter by adding the measurements and dividing the sum by the total number of
values. It is at this value that the line of symmetry intersects the x-axis. In other words, the mean of a
normal distribution is the center, or balance point, of the distribution.


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You can see that the two colors form a peak at the top of the line of symmetry and then spread out to
the left and to the right from the line of symmetry. The shape of the bell flattens out the further it moves
away from the line of symmetry. In other words, the data spreads out in both directions away from the
mean. This spread of the data is measured by the standard deviation, and it describes exactly how the
data moves away from the mean. You will learn more about standard deviation in the next lesson. For
now, that is all you have to know about standard deviation−it is a measure of the spread of the data away
from the mean.
Now you should be able to complete the statement that was presented earlier in this lesson.
“The typical measurement of the diameter is approximately 9.4 inches, give or take 0.2 inches.”
This statement assumes that the mean of the measurements was 9.4 inches and the standard deviation of
the measurements was 0.2 inches.
Example 1
For each of the following graphs, complete the statement. Fill in the first blank in each statement with
the mean and the second blank in each statement with the standard deviation. Assume that the standard
deviation is the difference between the mean and the first tick mark to the left of the mean.
a) “The typical measurement is approximately ______ in the bank, give or take ______.”


Solution
“The typical measurement is approximately $500 in the bank, give or take $50.”
b) “The typical measurement is approximately ______ goals scored, give or take ______ goals.”


Solution
“The typical measurement is approximately 64 goals scored, give or take 6 goals.”
Lesson Summary
In this lesson, you learned what was meant by normal distribution. You also learned about the smooth bell
curve that is used to represent a data set that is normally distributed. In addition, you learned that when
data is plotted on a bell curve, you can estimate the mean by using the value where the line of symmetry
crosses the x-axis. Finally, the spread of data in a normal distribution was represented by using a give or
take statement.
Points to Consider


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• Is there a way to determine the actual values for a give or take statement?
• Can a give or take statement go beyond a single give or take?
• Can all the actual values be represented on a bell curve?


7.2 Calculating the Standard Deviation
Learning Objectives


• Understand the meaning of standard deviation.
• Understanding the percents associated with standard deviation.
• Calculate the standard deviation for a normally distributed random variable.


This semester you decided to join the school’s bowling league for the first time. Having never bowled
previously, you are very anxious to find out what your bowling average is for the semester. If your average
is comparable to that of the other students, you will join the league again next semester. Your coach has
told you that your mean score is 70, and now you want to find out how your results compare to that of
the other members of the league.
Your coach has decided to let you figure this out for yourself. He tells you that the scores were normally
distributed and provides you with a list of the other mean scores. These average scores are in no particular
order. In other words, they are random.


54 88 49 44 96 72 46 58 79
92 44 50 102 80 72 66 64 61
60 56 48 52 54 60 64 72 68
64 60 56 52 55 60 62 64 68


We will discover how your mean bowling score compares to that of the other bowlers later in the lesson.
Standard Deviation
In the previous lesson, you learned that standard deviation is a measure of the spread of a set of data away
from the mean of the data.


In a normal distribution, on either side of the line of symmetry, the curve appears to change its shape
from being concave down (looking like an upside-down bowl) to being concave up (looking like a right-
side-up bowl). Where this happens is called an inflection point of the curve. If a vertical line is drawn
from an inflection point to the x-axis, the difference between where the line of symmetry goes through the
x-axis and where this line goes through the x-axis represents one standard deviation away from the mean.
Approximately 68% of all the data is located within one standard deviation of the mean.
To emphasize this fact and the fact that the mean is the middle of the distribution, let’s play a game of
Simon Says. Using color paper and two types of shapes, arrange the pattern of the shapes on the floor as
shown below. Randomly select seven students from your class to play the game. You will be Simon, and
you are to give orders to the selected students. Only when Simon Says are the students to obey the given
order. The orders can be given in many ways, but one suggestion is to deliver the following orders:


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• “Simon Says for Frank to stand on the rectangle.”
• “Simon Says for Joey to stand on the closest oval to the right of Frank.”
• “Simon Says for Liam to stand on the closest oval to the left of Frank.”
• “Simon Says for Mark to stand on the farthest oval to the right of Frank.”
• “Simon Says for Juan to stand on the farthest oval to the left of Frank.”
• “Simon Says for Jacob to stand on the middle oval to the right of Frank.”
• “Simon Says for Sean to stand on the middle oval to the left of Frank.”


Once the students are standing in the correct places, pose questions about their positions with respect
to Frank. The members of the class who are not playing the game should be asked to respond to these
questions about the position of their classmates. Some questions that should be asked are:


• “What two students are standing closest to Frank?”
• “Are Joey and Liam both the same distance away from Frank?”
• “What two students are furthest away from Frank?”
• “Are Mark and Juan both the same distance away from Frank?”


When the students have completed playing Simon Says, they should have an understanding of the concept
that the mean is the middle of the distribution and the remainder of the distribution is evenly spread out
on either side of the mean.
The picture below is a simplified form of the game you have just played. The yellow rectangle is the mean,
and the remaining rectangles represent three steps to the right of the mean and three steps to the left of
the mean.


If we consider the spread of the data away from the mean, which is measured using standard deviation,
as being a stepping process, then one step to the right or one step to the left is considered one standard
deviation away from the mean. Two steps to the left or two steps to the right are considered two standard
deviations away from the mean. Likewise, three steps to the left or three steps to the right are considered
three standard deviations away from the mean. The standard deviation of a data set is simply a value, and
in relation to the stepping process, this value would represent the size of your footstep as you move away
from the mean. Once the value of the standard deviation has been calculated, it is added to the mean
for moving to the right and subtracted from the mean for moving to the left. If the value of the yellow


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mean tile was 58, and the value of the standard deviation was 5, then you could put the resulting sums
and differences on the appropriate tiles.


For a normal distribution, 68% of the data values would be located within one standard deviation of the
mean, which is between 53 and 63. Also, 95% of the data values would be located within two standard
deviations of the mean, which is between 48 and 68. Finally, 99.7% of the data values would be located
within three standard deviations of the mean, which is between 43 and 73. The percentages mentioned
here make up what statisticians refer to as the 68-95-99.7 Rule. These percentages remain the same for
all data that can be assumed to be normally distributed. The following diagram represents the location of
these values on a normal distribution curve.


Now that you understand the distribution of the data and exactly how it moves away from the mean, you
are ready to calculate the standard deviation of a data set. For the calculation steps to be organized, a
table is used to record the results for each step. The table will consist of three columns. The first column
will contain the data and will be labeled x. The second column will contain the differences between the
data values and the mean of the data set. This column will be labeled (x − x). The final column will be
labeled (x − x)2, and it will contain the square of each of the values recorded in the second column.
Example 2
Calculate the standard deviation of the following numbers:


2, 7, 5, 6, 4, 2, 6, 3, 6, 9


Solution
Step 1: It is not necessary to organize the data. Create a table and label each of the columns appropriately.
Write the data values in column x.
Step 2: Calculate the mean of the data values.


µ = 2 + 7 + 5 + 6 + 4 + 2 + 6 + 3 + 6 + 9
10


= 50
10


= 5.0


Step 3: Calculate the differences between the data values and the mean. Enter the results in the second
column.


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Table 7.1:


x (x − µ)
2 −3
7 2
5 0
6 1
4 −1
2 −3
6 1
3 −2
6 1
9 4


Step 4: Calculate the values for column 3 by squaring each result in the second column.
Step 5: Calculate the mean of the third column and then take the square root of the answer. This value
is the standard deviation (σ) of the data set.


Table 7.2:


(x − µ)2


9
4
0
1
1
9
1
4
1
16


σ


2 = 9 + 4 + 0 + 1 + 1 + 9 + 1 + 4 + 1 + 16
10


= 46
10


= 4.6


σ =

4.6 ≈ 2.1


Step 5 can be written using the formula σ =




(x−µ)2
n .


The standard deviation of the data set is approximately 2.1.
Now that you have completed all the steps, here is the table that was used to record the results. The table
was separated as the steps were completed. Now that you know the process involved in calculating the
standard deviation, there is no need to work with individual columns−work with an entire table.


Table 7.3:


x (x − µ) (x − µ)2


2 −3 9


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Table 7.3: (continued)


x (x − µ) (x − µ)2


7 2 4
5 0 0
6 1 1
4 −1 1
2 −3 9
6 1 1
3 −2 4
6 1 1
9 4 16


Example 3
A company wants to test its exterior house paint to determine how long it will retain its original color
before fading. The company mixes two brands of paint by adding different chemicals to each brand. Six
one-gallon cans are made for each paint brand, and the results are recorded for every gallon of each brand
of paint. The following are the results obtained in the laboratory:


Table 7.4:


Brand A (Time in months) Brand B (Time in months)
15 40
65 50
55 35
35 40
45 45
25 30


Calculate the standard deviation for each brand of paint.
Solution
Brand A:


Table 7.5:


x (x − µ) (x − µ)2


15 −25 625
65 25 625
55 15 225
35 −5 25
45 5 25
25 −15 225


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µ = 15 + 65 + 55 + 35 + 45 + 25
6


= 240
6


= 40


σ =




(x − µ)2
n


σ =


625 + 625 + 225 + 25 + 25 + 225
6


σ =


1750
6




291.66 ≈ 17.1


The standard deviation for Brand A is approximately 17.1.
Brand B:


Table 7.6:


x (x − µ) (x − µ)2


40 0 0
50 10 100
35 −5 25
40 0 0
45 5 25
30 −10 100


µ = 40 + 50 + 35 + 40 + 45 + 30
6


= 240
6


= 40


σ =




(x − µ)2
n


σ =


0 + 100 + 25 + 0 + 25 + 100
6


σ =


250
6




41.66 ≈ 6.5


The standard deviation for Brand B is approximately 6.5.
Note: The standard deviation for Brand A (17.1) was much larger than that for Brand B (6.5). However,
the means of both brands were the same. When the means are equal, the larger the standard deviation is,
the more variable are the data.
To find the standard deviation, you subtract the mean from each data value to determine how much each
data value varies from the mean. The result is a positive value when the data value is greater than the
mean, a negative value when the data value is less than the mean, and zero when the data value is equal
to the mean.
If we were to add the variations found in the second column of the table, the total would be zero. This
result of zero implies that there is no variation between the data value and the mean. In other words, if
we were conducting a survey of the number of hours that students use a cell phone in one day, and we
relied upon the sum of the variations to give us some pertinent information, the only thing that we would
learn is that all the students who participated in the survey use a cell phone for the exact same number of
hours each day. We know that this is not true, because the survey does not show all the responses as being


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the same. In order to ensure that these variations do not lose their significance when added, the variation
values are squared prior to calculating their sum.
What we need for a normal distribution is a measure of spread that is proportional to the scatter of the
data, independent of the number of values in the data set and independent of the mean. The spread will
be small when the data values are consistent, but large when the data values are inconsistent. The reason
that the measure of spread should be independent of the mean is because we are not interested in this
measure of central tendency, but rather, only in the spread of the data. For a normal distribution, both
the variance and the standard deviation fit the above profile for an appropriate measure of spread, and
both values can be calculated for the set of data.
To calculate the variance (σ2) for a population of normally distributed data:
Step 1: Determine the mean of the data values.
Step 2: Subtract the mean of the data from each value in the data set to determine the difference between
the data value and the mean: (x − µ).
Step 3: Square each of these differences and determine the total of these positive, squared results.
Step 4: Divide this sum by the number of values in the data set.
These steps for calculating the variance of a data set for a population can be summarized in the following
formula:


σ


2 =


(x − µ)2
n


where:
x is a data value.
µ is the population mean.
n is number of data values (population size).
These steps for calculating the variance of a data set for a sample can be summarized in the following
formula:


s2 =


(x − x)2
n − 1


where:
x is a data value.
x is the sample mean.
n is number of data values (sample size).
The only difference in the formulas is the number by which the sum is divided. For a population, it is
divided by n, and for a sample, it is divided by n − 1.
Example 4
Calculate the variance of the two brands of paint in Example 3. These are both small populations.


Table 7.7:


Brand A (Time in months) Brand B (Time in months)
15 40
65 50


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Table 7.7: (continued)


Brand A (Time in months) Brand B (Time in months)
55 35
35 40
45 45
25 30


Solution
Brand A


Table 7.8:


x (x − µ) (x − µ)2


15 −25 625
65 25 625
55 15 225
35 −5 25
45 5 25
25 −15 225


µ = 15 + 65 + 55 + 35 + 45 + 25
6


= 240
6


= 40


σ


2 =


(x − µ)2
n


σ


2 = 625 + 625 + 225 + 25 + 25 + 225
6


= 1750
6


≈ 291.66


Brand B
Table 7.9:


x (x − µ) (x − µ)2


40 0 0
50 10 100
35 −5 25
40 0 0
45 5 25
30 −10 100


µ = 40 + 50 + 35 + 40 + 45 + 30
6


= 240
6


= 40


σ


2 =


(x − µ)2
n


σ


2 = 0 + 100 + 25 + 0 + 25 + 100
6


= 250
6


≈ 41.66


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From the calculations done in Example 3 and in Example 4, you should have noticed that the square root
of the variance is the standard deviation, and the square of the standard deviation is the variance. Taking
the square root of the variance will put the standard deviation in the same units as the given data. The
variance is simply the average of the squares of the distance of each data value from the mean. If these
data values are close to the value of the mean, the variance will be small. This was the case for Brand B.
If these data values are far from the mean, the variance will be large, as was the case for Brand A.
The variance and the standard deviation of a data set are always positive values.
Example 5
The following data represents the morning temperatures (◦C) and the monthly rainfall (mm) in July for
all the Canadian cities east of Toronto:
Temperature (◦C)


11.7 13.7 10.5 14.2 13.9 14.2 10.4 16.1 16.4
4.8 15.2 13.0 14.4 12.7 8.6 12.9 11.5 14.6


Precipitation (mm)


18.6 37.1 70.9 102 59.9 58.0 73.0 77.6 89.1
86.6 40.3 119.5 36.2 85.5 59.2 97.8 122.2 82.6


Which data set is more variable? Calculate the standard deviation for each data set.
Solution
Temperature (◦C)


Table 7.10:


x (x − µ) (x − µ)2


11.7 −1 1
13.7 1 1
10.5 −2.2 4.84
14.2 1.5 2.25
13.9 1.2 1.44
14.2 1.5 2.25
10.4 −2.3 5.29
16.1 3.4 11.56
16.4 3.7 13.69
4.8 −7.9 62.41
15.2 2.5 6.25
13.0 0.3 0.09
14.4 1.7 2.89
12.7 0 0
8.6 −4.1 16.81
12.9 0.2 0.04
11.5 −1.2 1.44
14.6 1.9 3.61


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µ =


x
n


= 228.6
18


≈ 12.7


σ


2 =


(x − µ)2
n


σ =




(x − µ)2
n


σ


2 = 136.86
18


≈ 7.6 σ =


136.86
18


≈ 2.8


The variance of the data set is approximately 7.6 ◦C, and the standard deviation of the data set is
approximately 2.8 ◦C.
Precipitation (mm)


Table 7.11:


x (x − µ) (x − µ)2


18.6 −54.5 2970.3
37.1 −36.0 1296
70.9 −2.2 4.84
102.0 28.9 835.21
59.9 −13.2 174.24
58.0 −15.1 228.01
73.0 −0.1 0.01
77.6 4.5 20.25
89.1 16.0 256
86.6 13.5 182.25
40.3 −32.8 1075.8
119.5 46.4 2153
36.2 −36.9 1361.6
85.5 12.4 153.76
59.2 −13.9 193.21
97.8 24.7 610.09
122.2 49.1 2410.8
82.6 9.5 90.25


µ =


x
n


= 1316.1
18


≈ 73.1


σ


2 =


(x − µ)2
n


σ =




(x − x)2
n


σ


2 = 14016
18


≈ 778.66 σ =


14016
18


≈ 27.9


The variance of the data set is approximately 778.66 mm, and the standard deviation of the data set is
approximately 27.9 mm.
Therefore, the data values for the precipitation are more variable. This is indicated by the large variance
of the data set.
Example 6


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Now that you know how to calculate the variance and the standard deviation of a set of data, let’s apply
this to a normal distribution by determining how your bowling average compared to those of the other
bowlers in your league. This time technology will be used to determine both the variance and the standard
deviation of the data.


54 88 49 44 96 72 46 58 79
92 44 50 102 80 72 66 64 61
60 56 48 52 54 60 64 72 68
64 60 56 52 55 60 62 64 68


Solution


From the list, you can see that the mean of the bowling averages is approximately 63.7 and that the
standard deviation is approximately 14.1.
To use technology to calculate the variance involves naming the lists according to the operations that you
need to do in order to determine the correct values. In addition, you can use the Catalog function of the
calculator to determine the sum of the squared variations. All of the same steps used to calculate the
standard deviation of the data are applied to give the mean of the data set. You could also use the Catalog
function to find the mean of the data, but since you are now familiar with 1-Var Stats, you can use this
method.


The mean of the data is approximately 63.7. L2 will now be renamed L1 − 63.7 to compute the values for
(x − x).
Likewise, L3 will be renamed L22.


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The sum of the values in L3 divided by the number of data values (36) is the variance of the bowling
averages.
Lesson Summary
In this lesson, you learned that the standard deviation of a set of data is a value that represents a measure
of the spread of the data from the mean. You also learned that the variance of the data from the mean is
the square of the standard deviation. Calculating the standard deviation manually and calculating it by
using technology were additional topics you learned in this lesson.
Points to Consider


• Does the value of standard deviation stand alone, or can it be displayed with a normal distribution?
• Are there defined increments for how data spreads away from the mean?
• Can the standard deviation of a set of data be applied to real-world problems?


7.3 Connecting the Standard Deviation and Nor-
mal Distribution


Learning Objectives


• Represent the standard deviation of a normal distribution on a bell curve.
• Use the percentages associated with normal distributions to solve problems.


Introduction
In the problem presented in the first lesson regarding your bowling average, your teacher told you that the
bowling averages were normally distributed. In the previous lesson, you calculated the standard deviation


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of the averages by using the TI-83 calculator. Later in this lesson, you will be able to represent the value
of the standard deviation in a normal distribution.
You have already learned that 68% of the data lies within one standard deviation of the mean, 95% of the
data lies within two standard deviations of the mean, and 99.7% of the data lies within three standard
deviations of the mean. To accommodate these percentages, there are defined values in each of the regions
to the left and to the right of the mean.


These percentages are used to answer real-world problems when both the mean and the standard deviation
of a data set are known.
Example 7
The lifetimes of a certain type of light bulb are normally distributed. The mean life is 400 hours, and the
standard deviation is 75 hours. For a group of 5000 light bulbs, how many are expected to last each of the
following times?
a) between 325 hours and 475 hours
b) more than 250 hours
c) less than 250 hours
Solution


a) 68% of the light bulbs are expected to last between 325 hours and 475 hours. This means that 5000 ×
0.68 = 3400 light bulbs are expected to last between 325 and 475 hours.
b) 95% + 2.35% = 97.35% of the light bulbs are expected to last more than 250 hours. This means that
5000 × 0.9735 = 4867.5 ≈ 4868 of the light bulbs are expected to last more than 250 hours.
c) Only 2.35% of the light bulbs are expected to last less than 250 hours. This means that 5000× 0.0235 =
117.5 ≈ 118 of the light bulbs are expected to last less than 250 hours.
Example 8
A bag of chips has a mean mass of 70 g, with a standard deviation of 3 g. Assuming a normal distribution,
create a normal curve, including all necessary values.
a) If 1250 bags of chips are processed each day, how many bags will have a mass between 67 g and 73 g?
b) What percentage of the bags of chips will have a mass greater than 64 g?
Solution


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a) Between 67 g and 73 g lies 68% of the data. If 1250 bags of chips are processed, 850 bags will have a
mass between 67 g and 73 g.
b) 97.35% of the bags of chips will have a mass greater than 64 grams.
Now you can represent the data that your teacher gave to you for the bowling averages of the players in
your league on a normal distribution curve. The mean bowling score was 63.7, and the standard deviation
was 14.1.


From the normal distribution curve, you can see that your average bowling score of 70 is within one
standard deviation of the mean. You can also see that 68% of all the data is within one standard deviation
of the mean, so you did very well bowling this semester. You should definitely return to the league next
semester.
Lesson Summary
In this chapter, you have learned what is meant by a set of data being normally distributed and the
significance of standard deviation. You are now able to represent data on a bell-curve and to interpret
a given normal distribution curve. In addition, you can calculate the standard deviation of a given data
set both manually and by using technology. All of this knowledge can be applied to real-world problems,
which you are now able to answer.
Points to Consider


• Is the normal distribution curve the only way to represent data?
• The normal distribution curve shows the spread of the data, but it does not show the actual data
values. Do other representations of data show the actual data values?


Vocabulary


Normal distribution A symmetric bell-shaped curve with tails that extend infinitely in both directions
from the mean of a data set.


Standard deviation A measure of spread of a data set equal to the square root of the sum of the squared
variances divided by the number of data values.


Variance A measure of spread of a data set equal to the mean of the squared variations of each data
value from the mean of the data set.


68-95-99.7 Rule The rule that includes the percentages of data that are within one, two, and three
standard deviations of the mean of a set of data.


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7.4 Review Questions


Part A. For each question, circle the most appropriate answer.


1. If the standard deviation of a population is 6, the population variance is:
(a) 2.44
(b) 3
(c) 6
(d) 36


2. What is the sample standard deviation of the following data values?
3.3 2.9 8.5 11.5


(a) 3.61
(b) 4.17
(c) 6.55
(d) 13.52


3. Suppose data are normally distributed, with a mean of 100 and a standard deviation of 20. Between
what two values will approximately 68% of the data fall?
(a) 60 and 140
(b) 80 and 120
(c) 20 and 100
(d) 100 and 125


4. The sum of all of the deviations about the mean of a set of data is always going to be equal to:
(a) positive
(b) the mode
(c) the standard deviation total
(d) zero


5. What is the population variance of the following data values?
40 38 42 47 35


(a) 4.03
(b) 4.51
(c) 15.24
(d) 20.34


6. Suppose data are normally distributed, with a mean of 50 and a standard deviation of 10. Between
what two values will approximately 95% of the data fall?
(a) 40 and 60
(b) 30 and 70


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(c) 20 and 80
(d) 10 and 95


7. Suppose data are normally distributed, with a mean of 50 and a standard deviation of 10. What
would be the variance?
(a) 10
(b) 40
(c) 50
(d) 100


8. If data are normally distributed, what percentage of the data should lie within the range of µ ± 3σ?
(a) 34%
(b) 68%
(c) 95%
(d) 99.7%


9. If a normally distributed population has a mean of 75 and a standard deviation of 15, what proportion
of the values would be expected to lie between 45 and 105?
(a) 34%
(b) 68%
(c) 95%
(d) 99.7%


10. If a normally distributed population has a mean of 25 and a standard deviation of 5.5, what proportion
of the values would be expected to lie between 19.5 and 30.5?
(a) 34%
(b) 68%
(c) 95%
(d) 99.7%


Part B. Answer the following questions and show all work (including diagrams) to create a complete
answer.


11. In the United States, cola can normally be bought in 8 oz cans. A survey was conducted where 250
cans of cola were taken from a manufacturing warehouse and the volumes were measured. It was
found that the mean volume was 7.5 oz, and the standard deviation was 0.1 oz. Draw a normal
distribution curve to represent this data and then answer the following questions.
(b) 68% of the volumes can be found between __________ and _________.
(c) 95% of the volumes can be found between __________ and _________.
(d) 99.7% of the volumes can be found between __________ and _________.


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12. The mean height of the fourth graders in a local elementary school was found to be 4’8”, or 56”. The
standard deviation was found to be 5”. Draw a normal distribution curve to represent this data and
then answer the following questions.
(b) 68% of the volumes can be found between __________ and _________.
(c) 95% of the volumes can be found between __________ and _________.
(d) 99.7% of the volumes can be found between __________ and _________.


13. The following data was collected.
5 8 9 10 4 3 7 5
Fill in the chart below and calculate the standard deviation and the variance.


Table 7.12:


Data (x) Mean (µ) Mean − Data
(µ − x)


Square of Mean
− Data (µ − x)2




14. The following data was collected.
11 15 16 12 19 17 14 18 15 10
Fill in the chart below and calculate the standard deviation and the variance.


Table 7.13:


Data (x) Mean (µ) Mean − Data
(µ − x)


Square of Mean
− Data (µ − x)2




15. Mrs. Meery has recorded her exam results for the current mathematics exam. The results are shown
below.


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64 98 78 76 56 48 89 78 69 90 89
97 67 58 59 50 78 89 68 83 72 91


(b) Determine the mean for this data.
(c) Determine the standard deviation for this data.
(d) Determine the variance for this data.
(e) Draw a normal distribution curve to represent the data Mrs. Meery found in her class.


16. Mrs. Landry decided to do the same analysis as Mrs. Meery for her math class. She has recorded
her exam results for the current mathematics exam. The results are shown below.89 87 81 84 76 72 67 49 55 38 67 90 59


87 89 69 92 90 79 84 69 93 85 70 87 80


(b) Determine the mean for this data.
(c) Determine the standard deviation for this data.
(d) Determine the variance for this data.
(e) Draw a normal distribution curve to represent the data Mrs. Landry found in her class.


17. 200 senior high students were asked how long they had to wait in the cafeteria line for lunch. Their
responses were found to be normally distributed, with a mean of 15 minutes and a standard deviation
of 3.5 minutes. Copy the following bell curve onto your paper and answer the questions below.


(b) How many students would you expect to wait more than 11.5 minutes?
(c) How many students would you expect to wait more than 18.5 minutes?
(d) How many students would you expect to wait between 11.5 and 18.5 minutes?


18. 350 babies were born at Neo Hospital in the past six months. The average weight for the babies was
found to be 6.8 lbs, with a standard deviation of 0.5 lbs. Copy the following bell curve onto your
paper and answer the questions below.


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(b) How many babies would you expect to weigh more than 7.3 lbs?
(c) How many babies would you expect to weigh more than 7.8 lbs?
(d) How many babies would you expect to weigh between 6.3 and 7.8 lbs?


19. Sudoku is a very popular logic game of number combinations. It originated in the late 1800s by
the French press, Le Siècle. The average times (in minutes) it takes those in a senior math class to
complete a Sudoku puzzle are found below. Draw a normal distribution curve to represent this data.
Determine what time a student must complete a Sudoku puzzle in to be in the top0.13%.


20 15 21 24 7 19 10 17 15 22 31 19 20 21
21 9 12 26 24 28 19 16 24 11 17 31 25 13
16 18 22 32 9 15 19 27 14 25 32 29


20. Sheldon has planted seedlings in his garden in the back yard. After 30 days, he measures the heights
of the seedlings to determine how much they have grown. The differences in heights can be seen in
the table below. The heights are measured in inches. Draw a normal distribution curve to represent
the data. Determine what the differences in heights of the seedlings are for 68% of the data.


10 3 8 4 7 12 8 5 4 9 3 8
6 10 7 10 11 8 12 9 10 7 8 11


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Chapter 8


Organizing and Displaying
Distributions of Data


Introduction
The local arena is trying to attract as many participants as possible to attend the community’s “Skate for
Scoliosis” event. Participants pay a fee of $10.00 for registering, and, in addition, the arena will donate
$3.00 for each hour a participant skates, up to a maximum of six hours. Create a table of values and draw
a graph to represent a participant who skates for the entire six hours. How much money can a participant
raise for the community if he/she skates for the maximum length of time?
This problem will be revisited later in the chapter.
When data is collected from surveys or experiments, it is often displayed in charts, tables, or graphs in
order to produce a visual image that is helpful in interpreting the results. From a graph or table, an
observer is able to detect any patterns or trends that may exist. The most common graphs that are used
in statistics are line graphs, scatter plots, bar graphs, histograms, frequency polygons, circle graphs, and
box-and-whisker plots.


8.1 Line Graphs and Scatter Plots
Learning Objectives


• Represent data that has a linear pattern on a graph.
• Represent data using a broken-line graph.
• Understand the difference between continuous data and discrete data as it applies to a line graph.
• Represent data that has no definite pattern as a scatter plot.
• Draw a line of best fit on a scatter plot.
• Use technology to create both line graphs and scatter plots.


Before you continue to explore the concept of representing data graphically, it is very important to under-
stand the meaning of some basic terms that will often be used in this lesson. The first such definition is that
of a variable. In statistics, a variable is simply a characteristic that is being studied. This characteristic
assumes different values for different elements, or members, of the population, whether it is the entire
population or a sample. The value of the variable is referred to as an observation, or a measurement. A
collection of these observations of the variable is the data set.


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Variables can be quantitative or qualitative. A quantitative variable is one that can be measured
numerically. Some examples of a quantitative variable are wages, prices, weights, numbers of vehicles, and
numbers of goals. All of these examples can be expressed numerically. A quantitative variable can be
classified as discrete or continuous. A discrete variable is one whose values are all countable and does
not include any values between two consecutive values of a data set. An example of a discrete variable
is the number of goals scored by a team during a hockey game. A continuous variable is one that can
assume any countable value, as well as all the values between two consecutive numbers of a data set. An
example of a continuous variable is the number of gallons of gasoline used during a trip to the beach.
A qualitative variable is one that cannot be measured numerically but can be placed in a category. Some
examples of a qualitative variable are months of the year, hair color, color of cars, a person’s status, and
favorite vacation spots. The following flow chart should help you to better understand the above terms.


Example 1
Select the best descriptions for the following variables and indicate your selections by marking an ‘x’ in
the appropriate boxes.


Table 8.1:


Variable Quantitative Qualitative Discrete Continuous
Number of mem-
bers in a family
A person’s marital
status
Length of a per-
son’s arm
Color of cars
Number of errors
on a math test


Solution


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Table 8.2:


Variable Quantitative Qualitative Discrete Continuous
Number of mem-
bers in a family


x x


A person’s marital
status


x


Length of a per-
son’s arm


x x


Color of cars x
Number of errors
on a math test


x x


Variables can also be classified as dependent or independent. When there is a linear relationship between
two variables, the values of one variable depend upon the values of the other variable. In a linear relation,
the values of y depend upon the values of x. Therefore, the dependent variable is represented by the
values that are plotted on the y-axis, and the independent variable is represented by the values that are
plotted on the x-axis.
Example 2
Sally works at the local ballpark stadium selling lemonade. She is paid $15.00 each time she works, plus
$0.75 for each glass of lemonade she sells. Create a table of values to represent Sally’s earnings if she sells
8 glasses of lemonade. Use this table of values to represent her earnings on a graph.
Solution
The first step is to write an equation to represent her earnings and then to use this equation to create a
table of values.
y = 0.75x + 15, where y represents her earnings and x represents the number of glasses of lemonade she
sells.


Table 8.3:


Number of Glasses of Lemonade Earnings
0 $15.00
1 $15.75
2 $16.50
3 $17.25
4 $18.00
5 $18.75
6 $19.50
7 $20.25
8 $21.00


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Table 8.3: (continued)


Number of Glasses of Lemonade Earnings


The dependent variable is the money earned, and the independent variable is the number of glasses of
lemonade sold. Therefore, money is on the y-axis, and the number of glasses of lemonade is on the x-axis.
From the table of values, Sally will earn $21.00 if she sells 8 glasses of lemonade.


Now that the points have been plotted, the decision has to be made as to whether or not to join them.
Between every two points plotted on the graph are an infinite number of values. If these values are
meaningful to the problem, then the plotted points can be joined. This type of data is called continuous
data. If the values between the two plotted points are not meaningful to the problem, then the points
should not be joined. This type of data is called discrete data. Since glasses of lemonade are represented
by whole numbers, and since fractions or decimals are not appropriate values, the points between two
consecutive values are not meaningful in this problem. Therefore, the points should not be joined. The
data is discrete.
Now it is time to revisit the problem presented in the introduction.
The local arena is trying to attract as many participants as possible to attend the community’s “Skate for
Scoliosis” event. Participants pay a fee of $10.00 for registering, and, in addition, the arena will donate
$3.00 for each hour a participant skates, up to a maximum of six hours. Create a table of values and draw
a graph to represent a participant who skates for the entire six hours. How much money can a participant
raise for the community if he/she skates for the maximum length of time?
Solution
y = 3x+ 10, where y represents the money made by the participant, and x represents the number of hours
the participant skates.


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Table 8.4:


Numbers of Hours Skating Money Earned
0 $10.00
1 $13.00
2 $16.00
3 $19.00
4 $22.00
5 $25.00
6 $28.00


The dependent variable is the money made, and the independent variable is the number of hours the
participant skated. Therefore, money is on the y-axis, and time is on the x-axis.


A participant who skates for the entire 6 hours can make $28.00 for the ”Skate for Scoliosis” event. The
points are joined, because the fractions and decimals between two consecutive points are meaningful for
this problem. A participant could skate for 30 minutes, and the arena would pay that skater $1.50 for the
time skating. The data is continuous.
Linear graphs are important in statistics when several data sets are used to represent information about a
single topic. An example would be data sets that represent different plans available for cell phone users.
These data sets can be plotted on the same grid. The resulting graph will show intersection points for the
plans. These intersection points indicate a coordinate where two plans are equal. An observer can easily
interpret the graph to decide which plan is best, and when. If the observer is trying to choose a plan to
use, the choice can be made easier by seeing a graphical representation of the data.
Example 3


The following graph represents three plans that are available to customers interested in hiring a maintenance
company to tend to their lawn. Using the graph, explain when it would be best to use each plan for lawn
maintenance.


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Solution
From the graph, the base fee that is charged for each plan is obvious. These values are found on the y-axis.
Plan A charges a base fee of $200.00, Plan C charges a base fee of $100.00, and Plan B charges a base fee
of $50.00. The cost per hour can be calculated by using the values of the intersection points and the base
fee in the equation y = mx+b and solving for m. Plan B is the best plan to choose if the lawn maintenance
takes less than 12.5 hours. At 12.5 hours, Plan B and Plan C both cost $150.00 for lawn maintenance.
After 12.5 hours, Plan C is the best deal, until 50 hours of lawn maintenance is needed. At 50 hours,
Plan A and Plan C both cost $300.00 for lawn maintenance. For more than 50 hours of lawn maintenance,
Plan A is the best plan. All of the above information was obvious from the graph and would enhance the
decision-making process for any interested client.
The above graphs represent linear functions, and are called linear (line) graphs. Each of these graphs has
a defined slope that remains constant when the line is plotted. A variation of this graph is a broken-line
graph. This type of line graph is used when it is necessary to show change over time. A line is used to join
the values, but the line has no defined slope. However, the points are meaningful, and they all represent
an important part of the graph. Usually a broken-line graph is given to you, and you must interpret the
given information from the graph.
Example 4


The following graph is an example of a broken-line graph, and it represents the time of a round-trip journey,
driving from home to a popular campground and back.


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a) How far is it from home to the picnic park?
b) How far is it from the picnic park to the campground?
c) At what two places did the car stop?
d) How long was the car stopped at the campground?
e) When does the car arrive at the picnic park?
f) How long did it take for the return trip?
g) What was the speed of the car from home to the picnic park?
h) What was the speed of the car from the campground to home?
Solution
a) It is 40 miles from home to the picnic park.
b) It is 60 miles from the picnic park to the campground.
c) The car stopped at the picnic park and at the campground.
d) The car was stopped at the campground for 15 minutes.
e) The car arrived at the picnic park at 11:00 am.
f) The return trip took 3 hours and 45 minutes.
g) The speed of the car from home to the picnic park was 40 mi/h.
h) The speed of the car from the campground to home was 100 mi/h.
Example 5
Sam decides to spend some time with his friend Aaron. He hops on his bike and starts off to Aaron’s house,
but on his way, he gets a flat tire and must walk the remaining distance. Once he arrives at Aaron’s house,
they repair the flat tire, play some poker, and then Sam returns home. On his way home, Sam decides to
stop at the mall to buy a book on how to play poker. The following graph represents Sam’s adventure.


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a) How far is it from Sam’s house to Aaron’s house?
b) How far is it from Aaron’s house to the mall?
c) At what time did Sam have a flat tire?
d) How long did Sam stay at Aaron’s house?
e) At what speed did Sam travel from Aaron’s house to the mall and then from the mall to home?
Solution
a) It is 25 km from Sam’s house to Aaron’s house.
b) It is 15 km from Aaron’s house to the mall.
c) Sam had a flat tire at 10:00 am.
d) Sam stayed at Aaron’s house for one hour.
e) Sam traveled at a speed of 30 km/h from Aaron’s house to the mall and then at a speed of 40 km/h
from the mall to home.
Often, when real-world data is plotted, the result is a linear pattern. The general direction of the data can
be seen, but the data points do not all fall on a line. This type of graph is called a scatter plot. A scatter
plot is often used to investigate whether or not there is a relationship or connection between two sets of
data. The data is plotted on a graph such that one quantity is plotted on the x-axis and one quantity
is plotted on the y-axis. The quantity that is plotted on the x-axis is the independent variable, and the
quantity that is plotted on the y-axis is the dependent variable. If a relationship does exist between the
two sets of data, it will be easy to see if the data is plotted on a scatter plot.
The following scatter plot shows the price of peaches and the number sold:


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The connection is obvious−when the price of peaches was high, the sales were low, but when the price was
low, the sales were high.
The following scatter plot shows the sales of a weekly newspaper and the temperature.


There is no connection between the number of newspapers sold and the temperature.
Another term used to describe two sets of data that have a connection or a relationship is correlation.
The correlation between two sets of data can be positive or negative, and it can be strong or weak. The
following scatter plots will help to enhance this concept.


If you look at the two sketches that represent a positive correlation, you will notice that the points are
around a line that slopes upward to the right. When the correlation is negative, the line slopes downward
to the right. The two sketches that show a strong correlation have points that are bunched together and
appear to be close to a line that is in the middle of the points. When the correlation is weak, the points
are more scattered and not as concentrated.
In the sales of newspapers and the temperature, there was no connection between the two data sets. The
following sketches represent some other possible outcomes when there is no correlation between data sets.


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Example 6
Plot the following points on a scatter plot, with m as the independent variable and n as the dependent
variable. Number both axes from 0 to 20. If a correlation exists between the values of m and n, describe
the correlation (strong negative, weak positive, etc.).


m 4 9 13 16 17 6 7 18 10
n 5 3 11 18 6 11 18 12 16


Solution


Example 7
Describe the correlation, if any, in the following scatter plot.


Solution
In the above scatter plot, there is a strong positive correlation.


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You now know that a scatter plot can have either a positive or a negative correlation. When this exists on
a scatter plot, a line of best fit can be drawn on the graph. The line of best fit must be drawn so that
the sums of the distances to the points on either side of the line are approximately equal and such that
there are an equal number of points above and below the line. Using a clear plastic ruler makes it easier to
meet all of these conditions when drawing the line. Another useful tool is a stick of spaghetti, since it can
be easily rolled and moved on the graph until you are satisfied with its location. The edge of the spaghetti
can be traced to produce the line of best fit. A line of best fit can be used to make estimations from the
graph, but you must remember that the line of best fit is simply a sketch of where the line should appear
on the graph. As a result, any values that you choose from this line are not very accurate−the values are
more of a ballpark figure.
Example 8
The following table consists of the marks achieved by 9 students on chemistry and math tests:


Table 8.5:


Student A B C D E F G H I
Chemistry
Marks


49 46 35 58 51 56 54 46 53


Math
Marks


29 23 10 41 38 36 31 24 ?


Plot the above marks on scatter plot, with the chemistry marks on the x-axis and the math marks on the
y-axis. Draw a line of best fit, and use this line to estimate the mark that Student I would have made in
math had he or she taken the test.
Solution


If Student I had taken the math test, his or her mark would have been between 32 and 37.
Scatter plots and lines of best fit can also be drawn by using technology. The TI-83 is capable of graphing
both a scatter plot and of inserting the line of best fit onto the scatter plot.
Example 9
Using the data from Example 8, create a scatter plot and draw a line of best fit with the TI-83.


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Table 8.6:


Student A B C D E F G H I
Chemistry
Marks


49 46 35 58 51 56 54 46 53


Math
Marks


29 23 10 41 38 36 31 24 ?


Solution


The calculator can now be used to determine a linear regression equation for the given values. The equation
can be entered into the calculator, and the line will be plotted on the scatter plot.


From the line of best fit, the calculated value for Student I’s math test mark was 33.6. Remember that
the mark that you estimated was between 32 and 37.
Lesson Summary


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In this lesson, you learned how to represent data by graphing a straight line of the form y = mx + b, and
also by using a scatter plot and a line of best fit. Interpreting a broken-line graph was also presented in
this lesson. You learned about correlation as it applies to a scatter plot and how to describe the correlation
of a scatter plot. You also learned how to draw a line of best fit on a scatter plot and to use this line to
make estimates from the graph. The final topic that was demonstrated in the lesson was how to use the
TI-83 calculator to produce a scatter plot and how to graph a line of best fit by using linear regression.
Points to Consider


• Can any of these graphs be used for comparing data?
• Can the equation for the line of best fit be used to calculate values?
• Is any other graphical representation of data used for estimations?


8.2 Circle Graphs, Bar Graphs, Histograms, and
Stem-and-Leaf Plots


Learning Objectives


• Construct a stem-and leaf plot.
• Understand the importance of a stem-and-leaf plot in statistics.
• Construct and interpret a circle graph.
• Construct and interpret a bar graph.
• Create a frequency distribution chart.
• Construct and interpret a histogram.
• Use technology to create graphical representations of data.


What is the puppet doing? She can’t be cutting a pizza, because the pieces are all different colors and
sizes. It seems like she is drawing some type of a display to show different amounts of a whole circle. The
colors must represent different parts of the whole. As you proceed through this lesson, refer back to this
picture so that you will be able to create a meaningful and detailed answer to the question, “What is the
puppet doing?”


Circle Graphs
Pie charts, or circle graphs, are used extensively in statistics. These graphs appear often in newspapers
and magazines. A pie chart shows the relationship of the parts to the whole by visually comparing the
sizes of the sections (slices). Pie charts can be constructed by using a hundreds disk or by using a circle.
The hundreds disk is built on the concept that the whole of anything is 100%, while the circle is built on
the concept that 360◦ is the whole of anything. Both methods of creating a pie chart are acceptable, and
both will produce the same result. The sections have different colors to enable an observer to clearly see


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the differences in the sizes of the sections. The following example will first be done by using a hundreds
disk and then by using a circle.
Example 10
The Red Cross Blood Donor Clinic had a very successful morning collecting blood donations. Within three
hours, people had made donations, and the following is a table showing the blood types of the donations:


Table 8.7:


Blood Type A B O AB
Number of
donors


7 5 9 4


Construct a pie graph to represent the data.
Solution
Step 1: Determine the total number of donors: 7 + 5 + 9 + 4 = 25.
Step 2: Express each donor number as a percent of the whole by using the formula Percent = fn · 100%,
where f is the frequency and n is the total number.


7
25


· 100% = 28% 5
25


· 100% = 20% 9
25


· 100% = 36% 4
25


· 100% = 16%


Step 3: Use the hundreds disk and simply count the correct number for each blood type (1 line = 1
percent).
Step 4: Graph each section. Write the name and correct percentage inside the section. Color each section
a different color.


The above pie graph was created by using a hundreds disk, which is a circle with 100 divisions in groups
of five. Each division (line) represents one percent. From the graph, you can see that more donations were
of Type O than any other type. The fewest number of donations of blood collected was of Type AB. If
the percentages had not been entered in each section, these same conclusions could have been made based
simply on the size of each section.
Solution
Step 1: Determine the total number of donors: 7 + 5 + 9 + 4 = 25.
Step 2: Express each donor number as the number of degrees of a circle that it represents by using the
formula Degrees = fn · 360◦, where f is the frequency and n is the total number.


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7
25


· 360◦ = 100.8◦ 5
25


· 360◦ = 72◦ 9
25


· 360◦ = 129.6◦ 4
25


· 360◦ = 57.6◦


Step 3: Using a protractor, graph each section of the circle.
Step 4: Write the name and correct percentage inside each section. Color each section a different color.


The above pie graph was created by using a protractor and graphing each section of the circle according to
the number of degrees needed. From the graph, you can see that more donations were of Type O than any
other type. The fewest number of donations of blood collected was of Type AB. Notice that the percentages
have been entered in each section of the graph and not the numbers of degrees. This is because degrees
would not be meaningful to an observer trying to interpret the graph. In order to create a pie graph by
using a circle, it is necessary to use the formula to calculate the number of degrees for each section, and in
order to create a pie graph by using a hundreds disk, it is necessary to use the formula to determine the
percentage for each section. In the end, however, both methods result in identical graphs.
Example 11
A new restaurant is opening in town, and the owner is trying very hard to complete the menu. He wants to
include a choice of 5 salads and has presented his partner with the following circle graph to represent the
results of a recent survey that he conducted of the town’s people. The survey asked the question, ”What
is your favorite kind of salad?”


Use the graph to answer the following questions:


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1. Which salad was the most popular choice?
2. Which salad was the least popular choice?
3. If 300 people were surveyed, how many people chose each type of salad?
4. What is the difference between the number of people who chose the spinach salad and the number
of people who chose the garden salad?


Solution
1. The most popular salad was the caesar salad
2. The least popular salad was the taco salad.
3. Caesar salad: 35% = 35100 = 0.35
(300)(0.35) = 105 people
Taco salad: 10% = 10100 = 0.10
(300)(0.10) = 30 people
Spinach salad: 17% = 17100 = 0.17
(300)(0.17) = 51 people
Garden salad: 13% = 13100 = 0.13
(300)(0.13) = 39 people
Chef salad: 25% = 25100 = 0.25
(300)(0.25) = 75 people
4. The difference between the number of people who chose the spinach salad and the number of people
who chose the garden salad is 51 − 39 = 12 people.
If we revisit the puppet who was introduced at the beginning of the lesson, you should now be able to
create a story that details what she is doing. An example would be that she is in charge of the student
body and is presenting to the students the results of a questionnaire regarding student activities for the
first semester. Of the five activities, the one that is orange in color is the most popular. The students have
decided that they want to have a winter carnival week more than any other activity.
Stem-and-Leaf Plots
In statistics, data is represented in tables, charts, and graphs. One disadvantage of representing data in
these ways is that the actual data values are often not retained. One way to ensure that the data values are
kept intact is to graph the values in a stem-and-leaf plot. A stem-and-leaf plot is a method of organizing
the data that includes sorting the data and graphing it at the same time. This type of graph uses a stem
as the leading part of a data value and a leaf as the remaining part of the value. The result is a graph that
displays the sorted data in groups, or classes. A stem-and-leaf plot is used most when the number of data
values is large.
Example 12


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At a local veterinarian school, the number of animals treated each day over a period of 20 days was
recorded. Construct a stem-and-leaf plot for the data set.


28 34 23 35 16
17 47 05 60 26
39 35 47 35 38
35 55 47 54 48


Solution
Step 1: Some people prefer to arrange the data in order before the stems and leaves are created. This
will ensure that the values of the leaves are in order. However, this is not necessary and can take a great
deal of time if the data set is large. We will first create the stem-and-leaf plot, and then we will organize
the values of the leaves.


The leading digit of a data value is used as the stem, and the trailing digit is used as the leaf. The numbers
in the stem column should be consecutive numbers that begin with the smallest class and continue to the
largest class. If there are no values in a class, do not enter a value in the leaf−just leave it blank.
Step 2: Organize the values in each leaf row.


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Now that the graph has been constructed, there is a great deal of information that can be learned from it.
The number of values in the leaf column should equal the number of data values that were given in the
table. The value that appears the most often in the same leaf row is the trailing digit of the mode of the
data set. The mode of this data set is 35. For 7 of the 20 days, the number of animals receiving treatment
was between 34 and 39. The veterinarian school treated a minimum of 5 animals and a maximum of 60
animals on any one day. The median of the data can be quickly calculated by using the values in the leaf
column to locate the value in the middle position. In this stem and leaf plot, the median is the mean of
the sum of the numbers represented by the 10th and the 11th leaves: 35+352 = 702 = 35.
Example 13
The following numbers represent the growth (in centimeters) of some plants after 25 days.
Construct a stem-and-leaf plot to represent the data, and list three facts that you know about the growth
of the plants.


18 10 37 36 61
39 41 49 50 52
57 53 51 57 39
48 56 33 36 19
30 41 51 38 60


Solution


Answers will vary, but the following are some possible responses:


• From the stem-and-leaf plot, the growth of the plants ranged from a minimum of 10 cm to a maximum


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of 61 cm.
• The median of the data set is the value in the 13th position, which is 41 cm.
• There was no growth recorded in the class of 20 cm, so there is no number in the leaf row.
• The data set is multimodal.


Bar Graphs
The different types of graphs that you have seen so far are plots to use with quantitative variables. A
qualitative variable can be plotted using a bar graph. A bar graph is a plot made of bars whose heights
(vertical bars) or lengths (horizontal bars) represent the frequencies of each category. There is one bar for
each category, with space between each bar, and the data that is plotted is discrete data. Each category
is represented by intervals of the same width. When constructing a bar graph, the category is usually
placed on the horizontal axis, and the frequency is usually placed on the vertical axis. These values can
be reversed if the bar graph has horizontal bars.
Example 14
Construct a bar graph to represent the depth of the Great Lakes:
Lake Superior – 1,333 ft.
Lake Michigan – 923 ft.
Lake Huron – 750 ft.
Lake Ontario – 802 ft.
Lake Erie – 210 ft.
Solution


Example 15
The following bar graph represents the results of a survey to determine the type of TV shows watched by
high school students.


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Use the bar graph to answer the following questions:


1. What type of show is watched the most?
2. What type of show is watched the least?
3. Approximately how many students participated in the survey?
4. Does the graph show the differences between the preferences of males and females?


Solution


1. Sit-coms are watched the most.
2. Quiz shows are watched the least.
3. Approximately 45 + 20 + 18 + 6 + 35 + 16 = 140 students participated in the survey.
4. No, the graph does not show the differences between the preferences of males and females.


If bar graphs are constructed on grid paper, it is very easy to keep the intervals the same size and to keep
the bars evenly spaced. In addition to helping in the appearance of the graph, grid paper also enables you
to more accurately determine the frequency of each class.
Example 16
The following bar graph represents the part-time jobs held by a group of grade-ten students:


Using the above bar graph, answer the following questions:


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1. What was the most popular part-time job?
2. What was the part-time job held by the least number of students?
3. Which part-time jobs employed 10 or more of the students?
4. Is it possible to create a table of values for the bar graph? If so, construct the table of values.
5. What percentage of the students worked as a delivery person?


Solution
1. The most popular part-time job was in the fast food industry.
2. The part-time job of tutoring was the one held by the least number of students.
3. The part-time jobs that employed ten or more students were in the fast food, delivery, lawn maintenance,
and grocery store businesses.
4. Yes, it’s possible to create a table of values for the bar graph.


Table 8.8:


Part-Time
Job


Baby Sit-
ting


Fast Food Delivery Lawn Care Grocery
Store


Tutoring


Number of
Students


8 14 12 10 13 5


5. The percentage of the students who worked as a delivery person was approximately 19.4%.


Histograms
An extension of the bar graph is the histogram. A histogram is a type of vertical bar graph in which the
bars represent grouped continuous data. While there are similarities between a bar graph and a histogram,
such as each bar being the same width, a histogram has no spaces between the bars. The quantitative
data is grouped according to a determined bin size, or interval. The bin size refers to the width of each
bar, and the data is placed in the appropriate bin.
The bins, or groups of data, are plotted on the x-axis, and the frequencies of the bins are plotted on the
y-axis. A grouped frequency distribution is constructed for the numerical data, and this table is used to
create the histogram. In most cases, the grouped frequency distribution is designed so there are no breaks
in the intervals. The last value of one bin is actually the first value counted in the next bin. This means
that if you had groups of data with a bin size of 10, the bins would be represented by the notation [0-10),
[10-20), [20-30), etc. Each bin appears to contain 11 values, which is one more than the desired bin size of
10. Therefore, the last digit of each bin is counted as the first digit of the following bin.
The first bin includes the values 0 through 9, and the next bin includes the values 9 through 19. This
makes the bins the proper size. Bin sizes are written in this manner to simplify the process of grouping the
data. The first bin can begin with the smallest number of the data set and end with the value determined
by adding the bin width to this value, or the bin can begin with a reasonable value that is smaller than
the smallest data value.
Example 17


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Construct a frequency distribution table with a bin size of 10 for the following data, which represents the
ages of thirty lottery winners:


38 41 29 33 40 74 66 45 60 55
25 52 54 61 46 51 59 57 66 62
32 47 65 50 39 22 35 72 77 49


Solution
Step 1: Determine the range of the data by subtracting the smallest value from the largest value.


Range: 77 − 22 = 55


Step 2: Divide the range by the bin size to ensure that you have at least 5 groups of data. A histogram
should have from 5 to 10 bins to make it meaningful: 5510 = 5.5 ≈ 6. Since you cannot have 0.5 of a bin,
the result indicates that you will have at least 6 bins.
Step 3: Construct the table:


Table 8.9:


Bin Frequency
[20-30) 3
[30-40) 5
[40-50) 6
[50-60) 8
[60-70) 5
[70-80) 3


Step 4: Determine the sum of the frequency column to ensure that all the data has been grouped.
3 + 5 + 6 + 8 + 5 + 3 = 30


When data is grouped in a frequency distribution table, the actual data values are lost. The table indicates
how many values are in each group, but it doesn’t show the actual values.
There are many different ways to create a distribution table and many different distribution tables that
can be created. However, for the purpose of constructing a histogram, the method shown works very well,
and it is not difficult to complete. When the number of data values is very large, another column is often
inserted in the distribution table. This column is a tally column, and it is used to account for the number
of values within a bin. A tally column facilitates the creation of the distribution table and usually allows
the task to be completed more quickly.
Example 18
The numbers of years of service for 75 teachers in a small town are listed below:


1, 6, 11, 26, 21, 18, 2, 5, 27, 33, 7, 15, 22, 30, 8
31, 5, 25, 20, 19, 4, 9, 19, 34, 3, 16, 23, 31, 10, 4
2, 31, 26, 19, 3, 12, 14, 28, 32, 1, 17, 24, 34, 16, 1,
18, 29, 10, 12, 30, 13, 7, 8, 27, 3, 11, 26, 33, 29, 20
7, 21, 11, 19, 35, 16, 5, 2, 19, 24, 13, 14, 28, 10, 31


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Using the above data, construct a frequency distribution table with a bin size of 5.
Solution


Range: 35 − 1 = 34
34
5


= 6.8 ≈ 7


You will have 7 bins.
For each value that is in a bin, draw a stroke in the Tally column. To make counting the strokes easier,
draw four strokes and cross them out with the fifth stroke. This process bundles the strokes in groups of
five, and the frequency can be readily determined.


Table 8.10:


Bin Tally Frequency
[0-5)




||||


|||| | 11
[5-10)




|||| |||| 9
[10-15)




||||


|||| || 12
[15-20)




||||


|||| |||| 14
[20-25)




|||| || 7
[25-30)




||||


|||| 10
[30-35)




||||


|||| || 12


11 + 9 + 12 + 14 + 7 + 10 + 12 = 75


Now that you have constructed the frequency table, the grouped data can be used to draw a histogram.
Like a bar graph, a histogram requires a title and properly labeled x- and y-axes.
Example 19
Use the data from Example 17 that displays the ages of the lottery winners to construct a histogram to
represent the given data:


Table 8.11:


Bin Frequency
[20-30) 3
[30-40) 5


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Table 8.11: (continued)


Bin Frequency
[40-50) 6
[50-60) 8
[60-70) 5
[70-80) 3


Solution
Use the data as it is represented in the distribution table:


From looking at the tops of the bars, you can see how many winners were in each category, and by adding
these numbers, you can determine the total number of winners. You can also determine how many winners
were within a specific category. For example, you can see that 8 winners were 60 years of age or older.
The graph can also be used to determine percentages. For example, it can answer the question, “What
percentage of the winners were 50 years of age or older?” as follows:
16
30 = 0.533 (0.533)(100%) ≈ 5.3%.
Example 20
a) Use the data and the distribution table that represent the ages of teachers from Example 18 to construct
a histogram to display the data.


Table 8.12:


Bin Tally Frequency
[0 − 5)




||||


|||| | 11
[5 − 10)




|||| |||| 9
[10 − 15)




||||


|||| || 12
[15 − 20)




||||


|||| |||| 14
[20 − 25)




|||| || 7
[25 − 30)




||||


|||| 10
[30 − 35)




||||


|||| || 12


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b) Now use the histogram to answer the following questions.
i) How many teachers teach in this small town?
ii) How many teachers have worked for less than 5 years?
iii) If teachers are able to retire when they have taught for 30 years or more, how many are eligible to
retire?
iv) What percentage of the teachers still have to teach for 10 years or fewer before they are eligible to
retire?
v) Do you think that the majority of the teachers are young or old? Justify your answer.
Solution
a)


b) i) 11 + 9 + 12 + 14 + 7 + 10 + 12 = 75
In this small town, 75 teachers are teaching.
ii) 11 teachers have taught for less than five years.
iii) 12 teachers are eligible to retire.
iv) 1775 = 0.2266 (0.2266)(100%) ≈ 2.3%
Approximately 2.3% of the teachers must teach for 10 years or fewer before they are eligible to retire.
v) The majority of the teachers are young, because 46 have taught for less than 20 years.
Technology can also be used to plot a histogram. The TI-83 can be used to create a histogram by using
the STAT function and the STAT PLOT function of the calculator.
Example 21
Scientists have invented a new dietary supplement that is supposed to increase the weight of a piglet within
its first three months of growth. Farmer John fed this supplement to his stock of piglets, and at the end
of three months, he recorded the weights of 50 randomly selected piglets.
The following table is the recorded weight (in pounds) of the 50 selected piglets:


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120 111 65 110 114 72 116 105 119 114
93 113 99 118 108 97 107 95 113 75
84 120 102 104 84 97 121 69 100 101
107 118 77 105 109 78 89 68 74 103
87 67 79 90 109 94 106 96 92 88


Using the above data set and the TI-83, construct a histogram to represent the data.
Solution


Using the TRACE function will give you information about the data in each bar of the histogram.


The TRACE function tells you that in the first bin, which is [60-70), there are 4 values.


The TRACE function tells you that in the second bin, which is [70-80), there are 6 values.
To advance to the next bin, or bar, of the histogram, use the cursor and advance to the right. The
information obtained by using the TRACE function will enable you to create a frequency table and to
draw the histogram on paper.


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The shape of a histogram can tell you a lot about the distribution of the data, as well as provide you with
information about the mean, median, and mode of the data set. The following are some typical histograms,
with a caption below each one explaining the distribution of the data, as well as the characteristics of the
mean, median, and mode. Distributions can have other shapes besides the ones shown below, but these
represent the most common ones that you will see when analyzing data.
a)


For a symmetric histogram, the values of the mean, median, and mode are all the same and are all
located at the center of the distribution.
b)


c)


For a histogram that is skewed to the right, the mean is located to the right on the distribution and is the
largest value of the measures of central tendency. The mean has the largest value because it is strongly
affected by the outliers on the right tail that pull the mean to the right. The mode is the smallest value,
and it is located to the left on the distribution. The mode always occurs at the highest point of the peak.


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The median is located between the mode and the mean.
d)


For a histogram that is skewed to the left, the mean is located to the left on the distribution and is the
smallest value of the measures of central tendency. The mean has the smallest value because it is strongly
affected by the outliers on the left tail that pull the mean to the left. The median is located between the
mode and the mean.
e)


In each of the above graphs, the distributions are not perfectly shaped, but are shaped enough to identify
an overall pattern. Figure a represents a bell-shaped distribution, which has a single peak and tapers off
to both the left and to the right of the peak. The shape appears to be symmetric about the center of the
histogram. The single peak indicates that the distribution is unimodal. The highest peak of the histogram
represents the location of the mode of the data set. The mode is the data value that occurs the most often
in a data set.
Figure b represents a distribution that is approximately uniform and forms a rectangular, flat shape. The
frequency of each class is approximately the same.
Figure c represents a right-skewed distribution, which has a peak to the left of the distribution and
data values that taper off to the right. This distribution has a single peak and is also unimodal.
Figure d represents a left-skewed distribution, which has a peak to the right of the distribution and
data values that taper off to the left. This distribution has a single peak and is also unimodal.
Figure e has no shape that can be defined. The only defining characteristic about this distribution is that
it has 2 peaks of the same height. This means that the distribution is bimodal.


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Another type of graph that can be drawn to represent the same set of data as a histogram represents is
a frequency polygon. A frequency polygon is a graph constructed by using lines to join the midpoints
of each interval, or bin. The heights of the points represent the frequencies. A frequency polygon can be
created from the histogram or by calculating the midpoints of the bins from the frequency distribution
table. The midpoint of a bin is calculated by adding the upper and lower boundary values of the bin and
dividing the sum by 2.
Example 22
The following histogram represents the marks made by 40 students on a math 10 test.
Use the histogram to construct a frequency polygon to represent the data.


Solution


There is no data value greater than zero and less than 20. The jagged line that is inserted on the x-axis
is used to represent this fact. The area under the frequency polygon is the same as the area under the
histogram and is, therefore, equal to the frequency values that would be displayed in a distribution table.
The frequency polygon also shows the shape of the distribution of the data, and in this case, it resembles
a bell curve.
Example 23


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The following distribution table represents the number of miles run by 20 randomly selected runners during
a recent road race.


Table 8.13:


Bin Frequency
[5.5-10.5) 1
[10.5-15.5) 3
[15.5-20.5) 2
[20.5-25.5) 4
[25.5-30.5) 5
[30.5-35.5) 3
[35.5-40.5) 2


Using this table, construct a frequency polygon.
Solution
Step 1: Calculate the midpoint of each bin by adding the two numbers of the interval and dividing the
sum by 2.


Midpoints: 5.5 + 10.5
2


= 16
2


= 8 10.5 + 15.5
2


= 26
2


= 13 15.5 + 20.5
2


= 36
2


= 18
20.5 + 25.5


2
= 46


2
= 23 25.5 + 30.5


2
= 56


2
= 28 30.5 + 35.5


2
= 66


2
= 33


35.5 + 40.5
2


= 76
2


= 38


Step 2: Plot the midpoints on a grid, making sure to number the x-axis with a scale that will include the
bin sizes. Join the plotted midpoints with lines.


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A frequency polygon usually extends one unit below the smallest bin value and one unit beyond the greatest
bin value. This extension gives the frequency polygon an appearance of having a starting point and an
ending point, which provides a view of the distribution of data. If the data set were very large so that the
number of bins had to be increased and the bin size decreased, the frequency polygon would appear as a
smooth curve.
Lesson Summary
In this lesson, you learned how to represent data that was presented in various forms. Data that could be
represented as percentages was displayed in a circle graph, or pie graph. Discrete data that was qualitative
was displayed on a bar graph. Finally, continuous data that was grouped was graphed on a histogram or
on a frequency polygon. You also learned to detect characteristics of a distribution by simply observing the
shape of a histogram. Once again, technology was shown to be an asset when constructing a histogram.
Points to Consider


• Can any of these graphs be used for comparing data?
• Can these graphs be used to display solutions to problems in everyday life?
• How do these graphs compare to ones presented in previous lessons?


8.3 Box-and-Whisker Plots
Learning Objectives


• Construct a box-and-whisker plot.
• Construct and interpret a box-and-whisker plot.
• Use technology to create box-and-whisker plots.


Bob, a financial planner, recorded the number of clients he saw each day over an eleven-day period. The
numbers of clients he saw are shown below:


31, 33, 29, 40, 51, 27, 30, 43, 38, 23, 42


After Bob reviewed the data, he was satisfied with the results and thought that he had met with a sufficient
number of clients. Display the set of data in order to explain whether the claim made by Bob is true or
false. Use the display to justify your answer.
We will revisit this problem later in the lesson to explain whether or not Bob’s claim was true or false.
Box-and Whisker Plots


In traditional statistics, data is organized by using a frequency distribution. The results of the frequency
distribution can then be used to create various graphs, such as a histogram or a frequency polygon, which
indicate the shape or nature of the distribution. The shape of the distribution will allow you to confirm
various conjectures about the nature of the data.
To examine data in order to identify patterns, trends, or relationships, exploratory data analysis is used. In
exploratory data analysis, organized data is displayed in order to make decisions or suggestions regarding


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further actions. A box-and-whisker plot (often called a box plot) can be used to graphically represent
the data set, and the graph involves plotting five specific values. The five specific values are often referred
to as a five-number summary of the organized data set. The five-number summary consists of:


1. The lowest number in the data set (minimum value)
2. The median of the lower quartile: Q1 (median of the first half of the data set)
3. The median of the entire data set (median)
4. The median of the upper quartile: Q3 (median of the second half of the data set)
5. The highest number in the data set (maximum value)


The display of the five-number summary produces a box-and-whisker plot:


The above model of a box-and-whisker plot shows two horizontal lines (the whiskers) that each contain
25% of the data and are of the same length. In addition, it shows that the median of the data set is in
the middle of the box, which contains 50% of the data. The lengths of the whiskers and the location of
the median with respect to the center of the box are used to describe the distribution of the data. It’s
important to note that this is just an example. Not all box-and-whisker plots have the median in the
middle of the box and whiskers of the same size.
Information about the data set that can be determined from the box-and-whisker plot with respect to the
location of the median is:
a) If the median is located in the center or near the center of the box, the distribution is approximately
symmetric.
b) If the median is located to the left of the center of the box, the distribution is positively skewed.
c) If the median is located to the right of the center of the box, the distribution is negatively skewed.
Information about the data set that can be determined from the box-and-whisker plot with respect to the
length of the whiskers is:
a) If the whiskers are the same or almost the same length, the distribution is approximately symmetric.
b) If the right whisker is longer than the left whisker, the distribution is positively skewed.
c) If the left whisker is longer than the right whisker, the distribution is negatively skewed.
The length of the whiskers also gives you information about how spread out the data is.
A box-and-whisker plot is often used when the number of data values is large. The center of the distribution,
the nature of the distribution, and the range of the data are very obvious from the graph. The five-number
summary divides the data into quarters by use of the medians of the upper and lower halves of the data.
Remember that, unlike the mean, the median of the entire data set is not affected by outliers, so it is the
measure of central tendency that is most often used in exploratory data analysis.
Example 23
For the following data sets, determine the five-number summary.


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a) 12, 16, 36, 10, 31, 23, 58
b) 144, 240, 153, 629, 540, 300
Solution
a) The first step is to organize the values in the data set:


12, 16, 36, 10, 31, 23, 58
10, 12, 16, 23, 31, 36, 58


Now complete the following list:
Minimum value → 10
Q1 → 12
Median → 23
Q3 → 36
Maximum value → 58
b) The first step is to organize the values in the data set:


144, 240, 153, 629, 540, 300
144, 153, 240, 300, 540, 629


Now complete the following list:
Minimum value → 144
Q1 → 153
Median → 270
Q3 → 540
Maximum value → 629
Example 24
Use the data set for Example 1 part a) and the five-number summary to construct a box-and-whisker plot
to model the data set.
Solution


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The five-number summary can now be used to construct a box-and-whisker plot. Be sure to provide a
scale on the number line that includes the range from the minimum value to the maximum value.
a) Minimum value → 10
Q1 → 12
Median → 23
Q3 → 36
Maximum value → 58


It is very visible that the right whisker is much longer than the left whisker. This indicates that the
distribution is positively skewed.
Example 25
For each box-and-whisker plot, list the five-number summary and describe the distribution based on the
location of the median.


Solution
a) Minimum value → 4
Q1 → 6
Median → 9
Q3 → 10
Maximum value → 12
The median of the data set is located to the right of the center of the box, which indicates that the
distribution is negatively skewed.
b) Minimum value → 225
Q1 → 250


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Median → 300
Q3 → 325
Maximum value → 350
The median of the data set is located to the right of the center of the box, which indicates that the
distribution is negatively skewed.
c) Minimum value → 60
Q1 → 70
Median → 75
Q3 → 95
Maximum value → 100
The median of the data set is located to the left of the center of the box, which indicates that the distribution
is positively skewed.
Example 26
The numbers of square feet (in 100s) of 10 of the largest museums in the world are shown below:
650, 547, 204, 213, 343, 288, 222, 250, 287, 269
Construct a box-and-whisker plot for the above data set and describe the distribution.
Solution
The first step is to organize the data values:


20, 400 21, 300 22, 200 25, 000 26, 900 28, 700 28, 800 34, 300 54, 700 65, 000


Now calculate the median, Q1, and Q3.


20, 400 21, 300 22, 200 25, 000 26, 900 28, 700 28, 800 34, 300 54, 700 65, 000


Median→ 26, 900 + 28, 700
2


= 55, 600
2


= 27, 800


Q1 = 22, 200
Q3 = 34, 300
Next, complete the following list:
Minimum value → 20, 400
Q1 → 22, 200
Median → 27, 800
Q3 → 34, 300
Maximum value → 65, 000


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The right whisker is longer than the left whisker, which indicates that the distribution is positively skewed.
The TI-83 or TI-84 can also be used to create a box-and whisker plot. In the following examples, the
TI-83 is used. In the next chapter, key strokes using the TI-84 will be presented to you. The five-number
summary values can be determined by using the TRACE function of the calculator or by using the CALC
function and one variable stats.
Example 27
The following numbers represent the number of siblings in each family for fifteen randomly selected students:


4, 1, 2, 2, 5, 3, 4, 2, 6, 4, 6, 1, 7, 8, 4


Use technology to construct a box-and-whisker plot to display the data. List the five-number summary
values.
Solution


The five–number summary can be obtained from the calculator in two ways.
1. The following results are obtained by simply using the TRACE function.


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The values at the bottom of each screen are the five-number summary.
2. The second method involves using the CALC function and one variable stats for L1.


Many data sets contain values that are either extremely high values or extremely low values compared to
the rest of the data values. These values are called outliers. There are several reasons why a data set
may contain an outlier:


1. The value may be the result of an error made in measurement or in observation. The researcher may
have measured the variable incorrectly.


2. The value may simply be an error made by the researcher in recording the value. The value may
have been written or typed incorrectly.


3. The value could be a result obtained from a subject not within the defined population. A researcher
recording marks from a math 12 examination may have recorded a mark by a student in grade 11
who was taking math 12.


4. The value could be one that is legitimate but is extreme compared to the other values in the data
set. (This rarely occurs, but it is a possibility)


If an outlier is present because of an error in measurement, observation, or recording, then either the error
should be corrected, or the outlier should be omitted from the data set. If the outlier is a legitimate value,
then the statistician must make a decision as to whether or not to include it in the set of data values.
There is no rule that tells you what to do with an outlier in this case.
One method for checking a data set for the presence of an outlier is to follow this procedure:


1. Organize the given data set and determine the values of Q1 and Q3.
2. Calculate the difference between Q1 and Q3. This difference is called the interquartile range


(IQR): IQR = Q3 − Q1.
3. Multiply the difference by 1.5, subtract this result from Q1, and add it to Q3.
4. The results from Step 3 will be the range into which all values of the data set should fit. Any values
that are below or above this range are considered outliers.


Example 28


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Using the procedure outlined above, check the following data sets for outliers:
a) 18, 20, 24, 21, 5, 23, 19, 22
b) 13, 15, 19, 14, 26, 17, 12, 42, 18
Solution
a) Organize the given data set:


18, 20, 24, 21, 5, 23, 19, 22
5, 18, 19, 20, 21, 22, 23, 24


Determine the values for Q1 and Q3.


5, 18, 19 , 20, 21, 22, 23 , 24


Q1 =
18 + 19


2
= 37


2
= 18.5 Q3 =


22 + 23
2


= 45
2


= 22.5


Calculate the difference between Q1 and Q3: Q3 − Q1 = 22.5 − 18.5 = 4.0.
Multiply this difference by 1.5: (4.0)(1.5) = 6.0.
Compute the range.


Q1 − 6.0 = 18.5 − 6.0 = 12.5


Q3 + 6.0 = 22.5 + 6.0 = 28.5


.
Are there any data values below 12.5? Yes, the value of 5 is below 12.5 and is, therefore, an outlier.
Are there any values above 28.5? No, there are no values above 28.5.
b) Organize the given data set:


13, 15, 19, 14, 26, 17, 12, 42, 18
12, 13, 14, 15, 17, 18, 19, 26, 42


Determine the values for Q1 and Q3.


12, 13, 14 , 15, 17 , 18, 19, 26 , 42


Q1 =
13 + 14


2
= 27


2
= 13.5 Q3 =


19 + 26
2


= 45
2


= 22.5


Calculate the difference between Q1 and Q3: Q3 − Q1 = 22.5 − 13.5 = 9.0.
Multiply this difference by 1.5: (9.0)(1.5) = 13.5.
Compute the range.


Q1 − 13.5 = 13.5 − 13.5 = 0


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Q3 + 13.5 = 22.5 + 13.5 = 36.0


Are there any data values below 0? No, there are no values below 0.
Are there any values above 36.0? Yes, the value of 42 is above 36.0 and is, therefore, an outlier.
Lesson Summary
You have learned the significance of the median as it applies to dividing a set of data values into quartiles.
You have also learned how to apply these values to the five-number summary needed to construct a box-
and-whisker plot. In addition, you have learned how to construct a box-and-whisker plot and how to obtain
the five-number summary by using technology. The last topic that you learned about in this chapter was
the meaning of the term outlier. Some reasons why an outlier might exist in a data set and the procedure
for determining whether or not a data set contains an outlier were also discussed.
Points to Consider


• Are there still other ways to represent data graphically?
• Are there other uses for a box-and-whisker plot?
• Can box-and-whisker plots be used for comparing data sets?


Vocabulary


Bar graph A plot made of bars whose heights (vertical bars) or lengths (horizontal bars) represent the
frequencies of each category.


Bins Quantitative or qualitative categories. Bins are also known as classes.


Box-and-whisker plot A graph of a data set in which the five-number summary is plotted. Fifty percent
of the data values are in the box and the remaining fifty percent is divided equally on the whiskers.


Broken-line graph A graph that is used when it is necessary to show change over time. A line is used
to join the values, but the line has no defined slope.


Continuous data Data for which the plotted points can be joined.


Continuous variable A variable that can assume all values between two consecutive values of a data
set.


Correlation A statistical method used to determine whether or not there is a linear relationship between
two variables.


Data set A collection of observations of a variable.


Dependent variable The variable represented by the values that are plotted on the y-axis.


Discrete data Data for which the plotted points cannot be joined.


Discrete variable A variable that can only assume values that can be counted.


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Five-number summary Five values for a data set that include the smallest value, the lower quartile,
the median, the upper quartile, and the largest value.


Frequency distribution A table that lists all of the classes and the number of data values that belong
to each of the classes.


Frequency polygon A graph that uses lines to join the midpoints of the tops of the bars of a histogram
or to join the midpoints of the classes.


Histogram A graph in which the classes or bins are on the horizontal axis and the frequencies are plotted
on the vertical axis. The frequencies are represented by vertical bars that are drawn adjacent to each
other.


Independent variable The variable represented by the values that are plotted on the x-axis.


Interquartile range (IQR) The difference between the third quartile and the first quartile.


Left-skewed distribution A distribution in which most of the data values are located to the right of
the mean.


Line of best fit A straight line drawn on a scatter plot such that the sums of the distances to points
on either side of the line are approximately equal and such that there are an equal number of points
above and below the line.


Midpoint The value obtained by adding the lower and upper limit of a class and dividing the sum by 2.


Pie chart A circle that is divided into sections (slices) according to the percentage of the frequencies in
each class.


Qualitative variable A variable that can be placed into specific categories according to some defined
characteristic.


Quantitative variable A variable that is numerical in nature and that can be ordered.


Right-skewed distribution A distribution in which most of the data values are located to the left of
the mean.


Scatter plot A graph used to investigate whether or not there is a relationship between two sets of data.
The data is plotted on a graph such that one quantity is plotted on the x-axis and one quantity is
plotted on the y-axis.


Stem-and-leaf plot A method of organizing data that includes sorting the data and graphing it at the
same time. This type of graph uses the stem as the leading part of the data value and the leaf as the
remaining part of the value.


Symmetric histogram A histogram for which the values of the mean, median, and mode are all the
same and are all located at the center of the distribution.


Variable A characteristic that is being studied.


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8.4 Review Questions
Line Graphs and Scatter Plots
Show all work necessary to answer each question.
Section A – All questions in this section are selected response.


1. What term is used to describe a data set in which all points between two consecutive points are
meaningful?
(a) discrete data
(b) continuous data
(c) random data
(d) fractional data


2. What is the correlation of a scatter plot that has few points that are not bunched together?
(a) strong
(b) no correlation
(c) weak
(d) negative


3. Which of the following calculations will create the line of best fit on the TI-83?
(a) quadratic regression
(b) cubic regression
(c) exponential regression
(d) linear regression (ax + b)


4. What type of variable is represented by the number of pets owned by families?
(a) qualitative
(b) quantitative
(c) independent
(d) continuous


5. What term is used to define the connection between two data sets?
(a) relationship
(b) scatter plot
(c) correlation
(d) discrete


6. What type of data, when plotted on a graph, does not have the points joined?
(a) discrete data
(b) continuous data
(c) random data
(d) independent data


7. What name is given to a graph that shows change over time, with points that are joined but have no
defined slope?
(a) linear graph
(b) broken-line graph
(c) scatter plot
(d) line of best fit


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Section B – All questions in this section are long answer questions. Be sure to show all of the work
necessary to arrive at the correct answer.


1. Select the best descriptions for the following variables and indicate your selections by marking an ‘x’
in the appropriate boxes.


Table 8.14:


Variable Quantitative Qualitative Discrete Continuous
Men’s favorite TV
shows
Salaries of baseball
players
Number of chil-
dren in a family
Favorite color of
cars
Number of hours
worked weekly


2. Describe the correlation of each of the following graphs:
(a)


(b)


(c)


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3. Answer the questions below for the following broken-line graph, which shows the distance, over time,
of a bus from the bus depot.


(a) What was the fastest speed of the bus?
(b) How many times did the bus stop on its trip? (Do not count the beginning and the end of the


trip.)
(c) What was the initial distance of the bus from the bus depot?
(d) What was the total distance traveled by the bus?


4. The following table represents the sales of Volkswagen Beetles in Iowa between 1994 and 2003:


Table 8.15:


Year 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003
Beetles
Sold


50 60 55 50 70 65 75 65 80 90


(a) Create a scatter plot and draw the line of best fit for the data. Hint: Let 0 = 1994, 1 = 1995, etc.
(b) Use the graph to predict the number of Beetles that will be sold in Iowa in the year 2007.
(c) Describe the correlation for the above graph.


5. You are selling your motorcycle, and you decide to advertise it on the Internet on Walton’s Web Ads.
He has three plans from which you may choose. The plans are shown on the following graph. Use
the graph and explain when it is best to use each plan.


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6. The data below gives the gasoline used by cars with the same-sized engines when driven at various
speeds.


Speed (m/h) 32 64 77 42 82 57 72
Gasoline Used (m/gal) 40 27 24 37 22 36 28


(a) Draw a scatter plot and a line of best fit. (You may use technology.)
(b) If a car were traveling at a speed of 47 m/h, estimate the amount of gasoline that would be


used.
(c) If a car uses 29 m/gal of gasoline, estimate the speed of the car.


7. For the following broken-line graph, write a story to accompany the graph, and provide a detailed
description of the events that are occurring.


8. Plot the following points on a scatter graph, with m as the independent variable and n as the
dependent variable. Number both axes from 0 to 20. If a correlation exists between the values of m
and n, describe the correlation (strong negative, weak positive, etc.).


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(a) m 5 14 2 10 16 4 18 2 8 11
n 6 13 4 10 15 7 16 5 8 12


(b) m 13 3 18 9 20 15 6 10 21 4
n 7 14 9 16 7 13 10 13 3 19


Circle Graphs, Bar Graphs, Histograms, and Stem-and-Leaf Plots
Section A – All questions in this section are selected response. Circle the correct answer.


1. In the following stem-and-leaf plot that represents the ages of 23 people waiting in line at Tim
Horton’s, how many people were older than 32?


(a) 4
(b) 12
(c) 14
(d) 11


2.


The above histogram shows data collected during a recent fishing derby. The number of fish caught is
being compared to the size of the fish caught. How many fish caught were between 20 cm and 29 cm in
length?
(a) 3
(b) 11
(c) 25
(d) 6


3. What name is given to a distribution that has two peaks of the same height?
(a) uniform


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(b) unimodal
(c) bimodal
(d) discrete


4. What is the midpoint of the bin [14.5-23.5)?


(a) 19
(b) 4.5
(c) 18.5
(d) 38


5. The following stem-and-leaf plot shows the cholesterol levels of a random number of students. These
values range from 2.0 to 9.0. What percentage of the students have levels between 5.0 and 7.0,
inclusive?


(a) 6 %
(b) 20%
(c) 24%
(d) 28%


6. What is the dependent variable in the following relationship?
The time it takes to run the 100 yard dash and the fitness level of the runner.


(a) fitness level
(b) time
(c) length of the track
(d) age of the runner


7. What name is given to the graph that uses lines to join the midpoints of the classes?


(a) bar graph
(b) stem-and-leaf
(c) histogram
(d) frequency polygon


8. What is the mode of the following data set displayed in the stem-and-leaf plot?


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(a) 32
(b) 41
(c) 7
(d) 23


9.


How many of the above people had less than $850 in the bank?
(a) 2
(b) 4
(c) 3
(d) 6


10. On a recent math test, 9 students out of the 25 who took the test scored above 85. What percentage
of the students scored above 85?
(a) 0.36 %
(b) 3.6 %
(c) 360 %
(d) 36 %


Section B – All questions in this section are long-answer questions. Be sure to show all of the work
necessary to arrive at the correct answer.


1. Construct a stem-and-leaf plot for the following data values.20 12 39 38 18 58 49 59 66 50
23 32 43 53 67 35 29 13 42 55
37 19 38 22 46 71 9 65 15 38


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2. The following pie chart represents the time spent doing various activities during one day. Using the
chart, supply a possible activity that may be represented by each of the percentages shown in the
graph.


3. Just like Presidents of the United States, Canadian Prime Ministers must be sworn into office. The
following data represents the ages of 22 Canadian Prime Ministers when they were sworn into office.
Construct a stem-and-leaf plot to represent the ages, and list four facts that you know from the
graph.


52 74 60 39 65 46 55 66 54 51 70 47 69 47 57 46
48 66 61 59 46 45


4. A questionnaire on the makes of people’s vehicles showed the following responses from 30 participants.
Construct a frequency distribution and a bar graph to represent the data.
(F = Ford, H = Honda, V = Volkswagen, M = Mazda)


F M M M V M F M F V H H F V F
H H F M M V H M V V F V H M F


5. The following histogram displays the heights of students in a classroom.


Use the information represented in the histogram to answer the following questions:


(a) How many students were in the class?
(b) How many students were over 60 inches in height?
(c) How many students had a height between 54 in and 62 in?
(d) Is the distribution unimodal or bimodal? How do you know?


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6. The following frequency polygon represents the weights of players who all participated in the same
sport. Use the polygon to answer the following questions:


(a) How many players played the sport?
(b) What was the most common weight for the players?
(c) What sport do you think the players may have been playing?
(d) What do the weights of 55 kg and 105 kg represent?
(e) What two weights have no recorded players weighing those amounts?


7. The following circle graph, which is incomplete, shows the extracurricular activities for 200 high
school students.


Use the circle graph to answer the following questions:


(a) What makes the above circle graph incomplete?
(b) How many students participated in sports?
(c) How many students do CMT?
(d) How many students participate in volunteer activities after school?
(e) Construct the circle graph so that it shows percentages and not degrees.


8. The following data represents the results of a test taken by a group of students:


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95 56 70 83 59 66 88 52 50 77 69 80
54 75 68 78 51 64 55 67 74 57 73 53


Construct a frequency distribution table using a bin size of 10 and display the results in a properly
labeled histogram.


9. Using the data for question 8, use technology to construct the histogram.
10. In a few sentences, explain the type of graph that you find most helpful for interpreting data.


Box-and-Whisker Plots
Section A – All questions in this section are selected response.


1. Which of the following is not a part of the five-number summary?
(a) Q1 and Q3
(b) the mean
(c) the median
(d) minimum and maximum values


2. What percent of the data is contained in the box of a box-and-whisker plot?
(a) 25 %
(b) 100 %
(c) 50 %
(d) 75 %


3. What name is given to the horizontal lines to the left and right of the box of a box-and-whisker plot?
(a) axis
(b) whisker
(c) range
(d) plane


4. What term describes the distribution of a data set if the median of the data set is located to the left
of the center of the box of a box-and-whisker plot?
(a) positively skewed
(b) negatively skewed
(c) approximately symmetric
(d) not skewed


5. What two values of the five-number summary are connected with two horizontal lines on a box-and-
whisker plot?
(a) Minimum value and the median
(b) Maximum value and the median
(c) Minimum and maximum values
(d) Q1 and Q3


Section B - Show all work necessary to answer each question.


1. For the following data sets, determine the five-number summary.
(a) 74, 69, 83, 79, 60, 75, 67, 71
(b) 6, 9, 3, 12, 11, 9, 15, 5, 7


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2. For each of the following box-and-whisker plots, list the five-number summary and comment on the
distribution of the data.
(a)


(b)


3. The following data represents the number of coins that 12 randomly selected people had in their
piggy banks.


35 58 29 44 104 39 72 34 50 41 64 54


Construct a box-and-whisker plot for the above data.
4. The following data represent the time (in minutes) that each of 20 people waited in line at a local
book store to purchase the latest Harry Potter book.


15 8 5 10 14 17 21 23 6 19 31 34 30 31
3 22 17 25 5 16


Construct a box-and-whisker plot for the above data. Are the data skewed in any direction?
5. Firman’s Fitness Factory is a new gym that offers reasonably-priced family packages. The following
table represents the number of family packages sold during the opening month.


24 21 31 28 29
27 22 27 30 32
26 35 24 22 34
30 28 24 32 27
32 28 27 32 23
20 32 28 32 34


Construct a box-and-whisker plot for the data. Are the data symmetric or skewed?
6. The following data represents the number of flat-screen televisions assembled at a local electronics
company for a sample of 28 days.


48 55 51 44 59 49 47
45 51 56 50 57 53 55
47 49 51 54 56 54 47
50 53 52 55 51 59 48


Using technology, construct a box-and-whisker plot for the data. What are the values for the five-
number summary?


7. Construct a box-and-whisker plot to represent the average number of sick days used by nine employees
of a large industrial plant. The numbers of sick days are as follows:


39 31 18 34 25 22 32 23 22


8. Shown below is the number of new stage shows that appeared in Las Vegas for each of the past several
years. Construct a box-and-whisker plot for the data and comment of the shape of the distribution.


31 29 34 30 38 40 36 38 32 39 35


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9. The following data represent the average snowfall (in centimeters) for 18 Canadian cities for the
month of January. Construct a box-and-whisker plot to model the data. Is the data skewed? Justify
your answer.


Table 8.16:


Name of City Amount of Snow(cm)
Calgary 123.4
Charlottetown 74.5
Edmonton 80.6
Fredericton 73.8
Halifax 64.0
Labrador City 110.4
Moncton 82.4
Montreal 63.6
Ottawa 48.9
Quebec City 53.8
Regina 35.9
Saskatoon 25.4
St. John’s 97.5
Sydney 44.2
Toronto 21.8
Vancouver 12.8
Victoria 8.3
Winnipeg 76.2


10. Using the procedure outlined in this chapter, check the following data sets for outliers:
(a) 25, 33, 55, 32, 17, 19, 15, 18, 21
(b) 149, 123, 126, 122, 129, 120


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Chapter 9


Organizing and Displaying
Data for Comparison


Introduction
Throughout this book, you have learned about variables. You have learned about random variables, dis-
crete variables, continuous variables, numerical (or quantitative) variables, and categorical (or qualitative)
variables. The various forms of graphical representations you have learned about in the previous chapters
can be added to your learning of variables. The graphic below may help to summarize what you have
learned.


Broken-line graphs, histograms, pie charts, stem-and-leaf plots, and box-and-whisker plots all represent
useful (often very useful) tools in determining trends. Broken-line graphs, for example, allow you to show
situations such as the distance traveled in specific time spans. Histograms use continuous grouped data
to show the frequency trend in the data. Bar charts are a little different from histograms in that they use
grouped discrete data, as do stem-and-leaf plots. Bar graphs, as you know, have gaps between the columns,
while histograms do not. Stem-and-leaf plots are excellent for giving you a quick visual representation of
data. Used for only smaller sets of data, stem-and-leaf plots are a good example of representations of
grouped discrete data. Box-and-whisker plots are a final visual way of representing grouped data that
you have learned about in the previous chapters. In a box-and-whisker plot, you are able to find the five


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number summary to describe the spread of the data.


9.1 Review
Learning Objectives


• Organize and describe distributions of data by using a number of different methods, including fre-
quency tables, histograms, standard line and bar graphs, stem-and-leaf displays, scatter plots, and
box-and-whisker plots.


In previous chapters, you learned about discrete and continuous data and were introduced to categorical
and numerical forms of displaying data. In this final chapter, you will learn how to display discrete and
continuous data in both categorical and numerical displays, but in a way that allows you to compare sets
of data.


Remember that discrete data is represented by exact values that result from counting, as in the number
of people in the households in your neighborhood. Continuous data is represented by a range of data that
results from measuring. For example, taking the average temperatures for each month during a year is an
example of continuous data. Also remember from an earlier chapter how you distinguished between these
types of data when you graphed them.


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The graph above shows discrete data. Remember that you know this because the data points are not
joined. The graph below represents the average temperatures during the months in 2009. This data is
continuous. You can easily tell this by looking at the graph and seeing the data points connected together.


Two newer terms used are the categorical and numerical data forms. Categorical data forms are just what
the term suggests. These are data forms that are in categories and describe characteristics, or qualities, of
a category. These data forms are more qualitative data and, therefore, are less numerical than they are
descriptive. Graphs such as pie charts and bar charts show descriptive data, or qualitative data. Below
are two examples of categorical data represented in these types of graphs:


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Numerical data is quantitative data. Numerical data involves measuring or counting a numerical
value. Therefore, when you talk about discrete and continuous data, you are talking about numerical
data. Line graphs, frequency polygons, histograms, and stem-and-leaf plots all involve numerical data, or
quantitative data, as is shown below:


Box-and-whisker plots are also considered numerical displays of data, as they are based on quantitative
data (the mean and median), as well as the maximum (upper) and minimum (lower) values found in the
data. The figure below is a typical box-and-whisker plot:


You will spend the remainder of the chapter learning about how to compare sets of categorical and numerical
data.


9.2 Double Line Graphs
Remember a line graph, by definition, can be the result of a linear function or can be simply a graph of
plotted points, where the points are joined together by line segments. Line graphs that are linear functions
are normally in the form y = mx + b, where m is the slope and b is the y-intercept. The graph below is an
example of a linear equation with a slope of 23 and a y-intercept of −2.


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The second type of line graph is known as a broken-line graph. In a broken-line graph, the slope represents
the rate of change, and the y-intercept is actually the starting point. The graph below is a broken-line
graph.


When the measurements began, the number of sales (the y-intercept) was 645. The graph shows a significant
increase in the number of sales from weeks 5 through 10 and a significant reduction in the number of sales
from weeks 16 through 20.
In this lesson, you will be learning about comparing two line graphs that each contain data points. In
statistics, when line graphs are in the form of broken-line graphs, they are of more use. Linear functions
(i.e., y = mx + b) are more for algebraic reasoning. Double line graphs, as with any double graphs, are
often called parallel graphs, due to the fact that they allow for the quick comparison of two sets of data.
In this chapter, you will see them referred to only as double graphs.
Example 1
Christopher and Jack are each opening businesses in their neighborhoods for the summer. Christopher is
going to sell lemonade for 50




c per glass. Jack is going to sell popsicles for $1.00 each. The following graph
represents the sales for each boy for the 8 weeks in the summer.


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a. Explain the slopes of the line segments for Christopher’s graph.
b. Explain the slopes of the line segments for Jack’s graph.
c. Are there any negative slopes? What does this mean?
d. Where is the highest point on Christopher’s graph? What does this tell you?
e. Where is the highest point on Jack’s graph? What does this tell you?
f. Can you provide some reasons for the shape of Jack’s graph?
g. Can you provide some reasons for the shape of Christopher’s graph?
Solution
a. The slope of the line segments for Christopher’s graph (red) are positive for the first five weeks, meaning
he was increasing his sales each week. This is also true from weeks 6 to 7. From weeks 5 to 6 and weeks 7
to 8, the slopes were decreasing, meaning there was a decrease in sales.
b. The same trend that is seen for Christopher’s graph (red) is also seen for Jack’s graph (blue). The slope
of the line segments for Jack’s graph (blue) are positive for the first five weeks, meaning he was increasing
his sales each week. This is also true from weeks 6 to 7. From weeks 5 to 6 and weeks 7 to 8, the slopes
were decreasing, meaning there was a decrease in sales.
c. Negative sales from weeks 5 to 6 and weeks 7 to 8 (for both boys) mean there was a decrease in sales
during these two-week periods.
d. The highest point on Christopher’s graph occurred in week 7, when he sold 65 glasses of lemonade.
This must have been a very good week−nice and hot!
e. The highest point on Jack’s graph occurred in week 5, when he sold 74 popsicles. This must have been
a very hot week as well!
f. Popsicles are a great food when you are warm and want a light snack. You can see how as the summer
became hotter, the sales increased. Even in the weeks where it looks like Jack had a decrease in sales
(maybe a few rainy days occurred, or it was not as hot), his sales still remained at a good level.
g. Lemonade is a very refreshing drink when you are warm. You can see how as the summer became hotter,
the sales increased. Even in the weeks where it looks like Christopher had a decrease in sales (maybe a few
rainy days occurred, or it was not as hot), his sales still remained at a good level, just as Jack’s sales did.
Example 2


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Thomas and Abby are training for the cross country meet at their school. Both students are in the 100
yard dash. The coach asks them to race 500 yards and time each 100 yard interval. The following graph
represents the times for both Thomas and Abby for each of the five 100 yard intervals.


a. Who won the race? How do you know?
b. Between what times did Thomas (blue) appear to slow down? How do you know?
c. Between what times was Abby (red) ahead of Thomas? How do you know?
d. At what time did Thomas pass Abby? How do you know?
Solution
a. Thomas (blue) won the race, because he finished the 500 yards in the least amount of time.
b. Between 20 and 40 seconds, Thomas (blue) seems to slow down, because the slope of the graph is less
steep.
c. Between 0 and 57 seconds, Abby (pink) is ahead of Thomas (blue). You can see this, because the pink
line is above the blue line.
d. At 57 seconds, Thomas (blue) passes Abby (pink). From this point onward, the blue line is above the
pink line, meaning Thomas is running faster 100 yard intervals.
Example 3
Brenda and Ervin are each planting corn in a section of garden in their back yard. Brenda says that they
need to put fertilizer on the plants 3 to 5 times per week. Ervin contradicts Brenda, saying that they need
to fertilize only 1 to 2 times per week. Each gardener plants his or her garden of corn and measures the
heights of their plants. The graph for the growth of their corn is found below.


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a. Who was right? How do you know?
b. Between what times did Brenda’s (pink) garden appear to grow more? How do you know?
c. Between what times were Ervin’s (blue) heights ahead of Brenda’s? How do you know?
Solution
a. Brenda (pink) is correct, because her plants grew more in the same amount of time.
b. Between 4 and 8 weeks, Brenda’s plants seemed to grow faster (taller) than Ervin’s plants. You can tell
this, because the pink line is above the blue line after the 4-week mark.
c. From 0 and 4 weeks, Ervin’s plants seemed to grow faster (taller) than Brenda’s plants. You can tell
this, because the blue line is above the pink line before the 4-week mark.
Example 4
Nicholas and Jordan went on holidays with their families. They decided to monitor the mileage they
traveled by keeping track of the time and the distance they were on the road. The boys collected the
following data.


Nicholas


Time (hr) 1 2 3 4 5 6
Distance (miles) 60 110 175 235 280 320


Jordan


Time (hr) 1 2 3 4 5 6
Distance (miles) 50 90 125 125 165 210


a. Draw a graph to show the trip for each boy.
b. What conclusions could you draw by looking at the graphs?
Solution


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a.


b. Looking at the speed of Nicholas’s family vehicle and the shape of the graph, it could be concluded that
Nicholas’s family was traveling on the highway going toward their family vacation destination. The family
did not stop and continued on at a pretty steady speed until they reached where they were going.
Jordan’s trip was more relaxed. The speed indicates they were probably not on a highway, but more on
country-type roads, and that they were traveling through a scenic route. In fact, from hours 3 to 4, the
family stopped for some reason (maybe lunch), and then they continued on their way.
You can also use TI technology to graph this data. First, you need to enter in all of the data Nicholas and
Jordan collected.


Now you need to graph the two sets of data.


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The resulting graph looks like the following.


9.3 Two-sided Stem-and-Leaf Plots
As you have learned in an earlier chapter, stem-and-leaf plots are an excellent tool for organizing data.
Remember that stem-and-leaf plots are a visual representation of grouped discrete data, but they can
also be referred to as a modal representation. This is because by looking at a stem-and-leaf plot, we
can determine the mode by quick visual inspection. In the last chapter, you learned about single-sided
stem-and-leaf plots. In this lesson, you will learn about two-sided stem-and-leaf plots, which are also
often called back-to-back stem-and-leaf plots.
Example 5
The girls and boys in one of BDF High School’s AP English classes are having a contest. They want to
see which group can read the most number of books. Mrs. Stubbard, their English teacher, says that the
class will tally the number of books each group has read, and the highest mode will be the winner. The
following data was collected for the first semester of AP English:


Girls 11 12 12 17 18 23 23 23 24 33 34 35 44 45 47 50 51 51
Boys 15 18 22 22 23 26 34 35 35 35 40 40 42 47 49 50 50 51


a. Draw a two-sided stem-and-leaf plot for the data.
b. Determine the mode for each group.
c. Help Mrs. Stubbard decide which group won the contest.
Solution


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a.


b. The mode for the girls is 23 books. It is the number in the girls column that appears most often. The
mode for the boys is 35 books. It is the number in the boys column that appears most often.
c. Mrs. Stubbard should decide that the boys group has won the contest.
Example 6
Mrs. Cameron teaches AP Statistics at GHI High School. She recently wrote down the class marks for
her current grade 12 class and compared it to the previous grade 12 class. The data can be found below.
Construct a two-sided stem-and-leaf plot for the data and compare the distributions.


2010 class 70 70 70 71 72 74 74 74 74 75 76 76 77 78 79 80 81
82 82 82 83 84 85 85 86 87 93 98 100


2009 class 76 76 76 76 77 78 78 78 79 80 80 82 82 83 83 83 85
85 88 91 95


Solution


There is a wide variation in the marks for both years in Mrs. Cameron’s AP Statistics Class. In 2009, her
class had marks anywhere from 76 to 95. In 2010, the class marks ranged from 70 to 100. The mode for
the 2009 class was 76, but for the 2010 class, it was 74. It would seem that the 2009 class had, indeed,
done slightly better than Mrs. Cameron’s current class.
Example 7
The following data was collected in a survey done by Connor and Scott for their statistics project. The
data represents the ages of people who entered into a new hardware store within its first half hour of
opening on its opening weekend. The M’s in the data represent males, and the F’s represent females.


12M 18F 15F 15M 10M 21F 25M 21M
26F 29F 29F 31M 33M 35M 35M 35M
41F 42F 42M 45M 46F 48F 51M 51M
55F 56M 58M 59M 60M 60F 61F 65M
65M 66M 70M 70M 71M 71M 72M 72F


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Construct a back-to-back stem-and-leaf plot showing the ages of male customers and the ages of female
customers. Compare the distributions.
Solution


For the male customers, the ages ranged from 10 to 72. The ages for the male customers were spread out
throughout this range, with the mode being age 35. In other words, for the males found to be at the store
in the first half hour of opening day, there was no real age category where a concentration of males could
be found.
For the female customers, the ages ranged from 15 to 72, but they were concentrated between 21 and 48.
The mode for the ages of the female customers was 29 years of age.


9.4 Double Bar Graphs
In Chapter 7, you studied both histograms and bar graphs. Remember that histograms have measurements
on the horizontal axis (x) and frequencies on the vertical axis (y). A bar chart, on the other hand, displays
categories on the horizontal (x) axis and frequencies on the vertical (y) axis. This means that bar charts
are more qualitative, and, therefore, display categorical data. The figure below shows one bar chart (on
the top) and one histogram (on the bottom).


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Let’s look at an example of double bar graphs.
Example 8
Kerry-Sue is surveying a random sample of students to determine which sports they would like to have set
up at the end-of-year Safe Grad event. She collects the following data.


Table 9.1:


Sports Girls Boys
Racquetball 6 3
Basketball 3 6
Volleyball 5 5
Swimming 7 8


Draw a double bar graph and help Kerry-Sue determine which two sports would be most equally-liked by
both boys and girls at the end-of-year Safe Grad event.
Solution


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According to the double bar graph, volleyball and swimming seem to be almost equally liked by both the
girls and the boys. Therefore, these two sports would be the ones Kerry-Sue should choose to set up at
the end-of-year Safe Grad event.
Example 9
Mrs. Smith teaches both academic and advanced math. She has been teaching these two courses for the
past 4 years. She decided she wanted to compare her grades to see how each class was doing over the past
few years and see if she has improved her class instruction at all. Her data can be found below.


Table 9.2:


Marks Academic Math Advanced Math
2007 61.3 74.7
2008 67.9 80.3
2009 50.9 86.8
2010 63.7 81.5


Draw a double bar graph and help Mrs. Smith determine if her class instruction has improved over the
past 4 years.
Solution


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Both the academic and advanced math marks went up and down over the past four years. If Mrs. Smith
looks at the difference between 2007 and 2010, she can see that there is an improvement in the final grades
for her students. Although there are many factors that can affect these grades, she can say that the change
in her instruction is making some difference in the results for her students. Other factors might have
contributed to the huge decline in grades for the academic math students in 2009. You can make this
conclusion for Mrs. Smith, as there was a marked improvement in her advanced math course. In 2009,
it seemed her instruction methods were working well with the advanced students, but other factors were
affecting the academic students.


9.5 Double Box-and-Whisker Plots
Double box-and-whisker plots give you a quick visual comparison of two sets of data, as was also
found with other double graph forms you learned about earlier in this chapter. The difference with double
box-and-whisker plots is that you are also able to quickly visually compare the means, the medians, the
maximums (upper range), and the minimums (lower range) of the data.
Example 10
Emma and Daniel are surveying the times it takes students to arrive at school from home. There are two
main groups of commuters who were in the survey. There were those who drove their own cars to school,
and there were those who took the school bus. They collected the following data:


Bus times (min) 14 18 16 22 25 12 32 16 15 18
Car times (min) 12 10 13 14 9 17 11 10 8 11


Draw a box-and-whisker plot for both sets of data on the same number line. Use the double box-and-whisker
plots to compare the times it takes for students to arrive at school either by car or by bus.
Solution
The box-and-whisker plots are plotted and look like the following:


Using the medians, 50% of the cars arrive at school in 11 minutes or less, whereas 50% of the students
arrive by bus in 17 minutes or less. The range for the car times is 17 − 8 = 9 minutes. For the bus times,
the range is 32 − 12 = 20 minutes. Since the range for the driving times is smaller, it means the times
to arrive by car are less spread out. This would, therefore, mean that the times are more predictable and
reliable.
Example 11
A new drug study was conducted by a drug company in Medical Town. In the study, 15 people were chosen
at random to take Vitamin X for 2 months and then have their cholesterol levels checked. In addition,
15 different people were randomly chosen to take Vitamin Y for 2 months and then have their cholesterol
levels checked. All 30 people had cholesterol levels between 8 and 10 before taking one of the vitamins.
The drug company wanted to see which of the two vitamins had the greatest impact on lowering people’s
cholesterol. The following data was collected:


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Vitamin X 7.2 7.5 5.2 6.5 7.7 10 6.4 7.6 7.7 7.8 8.1 8.3 7.2 7.1 6.5
Vitamin Y 4.8 4.4 4.5 5.1 6.5 8 3.1 4.6 5.2 6.1 5.5 4.2 4.5 5.9 5.2


Draw a box-and-whisker plot for both sets of data on the same number line. Use the double box-and-whisker
plots to compare the two vitamins and provide a conclusion for the drug company.
Solution
The box-and-whisker plots are plotted and look like the following:


Using the medians, 50% of the people in the study had cholesterol levels of 7.5 or lower after being on
Vitamin X for 2 months. Also 50% of the people in the study had cholesterol levels of 5.1 or lower after
being on Vitamin Y for 2 months. Knowing that the participants of the survey had cholesterol levels
between 8 and 10 before beginning the study, it appears that Vitamin Y had a bigger impact on lowering
the cholesterol levels. The range for the cholesterol levels for people taking Vitamin X was 10 − 5.2 = 4.8
points. The range for the cholesterol levels for people taking Vitamin Y was 8−3.1 = 4.9 points. Therefore,
the range is not useful in making any conclusions.
Drawing Double Box-and-Whisker Plots Using TI Technology
The above double box-and-whisker plots were drawn using a program called Autograph. You can also draw
double box-and-whisker plots by hand using pencil and paper or by using your TI-84 calculator. Follow
the key sequence below to draw double box-and-whisker plots.


After entering in the data into L1 and L2, the next step is to graph the data by using the STAT PLOT
menu.


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The resulting graph looks like the following:


You can then press TRACE and find the five-number summary. The five-number summary is shown
below for Vitamin X. By pressing the H button, you can get to the second box-and-whisker plot (for
Vitamin Y) and collect the five-number summary for this box-and-whisker plot.


Example 12
Two campus bookstores are having a price war on the prices of their first-year math books. James, a
first-year math major, is going into each store to try to find the cheapest books he can find. He looks
at five randomly chosen first-year books for first-year math courses in each store to determine where he
should buy the five textbooks he needs for his courses this coming year. He collects the following data.


Bookstore A prices($) 95 75 110 100 80
Bookstore B prices($) 120 60 89 84 100


Draw a box-and-whisker plot for both sets of data on the same number line. Use the double box-and-
whisker plots to compare the two bookstores’ prices, and provide a conclusion for James as to where to
buy his books for his first-year math courses.
Solution
The box-and-whisker plots are plotted and look like the following:


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Using the medians, 50% of the books at Bookstore A are likely to be in the price range of $95 or less,
whereas at Bookstore B, 50% of the books are likely to be around $89 or less. At first glance, you would
probably recommend to James that he go to Bookstore B. Let’s look at the range to see the spread of data.
For Bookstore A, the range is $110− $75 = $35. For Bookstore B, the range is $120− $60 = $60. With the
spread of the data much greater at Bookstore B than at Bookstore A, (i.e., the range for Bookstore B is
greater than that for Bookstore A), to say that it would be cheaper to buy James’s books at Bookstore A
would be more predictable and reliable. You would, therefore, suggest to James that he is probably better
off going to Bookstore A.
Points to Consider


• What is the difference between categorical and numerical data, and how does this relate to qualitative
and quantitative data?


• How is comparing double graphs (pie charts, broken-line graphs, box-and-whisker plots, etc.) useful
when doing statistics?


Vocabulary


Categorical data Data that are in categories and describe characteristics, or qualities, of a category.


Double bar graphs Two bar graphs that are graphed side-by-side.


Double box-and-whisker plots Two box-and-whisker plots that are plotted on the same number line.


Double line graphs Two line graphs that are graphed on the same coordinate grid. Double line graphs
are often called parallel graphs.


Quantitative data Numerical data, or data that is in the form of numbers.


Qualitative data Descriptive data, or data that describes categories.


Numerical data Data that involves measuring or counting a numerical value.


Two-sided stem-and-leaf plots Two stem-and-leaf plots that are plotted side-by-side. Two-sided stem-
and-leaf plots are also called back-to-back stem-and-leaf plots.


9.6 Review Questions
Answer the following questions and show all work (including diagrams) to create a complete answer.


1. In the table below, match the following types of graphs with the types of variables used to create the
graphs.


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Table 9.3:


Type of graph Type of Variable
a. Histogram _____ discrete
b. Stem-and-leaf plot _____ discrete
c. Broken-line graph _____ discrete
d. Bar chart _____ continuous
e. Pie chart _____ continuous


2. In the table below, match the following types of graphs with the types of variables used to create the
graphs.


Table 9.4:


Type of graph Type of Variable
a. Broken-line graph _____ qualitative
b. Bar chart _____ numerical
c. Pie chart _____ categorical
d. Stem-and-leaf plot _____ quantitative
e. Histogram _____ numerical


3. Jack takes a pot of water at room temperature (22◦C) and puts it on the stove to boil (100◦C), which
takes about 5 minutes. He then takes a cup of this water, adds a package of hot chocolate, and mixes
it up. He places the cup on the counter to cool for 10 minutes to 40◦C before having his first sip.
After 30 minutes, the hot chocolate is now at room temperature.
Thomas is making chocolate chip cookies. He mixes all of the ingredients together at room tem-
perature, which takes him about 5 minutes, and then places the cookies in the oven at 350◦C for
8 minutes. After cooking, he takes them off the pan and places them on a cooling rack. After 15
minutes, the cookies are still warm (about 30◦C), but he samples them for taste. After 30 minutes,
the cookies are at room temperature and ready to be served.
Draw a broken-line graph for both sets of data. Label the graph to show what is happening.


4. Scott is asked to track his daily video game playing. He gets up at 7 A.M. and plays for 1 hour.
He then eats his breakfast and gets ready for school. He runs to catch the bus at 8:25 A.M. On the
bus ride (about 35 minutes), he plays his IPOD until arriving for school. He is not allowed games at
school, so he waits for the bus ride home at 3:25 P.M. When he gets home, he does homework for 1
hour and plays games for 1 hour until dinner. There are no games in the evening.
Michael gets up at 7:15 A.M., eats breakfast, and gets ready for school. It takes him 30 minutes to
get ready. He then plays games until he goes to meet the bus with Scott. Michael is in Scott’s class,
but he has a free period from 11:00 A.M. until 11:45 A.M., when he goes outside to play a game.
He goes home and plays his 1 hour of games immediately, and he then works on his homework until
dinner. He, like Scott, is not allowed to play games in the evening.
Draw a broken-line graph for both sets of data. Label the graph to show what is happening.


5. The following graph shows the gasoline remaining in a car during a family trip east. Also found
on the graph is the gasoline remaining in a truck traveling west to deliver goods. Describe what is
happening for each graph. What other conclusions may you draw?


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6. Mr. Dugas, the senior high physical education teacher, is doing fitness testing this week in gym class.
After each test, students are required to take their pulse rate and record it on the chart in the front
of the gym. At the end of the week, Mr. Dugas looks at the data in order to analyze it. The data is
shown below.


Girls 70 88 80 76 76 77 89 72 72 76 72 75 77 80 76 68 68
82 78 60 64 64 65 81 84 84 79 78 70


Boys 76 88 87 86 85 70 76 70 70 79 80 82 82 82 83 84 85
85 78 81 85


Construct a two-sided stem-and-leaf plot for the data and compare the distributions.
7. Starbucks prides itself on its low line-up times in order to be served. A new coffee house in town
has also boasted that it will have your order in your hands and have you on your way quicker than
the competition. The following data was collected for the line-up times (in minutes) for both coffee
houses.


Starbucks 20 26 26 27 19 12 12 16 12 15 17 20 8 8 18
Just Us Coffee 17 16 15 10 16 10 10 29 20 22 22 12 13 24 15


Construct a two-sided stem-and-leaf plot for the data. Determine the median and mode using the
two-sided stem-and-leaf plot. What can you conclude from the distributions?


8. The boys and girls basketball teams had their heights measured at practice. The following data was
recorded for their heights (in inches).


Girls 171 170 176 176 177 179 162 172 160 157 155
168 178 174 170 155 155 154 164 145 171 161


Boys 168 170 162 153 176 167 158 180 181 176 172
168 167 165 159 185 184 173 177 167 169 177


Construct a two-sided stem-and-leaf plot for the data. Determine the median and mode using the
two-sided stem-and-leaf plot. What can you conclude from the distributions?


9. The grade 12 biology class did a survey to see what color eyes their classmates had and if there was
a connection between eye color and sex. The following data was recorded.


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Table 9.5:


Eye color Males Females
blue 5 5
green 6 8
brown 3 4
hazel 4 3


Draw a double bar chart to represent the data, and draw any conclusions that you can from the resulting
chart.


10. Robbie is in charge of the student organization for new food selections in the cafeteria. He designed
a survey to determine if four new food options would be good to put on the menu. The results are
shown.


Table 9.6:


Food Option Yes votes No votes
Fish burgers 10 5
Vegetarian pizza 7 18
Brown rice 23 9
Carrot soup 20 20


Draw a double bar chart to represent the data, and draw any conclusions that you can from the resulting
chart.


11. The guidance counselor at USA High School wanted to know what future plans the graduating class
had. She took a survey to determine the intended plans for both boys and girls in the school’s
graduating class. The following data was recorded.


Table 9.7:


Future Plans Boys Girls
University 35 40
College 27 22
Military 23 9
Employment 10 5
Other/unsure 5 10


Draw a double bar chart to represent the data, and draw any conclusions that you can from the resulting
chart.


12. International Baccalaureate has two levels of courses, which are standard level (SL) and higher level
(HL). Students say that study times are the same for both the standard level exams and the higher
level exams. The following data represents the results of a survey conducted to determine how many


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hours a random sample of students studied for their final exams at each level.
HL Exams 15 16 16 17 19 10 5 6 5 5 8 10 8 12 17
SL Exams 10 6 6 7 9 12 2 6 2 5 7 20 18 8 18


Draw a box-and-whisker plot for both sets of data on the same number line. Use the double box-
and-whisker plots to determine the 5-number summary for both sets of data. Compare the times
students prepare for each level of exam.


13. Students in the AP math class at BCU High School took their SATs for university entrance. The
following scores were obtained for the math and verbal sections.


Math 529 533 544 562 513 519 560 575 568 537 561 522 563 571
Verbal 499 509 524 530 550 499 545 560 579 524 478 487 482 570


Draw a box-and-whisker plot for both sets of data on the same number line. Use the double box-
and-whisker plots to determine the 5-number summary for both sets of data. Compare the data for
the two sections of the SAT using the 5-number summary data.


14. The following box-and-whisker plots were drawn to analyze the data collected in a survey of scores
for the doubles performances in the figure skating competitions at two Winter Olympic games. The
box-and-whisker plot on the top represents the scores obtained at the 2010 winter games in Whistler,
BC. The box-and-whisker plot on the bottom represents the scores obtained at the 2006 winter games
in Torino, Italy.


15. Use the double box-and-whisker plots to determine the 5-number summary for both sets of data.
Compare the scores obtained at each of the Winter Olympic games.


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