potential customers
11-1 Designing a Study
Determine whether each situation describes a
survey, an experiment, or an observational study.
Then identify the sample, and suggest a
population from which it may have been
selected.
1. SCHOOL A group of high school students is
randomly selected and asked to complete the form
shown.
ANSWER: observational study; sample: the participants in the
study; population: potential customers
CCSS ARGUMENTS Determine whether each situation calls for a survey, an experiment, or an
observational study. Explain your reasoning.
3. LITERACY A literacy group wants to determine
whether high school students that participated in a
recent national reading program had higher
standardized test scores than high school students
that did not participate in the program.
SOLUTION: Observation study; sample answer: The scores of the
participants are observed and compared without
them being affected by the study. ANSWER: Observation study; sample answer: The scores of the
participants are observed and compared without
them being affected by the study. SOLUTION: This is a survey because data are collected from
responses to the question.
sample: the students in the study; population: the student body
ANSWER: survey; sample: the students in the study; population:
the student body
2. DESIGN An advertising company wants to test a
new logo design. They randomly select 20
participants and watch them discuss the logo.
SOLUTION: This is an observational study because the company
is going to watch the reactions of the participants and
the participants are unaffected by the study.
sample: the participants in the study; population:
potential customers
ANSWER: observational study; sample: the participants in the
study; population: potential customers
CCSS ARGUMENTS Determine whether each situation calls for a survey, an experiment, or an
observational study. Explain your reasoning.
3. LITERACY A literacy group wants to determine
whether high school students that participated in a
recent national reading program had higher
standardized test scores than high school students
that did not participate in the program.
SOLUTION: eSolutions
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Observation
study; by
sample
answer:
The scores of the
participants are observed and compared without
them being affected by the study. 4. RETAIL The research department of a retail
company plans to conduct a study to determine
whether a dye used on a new T-shirt will begin
fading before 50 washes.
SOLUTION: Experiment; sample answer: A sample of dyed shirts
will need to be tested, which means that the
members of the sample will be affected by the study.
ANSWER: Experiment; sample answer: A sample of dyed shirts
will need to be tested, which means that the
members of the sample will be affected by the study.
Determine whether each survey question is
biased or unbiased. If biased, explain your
reasoning.
5. Which student council candidate’s platform do you
support?
SOLUTION: The question is not confusing, does not cause a
strong reaction, does not cause a certain response,
and addresses only one issue. So, it is unbiased.
ANSWER: unbiased
6. How long have you lived at your current address?
SOLUTION: The question is not confusing, does not cause a
strong reaction, does not cause a certain response,
and addresses only one issue. So, it is unbiased. Page 1
ANSWER: strong reaction, does not cause a certain response,
and addresses only one issue. So, it is unbiased.
ANSWER: 11-1unbiased
Designing a Study
6. How long have you lived at your current address?
SOLUTION: The question is not confusing, does not cause a
strong reaction, does not cause a certain response,
and addresses only one issue. So, it is unbiased.
ANSWER: unbiased
7. HYBRIDS A car manufacturer wants to determine
what the demand in the U.S. is for hybrid vehicles.
State the objective of the survey, suggest a
population, and write two unbiased survey questions.
SOLUTION: objective: to determine how many people in the U.S.
are interested in purchasing a hybrid vehicle
population: the people surveyed
sample survey questions:
Do you currently own a hybrid vehicle?
Are you planning on purchasing a hybrid vehicle?
The survey questions are unbiased because they are
not confusing, do not cause a strong reaction, do not
cause a certain response, and address only one issue.
ANSWER: objective: to determine how many people in the U.S.
are interested in purchasing a hybrid vehicle;
population: the people surveyed; sample survey
questions: Do you currently own a hybrid vehicle?
Are you planning on purchasing a hybrid vehicle?
8. Identify any flaws in the experiment design, and
describe how they could be corrected.
Experiment: A research company wants to determine
whether a new vitamin boosts energy levels and
decides to test the vitamin at a college campus. A
random sample is taken. The experimental group
consists of students who are given the vitamin, and
the control group consists of instructors who are
given a placebo.
Results: When given a physical test, the experimental
group outperformed the control group.
The company concludes that the vitamin is effective.
SOLUTION: Sample answer: The flaw is that the experimental
group consists of students, and the control group
consists of instructors. On average, college students
are younger than their instructors, and thus, are more
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likely
to score
higher
a physical test whether given
a vitamin or not.
objective: to determine how many people in the U.S.
are interested in purchasing a hybrid vehicle;
population: the people surveyed; sample survey
questions: Do you currently own a hybrid vehicle?
Are you planning on purchasing a hybrid vehicle?
8. Identify any flaws in the experiment design, and
describe how they could be corrected.
Experiment: A research company wants to determine
whether a new vitamin boosts energy levels and
decides to test the vitamin at a college campus. A
random sample is taken. The experimental group
consists of students who are given the vitamin, and
the control group consists of instructors who are
given a placebo.
Results: When given a physical test, the experimental
group outperformed the control group.
The company concludes that the vitamin is effective.
SOLUTION: Sample answer: The flaw is that the experimental
group consists of students, and the control group
consists of instructors. On average, college students
are younger than their instructors, and thus, are more
likely to score higher on a physical test whether given
a vitamin or not.
ANSWER: Sample answer: The flaw is that the experimental
group consists of students, and the control group
consists of instructors. On average, college students
are younger than their instructors, and thus, are more
likely to score higher on a physical test whether given
a vitamin or not.
9. SPORTS A research company wants to conduct an
experiment to test the claim of the protein shake
shown. State the objective of the experiment, suggest
a population, determine the experimental and control
groups, and describe a sample procedure.
SOLUTION: objective: to determine whether the protein shake
helps athletes recover from exercise; population: all
athletes; experiment group: athletes given the protein
shake; control group: athletes given a placebo;
sample procedure: The researchers could randomly
divide the athletes into two groups: an experimental
group given the protein shake and a control group
given the placebo. Next, they could have the athletes
exercise and then drink the protein shake or placebo.
Later, the researchers could interview the athletes to
see how they feel.
Page 2
ANSWER: objective: to determine whether the protein shake
group consists of students, and the control group
consists of instructors. On average, college students
are younger than their instructors, and thus, are more
11-1likely
Designing
a Study
to score
higher on a physical test whether given
a vitamin or not.
group given the protein shake and a control group
given the placebo. Next, they could have the athletes
exercise and then drink the protein shake or placebo.
Later, the researchers could interview the athletes to
see how they feel.
9. SPORTS A research company wants to conduct an
experiment to test the claim of the protein shake
shown. State the objective of the experiment, suggest
a population, determine the experimental and control
groups, and describe a sample procedure.
Determine whether each situation describes a
survey, an experiment, or an observational study.
Then identify the sample, and suggest a
population from which it may have been
selected.
10. FOOD A grocery store conducts an online study in
which customers are randomly selected and asked to
provide feedback on their shopping experience.
SOLUTION: This is a survey because data are collected from the
feedback on their shopping experiences.
SOLUTION: objective: to determine whether the protein shake
helps athletes recover from exercise; population: all
athletes; experiment group: athletes given the protein
shake; control group: athletes given a placebo;
sample procedure: The researchers could randomly
divide the athletes into two groups: an experimental
group given the protein shake and a control group
given the placebo. Next, they could have the athletes
exercise and then drink the protein shake or placebo.
Later, the researchers could interview the athletes to
see how they feel.
ANSWER: objective: to determine whether the protein shake
helps athletes recover from exercise; population: all
athletes; experiment group: athletes given the protein
shake; control group: athletes given a placebo;
sample procedure: The researchers could randomly
divide the athletes into two groups: an experimental
group given the protein shake and a control group
given the placebo. Next, they could have the athletes
exercise and then drink the protein shake or placebo.
Later, the researchers could interview the athletes to
see how they feel.
Determine whether each situation describes a
survey, an experiment, or an observational study.
Then identify the sample, and suggest a
population from which it may have been
selected.
10. FOOD A grocery store conducts an online study in
which customers are randomly selected and asked to
provide feedback on their shopping experience.
sample: customers that take the online survey;
population: all customers
ANSWER: survey; sample: customers that take the online
survey; population: all customers
11. GRADES A research group randomly selects 80
college students, half of whom took a physics course
in high school, and compares their grades in a college
physics course.
SOLUTION: This is an observational study because the study
group is going to observe the students’ performance
without directly affecting the students. The sample is
the 80 physics students because they are the ones
being observed. Both halves of the selected students
are included because one half is going to be
compared to the other half. The population is all
college students that take a physics course.
ANSWER: observational study; sample: physics students
selected; population: all college students that take a
physics course
12. HEALTH A research group randomly chooses 100
people to participate in a study to determine whether
eating blueberries reduces the risk of heart disease
for adults.
SOLUTION: This is a survey because data are collected from the
feedback on their shopping experiences.
SOLUTION: This is an experiment because the sample will be
divided into groups in which one group will be given
blueberries while the other group will be given a
placebo. The participants are affected by the study.
sample: customers that take the online survey;
population: all customers
sample: adults participating in the study; population:
all adults
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ANSWER: survey; sample: customers that take the online
Page 3
ANSWER: experiment; sample: adults participating in the study;
ANSWER: observational study; sample: physics students
11-1selected;
Designing
a Study all college students that take a
population:
physics course
12. HEALTH A research group randomly chooses 100
people to participate in a study to determine whether
eating blueberries reduces the risk of heart disease
for adults.
SOLUTION: This is an experiment because the sample will be
divided into groups in which one group will be given
blueberries while the other group will be given a
placebo. The participants are affected by the study.
ANSWER: Survey; sample answer: The data will be obtained
from opinions given by members of the sample
population.
15. TRAVEL A travel agency randomly calls 250 U.S.
citizens and asks them what their favorite vacation
destination is.
SOLUTION: Survey; sample answer: The data will be obtained
from opinions given by members of the sample
population.
sample: adults participating in the study; population:
all adults
ANSWER: Survey; sample answer: The data will be obtained
from opinions given by members of the sample
population.
ANSWER: experiment; sample: adults participating in the study;
population: all adults
16. FOOD Chee wants to examine the eating habits of
100 random students at lunch to determine how many
students eat in the cafeteria.
13. TELEVISION A television network mails a
questionnaire to randomly selected people across the
country to determine whether they prefer watching
sitcoms or dramas.
SOLUTION: Observational study; sample answer: The eating
habits of the participants will be observed and
compared without them being affected by the study.
SOLUTION: This is a survey because data are collected from the
responses to the questionnaire.
ANSWER: Observational study; sample answer: The eating
habits of the participants will be observed and
compared without them being affected by the study.
sample: people that receive the questionnaire;
population: all viewers
ANSWER: survey; sample: people that receive the questionnaire;
population: all viewers
Determine whether each situation calls for a
survey, an experiment, or an observational study.
Explain your reasoning.
14. FASHION A fashion magazine plans to poll 100
people in the U.S. to determine whether they would
be more likely to buy a subscription if given a free
issue.
SOLUTION: Survey; sample answer: The data will be obtained
from opinions given by members of the sample
population.
ANSWER: Survey; sample answer: The data will be obtained
from opinions given by members of the sample
population.
15. TRAVEL A travel agency randomly calls 250 U.S.
citizens
and- Powered
asks them
what their favorite vacation
eSolutions
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destination is.
SOLUTION: 17. ENGINEERING An engineer is planning to test 50
metal samples to determine whether a new titanium
alloy has a higher strength than a different alloy.
SOLUTION: Experiment; sample answer: Metal samples will need
to be tested, which means that the members of the
sample will be affected by the study.
ANSWER: Experiment; sample answer: Metal samples will need
to be tested, which means that the members of the
sample will be affected by the study.
Determine whether each survey question is
biased or unbiased. If biased, explain your
reasoning.
18. Do you think that the school needs a new gym and
football field?
SOLUTION: Biased; sample answer: The question is asking about
two issues: whether the school needs a new gym and
whether the school needs a new football field.
ANSWER: Biased; sample answer: The question is asking about
Page 4
two issues: whether the school needs a new gym and
whether the school needs a new football field.
ANSWER: Experiment; sample answer: Metal samples will need
11-1toDesigning
Studymeans that the members of the
be tested, awhich
sample will be affected by the study.
Determine whether each survey question is
biased or unbiased. If biased, explain your
reasoning.
18. Do you think that the school needs a new gym and
football field?
SOLUTION: Biased; sample answer: The question is asking about
two issues: whether the school needs a new gym and
whether the school needs a new football field.
ANSWER: Biased; sample answer: The question is asking about
two issues: whether the school needs a new gym and
whether the school needs a new football field.
19. Which is your favorite football team, the Dallas
Cowboys or the Pittsburgh Steelers?
SOLUTION: Biased; sample answer: The question only gives two
options, and thus encourages a certain response.
ANSWER: Biased; sample answer: The question only gives two
options, and thus encourages a certain response.
20. Do you play any extracurricular sports?
SOLUTION: The question is not confusing, does not cause a
strong reaction, does not cause a certain response,
and addresses only one issue.
ANSWER: unbiased
21. Don’t you agree that students should carpool to
school?
SOLUTION: Biased; sample answer: The question encourages a
certain response. The phrase “don’t you agree” suggests that the people surveyed should agree.
ANSWER: Biased; sample answer: The question encourages a
certain response. The phrase “don’t you agree” suggests that the people surveyed should agree.
22. COLLEGE A school district wants to conduct a
survey to determine the number of juniors in the
district who are planning to attend college after high
school. State the objective of the survey, suggest a
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population,
write
unbiased survey questions.
SOLUTION: objective: to determine the number of juniors in the
ANSWER: Biased; sample answer: The question encourages a
certain response. The phrase “don’t you agree” suggests that the people surveyed should agree.
22. COLLEGE A school district wants to conduct a
survey to determine the number of juniors in the
district who are planning to attend college after high
school. State the objective of the survey, suggest a
population, and write two unbiased survey questions.
SOLUTION: objective: to determine the number of juniors in the
district planning to attend college after high school
population: all juniors in the district
sample survey questions:
What grade are you in?
Do you plan on attending college after graduation?
The survey questions are unbiased because they are
not confusing, do not cause a strong reaction, do not
cause a certain response, and address only one issue.
ANSWER: objective: to determine the number of juniors in the
district planning to attend college after high school;
population: all juniors in the district; sample survey
questions: What grade are you in? Do you plan on
attending college after graduation?
23. Identify any flaws in the experiment design, and
describe how they could be corrected.
Experiment: A supermarket chain wants to determine
whether shoppers are more likely to buy sunscreen if
it is located near the checkout line. The experimental
group consists of a group of stores in the midwest in
which the sunscreen was moved next to the
checkout line, and the control group consists of stores
in Arizona in which the sunscreen was not moved.
Results: The Arizona stores sold more sunscreen
than the midwest stores. The company concluded
that moving the sunscreen closer to the checkout line
did not increase sales.
SOLUTION: Sample answer: The flaw is that the experimental
group consists of stores in the Midwest, and the
control group consists of stores in Arizona. On
average, the temperature is higher in Arizona than in
the Midwest, and people use more sunscreen.
Therefore, the sunscreen sales in stores located in
those regions would likely be different and should not
be compared in an experiment. ANSWER: Sample answer: The flaw is that the experimental
group consists of stores in the Midwest, and the
control group consists of stores in Arizona. On Page 5
average, the temperature is higher in Arizona than in
the Midwest, and people use more sunscreen.
objective: to determine the number of juniors in the
district planning to attend college after high school;
population: all juniors in the district; sample survey
11-1questions:
DesigningWhat
a Study
grade are you in? Do you plan on
attending college after graduation?
23. Identify any flaws in the experiment design, and
describe how they could be corrected.
Experiment: A supermarket chain wants to determine
whether shoppers are more likely to buy sunscreen if
it is located near the checkout line. The experimental
group consists of a group of stores in the midwest in
which the sunscreen was moved next to the
checkout line, and the control group consists of stores
in Arizona in which the sunscreen was not moved.
Results: The Arizona stores sold more sunscreen
than the midwest stores. The company concluded
that moving the sunscreen closer to the checkout line
did not increase sales.
SOLUTION: Sample answer: The flaw is that the experimental
group consists of stores in the Midwest, and the
control group consists of stores in Arizona. On
average, the temperature is higher in Arizona than in
the Midwest, and people use more sunscreen.
Therefore, the sunscreen sales in stores located in
those regions would likely be different and should not
be compared in an experiment. ANSWER: Sample answer: The flaw is that the experimental
group consists of stores in the Midwest, and the
control group consists of stores in Arizona. On
average, the temperature is higher in Arizona than in
the Midwest, and people use more sunscreen.
Therefore, the sunscreen sales in stores located in
those regions would likely be different and should not
be compared in an experiment. 24. CCSS ARGUMENTS In chemistry class, Pedro
learned that copper objects become dull over time
because the copper reacts with air to form a layer of
copper oxide. He plans to use the supplies shown
below to determine whether a mixture of lemon juice
and salt will remove copper oxide from pennies.
a. State the objective of the experiment, suggest a
population, determine the experimental and control
groups, and describe a sample procedure.
b. What factors do you think should be considered
whenManual
selecting
pennies
for the experiment? Explain
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your reasoning. SOLUTION: a. State the objective of the experiment, suggest a
population, determine the experimental and control
groups, and describe a sample procedure.
b. What factors do you think should be considered
when selecting pennies for the experiment? Explain
your reasoning. SOLUTION: a. objective: to determine whether a mixture of salt
and lemon juice will remove copper oxide from
copper objects; population: all copper objects;
experimental group: pennies that are submerged in
the mixture; control group: pennies that are
submerged in a placebo mixture; sample procedure:
Pedro could randomly assign the pennies into two
groups, and create the lemon and salt mixture. Next,
he could submerge the experimental group of pennies
in the mixture for a certain period of time, remove
them, and then visually compare the two groups of
pennies.
b. Sample answer: The pennies should be chosen so
that they have roughly the same amount of copper
oxide and are from the same time period. Using
these guidelines could eliminate bias regarding the
initial condition and composition, so that the
conditions of the experimental and control groups are
exactly the same at the start of the experiment.
ANSWER: a. objective: to determine whether a mixture of salt
and lemon juice will remove copper oxide from
copper objects; population: all copper objects;
experimental group: pennies that are submerged in
the mixture; control group: pennies that are
submerged in a placebo mixture; sample procedure:
Pedro could randomly assign the pennies into two
groups, and create the lemon and salt mixture. Next,
he could submerge the experimental group of pennies
in the mixture for a certain period of time, remove
them, and then visually compare the two groups of
pennies.
b. Sample answer: The pennies should be chosen so
that they have roughly the same amount of copper
oxide and are from the same time period. Using
these guidelines could eliminate bias regarding the
initial condition and composition, so that the
conditions of the experimental and control groups are
exactly the same at the start of the experiment.
25. REPORTS The graph shown is from a report on the
average number of minutes 8- to 18-year-olds in the
U.S. spend on cell phones each day.
Page 6
exactly the same at the start of the experiment.
25. REPORTS The graph shown is from a report on the
number
of minutes 8- to 18-year-olds in the
11-1average
Designing
a Study
U.S. spend on cell phones each day.
a. Describe the sample and suggest a population.
b. What type of sample statistic do you think was
calculated for this report?
c. Describe the results of the study for each age
group.
d. Who do you think would be interested in this type
of report? Explain your reasoning.
SOLUTION: a. sample: the 8- to 18-year-olds surveyed;
population: all 8- to 18-year-olds in the U.S.
b. average time
c. Sample answer: The 8- to 10-year-old group talked
for about 10 minutes a day and did not text at all. The
11- to 14-year old group talked for about 30 minutes
a day and texted for about 70 minutes a day. The 15to 18-year-old group talked for about 40 minutes a
day and texted for about 110 minutes a day.
d. Sample answer: A cell phone company might use
a report like this to determine which age group to
target in their ads.
ANSWER: a. sample: the 8- to 18-year-olds surveyed;
population: all 8- to 18-year-olds in the U.S.
b. average time
c. Sample answer: The 8- to 10-year-old group talked
for about 10 minutes a day and did not text at all. The
11- to 14-year old group talked for about 30 minutes
a day and texted for about 70 minutes a day. The 15to 18-year-old group talked for about 40 minutes a
day and texted for about 110 minutes a day.
d. Sample answer: A cell phone company might use
a report like this to determine which age group to
target in their ads.
26. CCSS PERSEVERANCE In 1936, the Literary
Digest reported the results of a statistical study used
to predict whether Alf Landon or Franklin D.
Roosevelt
win
presidential election that
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year. The sample consisted of 2.4 million Americans,
including subscribers to the magazine, registered
a report like this to determine which age group to
target in their ads.
26. CCSS PERSEVERANCE In 1936, the Literary
Digest reported the results of a statistical study used
to predict whether Alf Landon or Franklin D.
Roosevelt would win the presidential election that
year. The sample consisted of 2.4 million Americans,
including subscribers to the magazine, registered
automobile owners, and telephone users. The results
concluded that Landon would win 57% of the popular
vote. The actual election results are shown.
a. Describe the type of study performed, the sample
taken, and the population.
b. How do the predicted and actual results compare?
c. Do you think that the survey was biased? Explain
your reasoning.
SOLUTION: a. This is a survey because data are collected from
responses to the question about who would be
president. sample: the 2.4 million people polled
population: all U.S. citizens of voting age in 1936
b. According to the predicted results, Landon should
have won 57% of the popular vote. However, in the
actual election, Roosevelt won 60.8% of the popular
vote.
c. Yes; sample answer: The people sampled could
afford magazine subscriptions, automobiles, and
telephones, suggesting that they were wealthier than
the average American citizen. The sampling method
did not represent citizens that could not afford these
things, and therefore, was not representative of the
entire population.
ANSWER: a. survey; sample: the 2.4 million people polled,
population: all U.S. citizens of voting age in 1936
b. According to the predicted results, Landon should
have won 57% of the popular vote. However, in the
actual election, Roosevelt won 60.8% of the popular
vote.
c. Yes; sample answer: The people sampled could
afford magazine subscriptions, automobiles, and
telephones, suggesting that they were wealthier than
the average American citizen. The sampling method
did not represent citizens that could not afford these
things, and therefore, was not representative of the
entire population.
Page 7
27. MULTIPLE REPRESENTATIONS The results
of two experiments concluded that Product A is 70%
afford magazine subscriptions, automobiles, and
telephones, suggesting that they were wealthier than
the average American citizen. The sampling method
not represent
citizens that could not afford these
11-1did
Designing
a Study
things, and therefore, was not representative of the
entire population.
27. MULTIPLE REPRESENTATIONS The results
of two experiments concluded that Product A is 70%
effective and Product B is 80% effective.
a. NUMERICAL To simulate the experiment for
Product A, use the random number generator on a
graphing calculator to generate 30 integers between
0 and 9. Let 0–6 represent an effective outcome and
7–9 represent an ineffective outcome.
The probability that Product A was effective is
63.3%
≈ The probability that Product B was effective is
76.7%
c. 76.7% – 63.3% = 13.4%
≈ The probability that Product B is effective is 13.4%
higher than that of Product A. Yes, the difference
appears to be significant.
b. TABULAR Copy and complete the frequency
table shown using the results. from part a. Then use
the data to calculate the probability that Product A
was effective. Repeat to find the probability for
Product B.
d. Sample answer: It depends on what the product is
and how it is being used. For example, if the product
is a pencil sharpener, then the lower price may be
more important than the effectiveness, and therefore,
might not justify the price difference. However, if the
product is a life-saving medicine, the effectiveness
may be more important than the price, and therefore,
might justify the price difference. ANSWER: a. See students’ work.
b. Sample answer for Product A: ≈63.3%
c. ANALYTICAL Compare the probabilities that
you found in part b. Do you think that the difference
in the effectiveness of each product is significant
enough to justify selecting one product over the
other? Explain.
d. LOGICAL Suppose Product B costs twice as
much as Product A. Do you think the probability of
the product’s effectiveness justifies the price
difference to a consumer? Explain. Sample answer for Product B: ≈76.7%
SOLUTION: a. See students’ work.
b. Sample answers:
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c. Sample answer: Yes; the probability that Product
B is effective is 14.4% higher than that of Product
A.
d. Sample answer: It depends on what the product is
Page 8
and how it is being used. For example, if the product
is a pencil sharpener, then the lower price may be
more important than the effectiveness, and therefore,
Sample answer:
Yes; the probability that Product
11-1c.Designing
a Study
B is effective is 14.4% higher than that of Product
A.
d. Sample answer: It depends on what the product is
and how it is being used. For example, if the product
is a pencil sharpener, then the lower price may be
more important than the effectiveness, and therefore,
might not justify the price difference. However, if the
product is a life-saving medicine, the effectiveness
may be more important than the price, and therefore,
might justify the price difference. REASONING Determine whether each
statement is true or false . If false, explain.
28. To save time and money, population parameters are
used to estimate sample statistics.
does affect the group directly. Therefore, the
statement is true.
ANSWER: true
30. OPEN ENDED Design an observational study.
Identify the objective of the study, define the
population and sample, collect and organize the data,
and calculate a sample statistic.
SOLUTION: Sample answer:
objective: Determine the average amount of time
that students spend studying at the library.
population: All students that study at the library. sample: 30 randomly selected students studying at
the library during a given week.
SOLUTION: False; sample answer: A sample statistic is used to
estimate a population parameter.
ANSWER: False; sample answer: A sample statistic is used to
estimate a population parameter.
29. Observational studies and experiments can both be
used to study cause-and-effect relationships.
SOLUTION: Observation studies are used to study cause-andeffect relationships by observing or measuring the
reactions of a group to something that does not affect
the group directly. Experiments are used to study
cause-and-effect relationships by observing or
measuring the reactions of a group to something that
does affect the group directly. Therefore, the
statement is true.
mean: ≈26.1 min
ANSWER: Sample answer:
objective: Determine the average amount of time
that students spend studying at the library.
population: All students that study at the library. sample: 30 randomly selected students studying at
the library during a given week.
ANSWER: true
30. OPEN ENDED Design an observational study.
Identify the objective of the study, define the
population and sample, collect and organize the data,
and calculate a sample statistic.
SOLUTION: Sample answer:
objective: Determine the average amount of time
that students spend studying at the library.
population: All students that study at the library. sample: 30 randomly selected students studying at
the library during a given week.
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mean: ≈26.1 min
31. CHALLENGE What factors should be considered
when determining whether a given statistical study is
reliable?
SOLUTION: An invalid sampling method and type of sample can
produce bias. For example, if a sample is not random,
the person conducting the study can influence the
results by selecting a specific sample of people. Also,
if an experiment is used when an observational study
Page 9
is the more logical type of study to be used, the study
can be unreliable. For example, if someone wants to
analyze the speeds of vehicles on a specific stretch
11-1 Designing a Study
mean: ≈26.1 min
31. CHALLENGE What factors should be considered
when determining whether a given statistical study is
reliable?
SOLUTION: An invalid sampling method and type of sample can
produce bias. For example, if a sample is not random,
the person conducting the study can influence the
results by selecting a specific sample of people. Also,
if an experiment is used when an observational study
is the more logical type of study to be used, the study
can be unreliable. For example, if someone wants to
analyze the speeds of vehicles on a specific stretch
of highway and decides to place an empty police car
on the side of the road, the data will be affected by
the police car. The results of this study will show
lower speeds than are normally driven on the
highway. Biased survey questions and incorrect
procedures can affect the reliability of a study as
well. A survey question that is poorly written may
result in a response that does not accurately reflect
the opinion of the participant.
ANSWER: Sample answer: the sampling method used, the type
of sample that was selected, the type of study
performed, the survey question(s) that were asked or
procedures that were used
32. WRITING IN MATH Research each of the
following sampling methods. Then describe each
method and discuss whether using the method could
result in bias.
a. convenience sample b. self-selected sample
c. stratified sample
d. systematic sample
SOLUTION: a. Sample answer: In a convenience sample,
members are selected based on the convenience of
the researcher. One example is handing a survey to
shoppers as they walk out of the mall. This method
could result in bias if the members of the population
who are readily available to be sampled are not
representative of the entire population.
b. Sample answer: In a self-selected sample,
members volunteer to be in the sample. This method
could result in bias if certain groups of people in the
population choose not to volunteer.
c. Sample answer: In a stratified sample, the
population is first divided into similar, nonoverlapping
groups, and members are then randomly selected
eSolutions
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by Cognero
fromManual
each group.
This
method could result in bias if
the entire population is not represented when divided
into groups or if the members are not randomly
members volunteer to be in the sample. This method
could result in bias if certain groups of people in the
population choose not to volunteer.
c. Sample answer: In a stratified sample, the
population is first divided into similar, nonoverlapping
groups, and members are then randomly selected
from each group. This method could result in bias if
the entire population is not represented when divided
into groups or if the members are not randomly
selected from each group.
d. Sample answer: In a systematic sample, a rule is
used to select the members. This method could result
in bias if the rule does not include everyone in the
population.
ANSWER: a. Sample answer: In a convenience sample,
members are selected based on the convenience of
the researcher. One example is handing a survey to
shoppers as they walk out of the mall. This method
could result in bias if the members of the population
who are readily available to be sampled are not
representative of the entire population.
b. Sample answer: In a self-selected sample,
members volunteer to be in the sample. This method
could result in bias if certain groups of people in the
population choose not to volunteer.
c. Sample answer: In a stratified sample, the
population is first divided into similar, nonoverlapping
groups, and members are then randomly selected
from each group. This method could result in bias if
the entire population is not represented when divided
into groups or if the members are not randomly
selected from each group.
d. Sample answer: In a systematic sample, a rule is
used to select the members. This method could result
in bias if the rule does not include everyone in the
population.
33. GEOMETRY In
, BC > AB.
Which of the following must be true?
A AB = BC
B AC > AB
C a > 60
Da=b
SOLUTION: Since the side BC > the side AB,
Therefore, a > 60
Option C is the correct answer.
ANSWER: C
.
Page 10
selected from each group.
d. Sample answer: In a systematic sample, a rule is
used to select the members. This method could result
11-1inDesigning
Study
bias if the arule
does not include everyone in the
population.
The solution set is {1, –0.5}.
ANSWER: 35. SAT/ACT A pie is divided evenly between 3 boys
and a girl. If one boy gives one half of his share to
the girl and a second boy keeps two thirds of his
share and gives the rest to the girl, what portion will
the girl have in all?
F
33. GEOMETRY In
, BC > AB.
Which of the following must be true?
G H
A AB = BC
B AC > AB
C a > 60
Da=b
J
K
SOLUTION: Since the side BC > the side AB,
Therefore, a > 60
Option C is the correct answer.
SOLUTION: .
ANSWER: C
Each of them having
portion in all.
The girl getting one half of one boy’s share and one
third of the second boy’s share.
The girl will have
in all.
34. SHORT RESPONSE What is the solution set of
?
SOLUTION: Option G is the correct answer.
By the Zero Product Property:
ANSWER: G
36. Which equation represents a hyperbola?
2
2
A y = 49 – x
B y = 49 – x2
C y = 49x
2
D
The solution set is {1, –0.5}.
ANSWER: 35. SAT/ACT A pie is divided evenly between 3 boys
and a girl. If one boy gives one half of his share to
the girl and a second boy keeps two thirds of his
share and gives the rest to the girl, what portion will
the girl have in all?
eSolutions
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G SOLUTION: Option A represents a circle.
Option B represents a parabola.
Option C represents a parabola.
Option D represents a hyperbola.
Therefore, option D is the correct answer.
ANSWER: D
Page 11
37. Prove that the statement 9n – 1 is divisible by 8 is
true for all natural numbers.
Option D represents a hyperbola.
Therefore, option D is the correct answer.
Option G is the correct answer.
11-1ANSWER: Designing a Study
G
ANSWER: D
36. Which equation represents a hyperbola?
2
2
A y = 49 – x
B y = 49 – x2
C y = 49x
2
37. Prove that the statement 9n – 1 is divisible by 8 is
true for all natural numbers.
SOLUTION: 1
Step 1: 9 – 1 = 8, which is divisible by 8. The
statement is true for n = 1.
D
k
SOLUTION: Option A represents a circle.
Option B represents a parabola.
Option C represents a parabola.
Option D represents a hyperbola.
Therefore, option D is the correct answer.
Step 2: Assume that 9 – 1 is divisible by 8 for some
k
positive integer k. This means that 9 – 1 = 8r for
some whole number r.
Step 3:
ANSWER: D
37. Prove that the statement 9n – 1 is divisible by 8 is
true for all natural numbers.
SOLUTION: 1
Step 1: 9 – 1 = 8, which is divisible by 8. The
statement is true for n = 1.
k
Step 2: Assume that 9 – 1 is divisible by 8 for some
k
positive integer k. This means that 9 – 1 = 8r for
some whole number r.
Step 3:
Since r is a whole number, 9r + 1 is a whole number.
k
Thus, 9 + 1 – 1 is divisible by 8, so the statement is
n
true for n = k + 1. Therefore, 9 – 1 is divisible by 8
for all positive integers n.
ANSWER: 1
Step 1: 9 – 1 = 8, which is divisible by 8. The
statement is true for n = 1.
k
Step 2: Assume that 9 – 1 is divisible by 8 for some
k
positive integer k. This means that 9 – 1 = 8r for
some whole number r.
Step 3:
Since r is a whole number, 9r + 1 is a whole number.
k
Thus, 9 + 1 – 1 is divisible by 8, so the statement is
n
true for n = k + 1. Therefore, 9 – 1 is divisible by 8
for all positive integers n.
ANSWER: 1
Step 1: 9 – 1 = 8, which is divisible by 8. The
statement is true for n = 1.
k
Step 2: Assume that 9 – 1 is divisible by 8 for some
k
positive integer k. This means that 9 – 1 = 8r for
some whole number r.
Step 3:
eSolutions
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Since r is a whole number, 9r + 1 is a whole number.
k +1
Thus, 9
– 1 is divisible by 8, so the statement is
n
true for n = k + 1. Therefore, 9 – 1 is divisible by 8
for all positive integers n.
38. INTRAMURALS Ofelia is taking ten shots in the
intramural free-throw shooting competition. How
Page 12
many sequences of hits and misses are there that
result in her making eight shots and missing two?
Since r is a whole number, 9r + 1 is a whole number.
k +1
Thus, 9
– 1 is divisible by 8, so the statement is
ANSWER: n
for n = ka +
1. Therefore, 9 – 1 is divisible by 8
11-1true
Designing
Study
for all positive integers n.
38. INTRAMURALS Ofelia is taking ten shots in the
intramural free-throw shooting competition. How
many sequences of hits and misses are there that
result in her making eight shots and missing two?
SOLUTION: The number of sequences of hits and misses is
.
2
2
40. x + y = 36
y =x+2
SOLUTION: Substitute x + 2 for y in the first equation and solve
for x.
ANSWER: 45
Solve each system of equations.
39. y = x + 3
2
y = 2x
SOLUTION: Substitute x + 3 for y and solve for x.
Substitute the values of x in the second equation and
find the corresponding values of y.
Therefore, x = –1 and
.
Substitute the values of x in the first equation and
evaluate.
Therefore, the solutions are
.
ANSWER: 2
2
41. y + x = 9
y =7– x
Therefore, the solutions are (–1, 2) and
.
ANSWER: 2
SOLUTION: Substitute 7 – x for y in the first equation and solve
for x.
2
40. x + y = 36
y =x+2
SOLUTION: Substitute x + 2 for y in the first equation and solve
for x.
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Page 13
.
ANSWER: 11-1 Designing a Study
2
Since the radicand is negative there is no real solution
for the system.
ANSWER: no solution
2
2
41. y + x = 9
y =7– x
42. y + x = 3
2
2
x + 4y = 36
SOLUTION: Substitute 7 – x for y in the first equation and solve
for x.
SOLUTION: 2
Solve the first equation for x and substitute in the
second equation.
Since the radicand is negative there is no real solution
for the system.
ANSWER: no solution
Substitute the values of y in the first equation and
find the corresponding values of x.
2
42. y + x = 3
2
2
x + 4y = 36
SOLUTION: 2
Solve the first equation for x and substitute in the
second equation.
Therefore, the solutions are
.
ANSWER: 2
2
43. x + y = 64
2
2
x + 64y = 64
SOLUTION: eSolutions Manual - Powered by Cognero
2
Page 14
Solve the first equation for y and substitute in the
second equation.
Therefore, the solutions are
ANSWER: ANSWER: (±8, 0)
11-1 Designing a Study
2
2
2
43. x + y = 64
2
.
2
44. y = x – 25
2
2
x –y =7
2
x + 64y = 64
SOLUTION: SOLUTION: 2
2
2
Solve the first equation for y and substitute in the
second equation.
Substitute x – 25 for y in the second equation and
solve for x.
This is impossible. Therefore, there is no solution.
ANSWER: no solution
Find the distance between each pair of points
with the given coordinates.
45. (9, –2), (12, –14)
Substitute the values of x in the first equation and
find the corresponding values of y.
SOLUTION: Substitute the given points in the distance formula.
Therefore, the solutions are
.
ANSWER: (±8, 0)
2
ANSWER: 2
44. y = x – 25
2
2
x –y =7
46. (–4, –10), (–3, –11)
SOLUTION: 2
2
Substitute x – 25 for y in the second equation and
solve for x.
SOLUTION: Substitute the given points in the distance formula.
This is impossible. Therefore, there is no solution.
ANSWER: ANSWER: eSolutions
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Page 15
no solution
47. (1, –14), (–6, 10)
ANSWER: 11-1 Designing a Study
ANSWER: 13 units
46. (–4, –10), (–3, –11)
49. (2.3, –1.2), (–4.5, 3.7)
SOLUTION: Substitute the given points in the distance formula.
SOLUTION: Substitute the given points in the distance formula.
ANSWER: ANSWER: 47. (1, –14), (–6, 10)
50. (0.23, 0.4), (0.68, –0.2)
SOLUTION: Substitute the given points in the distance formula.
SOLUTION: Substitute the given points in the distance formula.
ANSWER: 25 units
ANSWER: 0.75 unit
48. (–4, 9), (1, –3)
SOLUTION: Substitute the given points in the distance formula.
Simplify. Assume that no variable equals 0.
2
4
51. (5cd )(–c d)
SOLUTION: ANSWER: 5 3
–5c d
3 –5
3
52. (7x y )(4xy )
SOLUTION: ANSWER: 13 units
49. (2.3, –1.2), (–4.5, 3.7)
ANSWER: SOLUTION: eSolutions
Manual the
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Substitute
givenby
points
in the
distance formula.
Page 16
ANSWER: 5 3
11-1–5c
Designing
a Study
d
3 –5
3
52. (7x y )(4xy )
ANSWER: 56. SOLUTION: SOLUTION: ANSWER: ANSWER: Write a quadratic equation with the given root
2
(s). Write the equation in the form ax + bx + c
= 0, where a, b, and c are integers.
57. –3, 9
53. SOLUTION: SOLUTION: ANSWER: an
ANSWER: 54. 2
x – 6x – 27 = 0
SOLUTION: 58. ANSWER: SOLUTION: 55. SOLUTION: ANSWER: ANSWER: 56. 2
12x + 13x + 3 = 0
SOLUTION: 59. 4, –5
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ANSWER: Page 17
ANSWER: 11-1 Designing a Study
2
12x + 13x + 3 = 0
59. 4, –5
SOLUTION: ANSWER: 2
x + x – 20 = 0
60. TESTS Ms. Bonilla’s class of 30 students took a
biology test. If 20 of her students had an average of
83 on the test and the other students had an average
score of 74, what was the average score of the
whole class?
SOLUTION: Total score of 20 students is 20 × 83 or 1660.
Total score of remaining 10 students is 10 × 74 or
740.
Total score of the whole class is 1660 + 740 or 2400.
The average score of the whole class is
.
ANSWER: 80
61. DRIVING During a 10-hour trip, Kwan drove 4
hours at 60 miles per hour and 6 hours at 65 miles
per hour. What was her average rate, in miles per
hour, for the entire trip?
SOLUTION: The distance traveled in the first four hours is 4 × 60
or 240 miles.
The distance traveled in the next six hours is 6 × 65
or 390 miles.
Total distance traveled is 240 + 390 or 630 miles.
Her average rate for the entire trip is
or 63 mph.
ANSWER: 63
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11-2 Distributions of Data
1. EXERCISE The amount of time that James ran on
a treadmill for the first 24 days of his workout is
shown.
a. Use a graphing calculator to create a histogram.
Then describe the shape of the distribution.
b. Describe the center and spread of the data using
either the mean and standard deviation or the fivenumber summary. Justify your choice.
SOLUTION: a. First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
ENTER and choose fl. Finally, adjust the window to
dimensions appropriate for the data.
Since the majority of the data is on the right and
there is a tail on the left, the distribution is negatively
skewed.
b. Sample answer: The distribution is skewed, so use
the five-number summary. Press STAT ► ENTER
ENTER and scroll down to display the statistics for
the data set.
The range is 7 to 30 minutes. The median is 22.5
minutes, and half of the data are between 15.5 and
26 minutes. ANSWER: a. Manual - Powered by Cognero
eSolutions
The range is 7 to 30 minutes. The median is 22.5
minutes, and half of the data are between 15.5 and
26 minutes. ANSWER: a.
negatively skewed
b. Sample answer: The distribution is skewed, so use
the five-number summary. The times range from 7 to
30 minutes. The median is 22.5 minutes, and half of
the data are between 15.5 and 26 minutes.
2. RESTAURANTS The total number of times that 20
random people either ate at a restaurant or bought
fast food in a month are shown.
a. Use a graphing calculator to create a box-andwhisker plot. Then describe the shape of the
distribution.
b. Describe the center and spread of the data using
either the mean and standard deviation or the fivenumber summary. Justify your choice.
SOLUTION: a. Enter the data as L1. Press 2ND [STAT PLOT]
ENTER ENTER and choose fl. Adjust the window
to the dimensions shown.
The data to the right of the median are distributed
over a wider range than the data to the left. There is
a longer tail to the right. Therefore, the distribution is
positively skewed.
b. The distribution is skewed, so use the five-number
summary. Press STAT ► ENTER ENTER and
scroll down to view the five-number summary. Page 1
The data to the right of the median are distributed
over a wider range than the data to the left. There is
a longer tail to the right. Therefore, the distribution is
skewed.
11-2positively
Distributions
of Data
b. The distribution is skewed, so use the five-number
summary. Press STAT ► ENTER ENTER and
scroll down to view the five-number summary.
The range is 3 to 22 times. The median number is 7
times, and half of the data are between 5 and 13
times.
a. Use a graphing calculator to create a histogram
for each data set. Then describe the shape of each
distribution.
b. Compare the distributions using either the means
and standard deviations or the five-number
summaries. Justify your choice.
SOLUTION: a.
Mrs. Johnson’s class:
First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
ENTER and choose „ . Finally, adjust the window to
the dimensions shown.
ANSWER: a.
positively skewed
b. Sample answer: The distribution is skewed, so use
the five-number summary. The data range from 3 to
22 times. The median number is 7 times, and half of
the data are between 5 and 13 times.
3. CCSS TOOLS The total fundraiser sales for the
students in two classes at Cantonville High School
are shown.
The majority of the data are on the left of the mean,
so the distribution is positively skewed.
Mr. Edmunds’ Class:
First, press STAT ENTER and enter each data
value in L2. Then, press 2ND [STAT PLOT]
ENTER ENTER and choose „ . Finally, adjust the
window to the dimensions shown.
The majority of the data are on the right of the mean,
so the distribution is negatively skewed.
eSolutions
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a. Use a graphing calculator to create a histogram
for each data set. Then describe the shape of each
b. The distributions are skewed, so use the fivenumber summaries.
For the first distribution, press STAT ► ENTER
ENTER and scroll down to view the five-number
Page 2
summary. For the second distribution, press STAT
► ENTER ENTER L2 and scroll down to view the
five-number summary.
The majority of the data are on the right of the mean,
so the distribution is negatively skewed.
The distributions
are skewed, so use the five11-2b.
Distributions
of Data
number summaries.
For the first distribution, press STAT ► ENTER
ENTER and scroll down to view the five-number
summary. For the second distribution, press STAT
► ENTER ENTER L2 and scroll down to view the
five-number summary.
Mrs. Johnson’s class:
Mr. Edmunds’ Class:
The range for both classes is the same. However,
the median for Mrs. Johnson’s class is 17 and the
median for Mr. Edmunds’ class is 28. The lower
quartile for Mr. Edmunds’ class is 20. Since this is
greater than the median for Mrs. Johnson’s class,
this means that 75% of the data from Mr. Edmunds’ class is greater than 50% of the data from Mrs.
Johnson’s class. Therefore, we can conclude that the
students in Mr. Edmunds’ class had slightly higher
sales overall than the students in Mrs. Johnson’s
class.
Mrs. Johnson’s class, positively skewed; Mr.
Edmunds’ class, negatively skewed
b. Sample answer: The distributions are skewed, so
use the five-number summaries. The range for both
classes is the same. However, the median for Mrs.
Johnson’s class is 17 and the median for Mr.
Edmunds’ class is 28. The lower quartile for Mr.
Edmunds’ class is 20. Since this is greater than the
median for Mrs. Johnson’s class, this means that
75% of the data from Mr. Edmunds’ class is greater
than 50% of the data from Mrs. Johnson’s class.
Therefore, we can conclude that the students in Mr.
Edmunds’ class had slightly higher sales overall than
the students in Mrs. Johnson’s class.
4. RECYCLING The weekly totals of recycled paper
for the junior and senior classes are shown.
ANSWER: a. a. Use a graphing calculator to create a box-andwhisker plot for each data set. Then describe the
shape of each distribution.
b. Compare the distributions using either the means
and standard deviations or the five-number
summaries. Justify your choice.
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SOLUTION: a.
Junior class:
Enter the data as L1. Press 2ND [STAT PLOT]
Page 3
ENTER ENTER and choose fl.
Senior Class:
and standard deviations or the five-number
summaries. Justify your choice.
11-2SOLUTION: Distributions of Data
a.
Junior class:
Enter the data as L1. Press 2ND [STAT PLOT]
ENTER ENTER and choose fl.
Senior Class:
Senior Class:
Enter the data as L2. Press 2ND [STAT PLOT]
Plot 2 ENTER ENTER and choose fl. Adjust the
window to the dimensions shown. The top plot is for
the Junior Class and the bottom plot if for the Senior
Class.
The median for the junior class is 15, and the median
for the senior class is 27.5. The minimum value for
the senior class is 18. This means that all of the
weekly totals for the senior class are greater than
50% of the weekly totals for the junior class.
Therefore, we can conclude that the senior class’s
weekly totals were far greater than the junior class’s
weekly totals.
ANSWER: a. For the Junior Class, the data to the right of the
median are distributed over a wider range than the
data to the left. There is a longer tail to the right.
Therefore, the distribution is positively skewed. For
the Senior class, the data are equally distributed to
the left and right of the median. Therefore, the
distribution is symmetric.
b. One of the distributions is skewed, so use the fivenumber summaries.
For the first distribution, press STAT ► ENTER
ENTER and scroll down to view the five-number
summary. For the second distribution, press STAT
► ENTER ENTER L2 and scroll down to view the
five-number summary.
Junior Class:
Senior Class:
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The median for the junior class is 15, and the median
for the senior class is 27.5. The minimum value for
junior class, positively skewed; senior class,
symmetric
b. Sample answer: One distribution is symmetric and
the other is skewed, so use the five-number
summaries. The median for the junior class is 15, and
the median for the senior class is 27.5. The minimum
value for the senior class is 18. This means that all of
the weekly totals for the senior class are greater than
50% of the weekly totals for the junior class.
Therefore, we can conclude that the senior class’s
weekly totals were far greater than the junior class’s
weekly totals.
For Exercises 5 and 6, complete each step.
a. Use a graphing calculator to create a
histogram and a box-and-whisker plot. Then
describe the shape of the distribution.
b. Describe the center and spread of the data
using either the mean and standard deviation or
the five-number summary. Justify your choice.
5. FANTASY The weekly total points of Kevin’s
fantasy football team are shown.
Page 4
b. Describe the center and spread of the data
using either the mean and standard deviation or
the five-number summary. Justify your choice.
5. Theofweekly
11-2FANTASY
Distributions
Data total points of Kevin’s
fantasy football team are shown.
SOLUTION: a.
Histogram:
First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
ENTER and choose „ . Finally, adjust the window to
the dimensions shown.
The range is 53 to 179 points. The median is 138.5
points, and half of the data are between 106.5 and
157 points.
ANSWER: a.
Box-and-whisker Plot: Press 2ND [STAT PLOT]
ENTER ENTER and choose fl. Adjust the window
to the dimensions shown.
The majority of the data are on the right of the mean
and the data to the left of the mean are distributed
over a wider range than the data to the right, so the
distribution is negatively skewed.
b. The distribution is skewed, so use the five-number
summary. Press STAT ► ENTER ENTER and
scroll down to view the five-number summary.
negatively skewed
b. Sample answer: The distribution is skewed, so use
the five-number summary. The points range from 53
to 179. The median is 138.5 points, and half of the
data are between 106.5 and 157 points.
6. MOVIES The students in one of Mr. Peterson’s
classes recorded the number of movies they saw
over the past month.
SOLUTION: a.
Histogram:
First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
ENTER and choose „ . Finally, adjust the window to
the dimensions shown.
The range is 53 to 179 points. The median is 138.5
between 106.5 and
157 points.
points,
and- half
of the
data are
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ANSWER: Page 5
Histogram:
First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
„ . Finally, adjust the window to
and choose
11-2 ENTER
Distributions
of Data
the dimensions shown.
The mean number of movies watched was about
10.7 with standard deviation of about 3 movies.
ANSWER: a.
Box-and-whisker Plot: Press 2ND [STAT PLOT]
ENTER ENTER and choose fl. Adjust the window
to the dimensions shown.
The date are equally distributed on each side of the
mean, so the distribution is symmetric.
b. The distribution is symmetric, so use the mean and
standard deviation. Press STAT ► ENTER
ENTER and scroll down to view the mean and
standard deviation.
symmetric
b. Sample answer: The distribution is symmetric, so
use the mean and standard deviation. The mean
number of movies watched was about 10.7 with
standard deviation of about 3 movies.
CCSS TOOLS Complete each step.
a. Use a graphing calculator to create a
histogram for each data set. Then describe the
shape of each distribution.
b. Compare the distributions using either the
means and standard deviations or the fivenumber summaries. Justify your choice.
7. SAT A group of students took the SAT their
sophomore year and again their junior year. Their
scores are shown.
The mean number of movies watched was about
10.7 with standard deviation of about 3 movies.
ANSWER: a.
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SOLUTION: a.
Sophomore Year:
First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
Page 6
ENTER and choose „ . Finally, adjust the window to
the dimensions shown.
SOLUTION: 11-2a.Distributions of Data
Sophomore Year:
First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
ENTER and choose „ . Finally, adjust the window to
the dimensions shown.
standard deviation. For the second distribution, press
STAT ► ENTER ENTER L2 and scroll down to
view the mean and standard deviation.
Sophomore Year:
Junior Year:
The data are evenly distributed, so the distribution is
symmetric.
Junior Year:
First, press STAT ENTER and enter each data
value in L2. Then, press 2ND [STAT PLOT]
ENTER ENTER and choose „ . Finally, adjust the
window to the dimensions shown.
The mean score for sophomore year is about 1552.9
with standard deviation of about 147.2. The mean
score for junior year is about 1753.8 with standard
deviation of about 159.1. We can conclude that the
scores and the variation of the scores from the mean
both increased from sophomore year to junior year.
ANSWER: a. The data are evenly distributed, so the distribution is
symmetric.
b. The distributions are symmetric, so use the means
and standard deviations.
For the first distribution, press STAT ► ENTER
ENTER and scroll down to view the mean and
standard deviation. For the second distribution, press
STAT ► ENTER ENTER L2 and scroll down to
view the mean and standard deviation.
Sophomore Year:
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Junior Year:
both symmetric
b. Sample answer: The distributions are symmetric,
so use the means and standard deviations. The mean
score for sophomore year is about 1552.9 with
standard deviation of about 147.2. The mean score
Page 7
for junior year is about 1753.8 with standard
deviation of about 159.1. We can conclude that the
scores and the variation of the scores from the mean
both symmetric
Sample answer:
The distributions are symmetric,
11-2b.
Distributions
of Data
so use the means and standard deviations. The mean
score for sophomore year is about 1552.9 with
standard deviation of about 147.2. The mean score
for junior year is about 1753.8 with standard
deviation of about 159.1. We can conclude that the
scores and the variation of the scores from the mean
both increased from sophomore year to junior year.
8. INCOME The total incomes for 18 households in
two neighboring cities are shown.
Applewood:
First, press STAT ENTER and enter each data
value in L2. Then, press 2ND [STAT PLOT]
ENTER ENTER and choose „ . Finally, adjust the
window to the dimensions shown.
The majority of the data are on the right of the mean
and the data to the left of the mean are distributed
over a wider range than the data to the right, so the
distribution is negatively skewed.
b. The distributions are skewed, so use the fivenumber summary.
For the first distribution, press STAT ► ENTER
ENTER and scroll down to view the five-number
summary. For the second distribution, press STAT
► ENTER ENTER L2 and scroll down to view the
five-number summary.
SOLUTION: a.
Yorkshire:
First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
ENTER and choose „ . Finally, adjust the window to
the dimensions shown.
Yorkshire:
Applewood:
The majority of the data are on the left of the mean
and the data to the right of the mean are distributed
over a wider range than the data to the left, so the
distribution is positively skewed.
Applewood:
First, press STAT ENTER and enter each data
value in L2. Then, press 2ND [STAT PLOT]
ENTER ENTER and choose „ . Finally, adjust the
window to the dimensions shown.
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The median for Yorkshire is 63.5, and the median for
Applewood is 60. The lower quartile for Applewood
is 52, while the minimum for Yorkshire is 53. This
means that 25% of the incomes for Applewood are
lower than any of the incomes for Yorkshire. Also,
the upper 25% of incomes for Yorkshire is between
72 and 103, while the upper 25% of incomes for
Applewood is between 65 and 87. We can conclude
that the incomes for the households in Yorkshire are
greater than the incomes for the households in
Applewood.
ANSWER: a.
Page 8
Applewood is between 65 and 87. We can conclude
that the incomes for the households in Yorkshire are
greater than the incomes for the households in
11-2Applewood.
Distributions of Data
ANSWER: a.
SOLUTION: a. Enter the tuitions for the public colleges as L1.
Graph these data as Plot1 by pressing 2ND [STAT
PLOT] ENTER ENTER and choosing fl. Enter the
tuitions for the private colleges as L2. Graph these
data as Plot2 by pressing 2ND [STAT PLOT]
ENTER ENTER and choosing fl.. For Xlist,
enter L2. Adjust the window to dimensions
appropriate for the data.
Yorkshire, positively skewed; Applewood, negatively
skewed
b. Sample answer: The distributions are skewed, so
use the five-number summaries. The median for
Yorkshire is 63.5, and the median for Applewood is
60. The lower quartile for Applewood is 52, while the
minimum for Yorkshire is 53. This means that 25%
of the incomes for Applewood are lower than any of
the incomes for Yorkshire. Also, the upper 25% of
incomes for Yorkshire is between 72 and 103, while
the upper 25% of incomes for Applewood is between
65 and 87. We can conclude that the incomes for the
households in Yorkshire are greater than the incomes
for the households in Applewood.
9. TUITION The annual tuitions for a sample of public
colleges and a sample of private colleges are shown.
Complete each step
a. Use a graphing calculator to create a box-andwhisker plot for each data set. Then describe the
shape of each distribution.
b. Compare the distributions using either the means
and standard deviations or the five-number
summaries. Justify your choice.
For both sets of data, the whiskers are approximately
equal, and the median is in the middle of the data.
The distributions are symmetric.
b. Sample answer: The distributions are symmetric,
so use the means and standard deviations. Press
STAT ► ENTER ENTER to display the statistics
for the public colleges.
Press STAT ► ENTER 2ND [L1] ENTER to
display the statistics for the private colleges.
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Page 9
The mean for the public colleges is $4037.50 with
standard deviation of about $621.93. The mean for
private colleges is about $12,803.11 with standard
11-2 Distributions of Data
The mean for the public colleges is $4037.50 with
standard deviation of about $621.93. The mean for
private colleges is about $12,803.11 with standard
deviation of about $2915.20. We can conclude that
not only is the average cost of private schools far
greater than the average cost of public schools, but
the variation of the costs from the mean is also much
greater.
ANSWER: a.
SOLUTION: a.
Boys:
Enter the data as L1. Press 2ND [STAT PLOT]
ENTER ENTER and choose fl.
Girls:
Enter the data as L2. Press 2ND [STAT PLOT]
Plot 2 ENTER ENTER and choose fl. Adjust the
window to the dimensions shown. The top plot is for
the boys and the bottom plot if for the girls.
both symmetric
b. Sample answer: The distributions are symmetric,
so use the means and standard deviations. The mean
for the public colleges is $4037.50 with standard
deviation of about $621.93. The mean for private
colleges is about $12,803.11 with standard deviation
of about $2915.20. We can conclude that not only is
the average cost of private schools far greater than
the average cost of public schools, but the variation
of the costs from the mean is also much greater.
10. DANCE The total amount of money that a random
sample of seniors spent on prom is shown. Complete
each step.
a. Use a graphing calculator to create a box-andwhisker plot for each data set. Then describe the
shape of each distribution.
b. Compare the distributions using either the means
and standard deviations or the five-number
summaries. Justify your choice.
For the boys, the data to the right of the median are
distributed over a wider range than the data to the
left. There is a longer tail to the right. Therefore, the
distribution is positively skewed. For the girls, the
data are equally distributed to the left and right of the
median. Therefore, the distribution is symmetric.
b. One of the distributions is skewed, so use the fivenumber summaries.
For the first distribution, press STAT ► ENTER
ENTER and scroll down to view the five-number
summary. For the second distribution, press STAT
► ENTER ENTER L2 and scroll down to view the
five-number summary.
Boys:
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SOLUTION: Girls:
Page 10
11-2 Distributions of Data
Girls:
The maximum value for the boys is 450. The lower
quartile for the girls is 453. This means that 75% of
the data for the girls is higher than any data for the
boys. We can conclude that 75% of the girls spent
more money on prom than any of the boys.
a. Use a graphing calculator to create a box-andwhisker plot. Describe the center and spread of the
data. b. Craig scored 0, 2, 1, and 0 points in the first four
games. Use a graphing calculator to create a boxand-whisker plot that includes the new data. Then
find the mean and median of the new data set.
c. What effect does adding the scores from the first
four games have on the shape of the distribution and
on how you should describe the center and spread?
SOLUTION: a. Enter the data as L1. Press 2ND [STAT PLOT]
ENTER ENTER and choose fl. Adjust the window
to the dimensions shown.
ANSWER: a.
The distribution is symmetric, so use the mean and
standard deviation.
boys, positively skewed; girls, symmetric
b. Sample answer: One distribution is skewed and
the other is symmetric, so use the five-number
summaries. The maximum value for the boys is 450.
The lower quartile for the girls is 453. This means
that 75% of the data for the girls is higher than any
data for the boys. We can conclude that 75% of the
girls spent more money on prom than any of the
boys.
The mean of the data is 18 with a sample standard
deviation of about 5.2 points.
11. BASKETBALL Refer to the beginning of the
lesson. The points that Craig scored in the remaining
games are shown.
a. Use a graphing calculator to create a box-andwhisker plot. Describe the center and spread of the
data. b. Craig scored 0, 2, 1, and 0 points in the first four
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games. Use a graphing calculator to create a boxand-whisker plot that includes the new data. Then
find the mean and median of the new data set.
b. Add the new data to L1. Press 2ND [STAT
PLOT] ENTER ENTER and choose fl. Adjust the
window to the dimensions shown.
Find the new mean and median:
Page 11
11-2 Distributions of Data
Find the new mean and median:
mean: 14.6; median: 17
c. Sample answer: Adding the scores from the first
four games causes the shape of the distribution to go
from being symmetric to being negatively skewed.
Therefore, the center and spread should be described
using the five-number summary.
12. SCORES Allison’s quiz scores are shown.
The new mean is 14.55 and the new median is 17.
c. Adding the scores from the first four games
causes the shape of the distribution to go from being
symmetric to being negatively skewed. Therefore,
the center and spread should be described using the
five-number summary.
ANSWER: a.
Sample answer: The distribution is symmetric, so use
the mean and standard deviation. The mean of the
data is 18 with sample standard deviation of about
5.2 points.
b. a. Use a graphing calculator to create a box-andwhisker plot. Describe the center and spread.
b. Allison’s teacher allows students to drop their two
lowest quiz scores. Use a graphing calculator to
create a box-and-whisker plot that reflects this
change. Then describe the center and spread of the
new data set.
SOLUTION: a. Enter the data as L1. Press 2ND [STAT PLOT]
ENTER ENTER and choose fl. Adjust the window
to the dimensions shown.
The majority of the data are on the right of the mean
and the data to the left of the mean are distributed
over a wider range than the data to the right, so the
distribution is negatively skewed.
Press STAT ► ENTER ENTER and scroll down
to view the five-number summary.
mean: 14.6; median: 17
c. Sample answer: Adding the scores from the first
four games causes the shape of the distribution to go
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from being symmetric to being negatively skewed.
Therefore, the center and spread should be described
using the five-number summary.
Page
The range is 57 to 96. The median is 86, and half
of 12
the data are between 76 and 89.
11-2 Distributions of Data
Sample answer: The distribution is negatively
skewed, so use the five-number summary. The
scores range from 57 to 96. The median is 86, and
half of the data are between 76 and 89.
b. The range is 57 to 96. The median is 86, and half of
the data are between 76 and 89.
b. Remove the 57 and 62 from L1. Press 2ND
[STAT PLOT] ENTER ENTER and choose fl.
Adjust the window to the dimensions shown.
Sample answer: The distribution is symmetric, so use
the mean and standard deviation. The mean is about
85.6 with standard deviation of about 5.9.
13. CHALLENGE Approximate the mean and median
for each distribution of data.
The data are equally distributed to the left and right
of the median. Therefore, the distribution is
symmetric.
Press STAT ► ENTER ENTER and scroll down
to view the mean and standard deviation.
a.
The distribution is symmetric, so use the mean and
standard deviation. The mean is about 85.6 with
standard deviation of about 5.9.
ANSWER: a.
Sample answer: The distribution is negatively
skewed, so use the five-number summary. The
scores range from 57 to 96. The median is 86, and
half of the data are between 76 and 89.
b. eSolutions Manual - Powered by Cognero
b.
c.
SOLUTION: a. Sample answer: Since the distribution is positively
skewed, the median will be to the left of the mean
Page
13
closer to the majority of the data. An estimate for
the
median is 10. The mean will be more affected by the
tail, and will be to the right of the majority of the
c.
11-2 Distributions of Data
SOLUTION: a. Sample answer: Since the distribution is positively
skewed, the median will be to the left of the mean
closer to the majority of the data. An estimate for the
median is 10. The mean will be more affected by the
tail, and will be to the right of the majority of the
data. An estimate for the mean is 14.
b. Sample answer: Since the distribution is negatively
skewed, the median will be to the right of the mean
closer to the majority of the data. An estimate for the
median is 24. The mean will be more affected by the
tail, and will be to the left of the majority of the data.
An estimate for the mean is 20.
c. Sample answer: Since the distribution is
symmetric, the mean and median will be
approximately equal near the middle of the data. An
estimate for the mean and median is 17.
ANSWER: a. Sample answer: mean = 14; median = 10
b. Sample answer: mean = 20; median = 24
c. Sample answer: mean = 17; median = 17
Instead, summarize the center and spread of each
cluster individually using its mean and standard
deviation.
15. OPEN ENDED Find a real-world data set that
appears to represent a symmetric distribution and one
that does not. Describe each distribution. Create a
visual representation of each set of data.
SOLUTION: Sample answer: The heights of the players on the
Pittsburgh Steelers roster appear to represent a
normal distribution.
The mean of the data is about 73.61 in. or 6 ft 1.61
in. The standard deviation is about 2.97 in.
14. CCSS ARGUMENTS Distributions of data are not
always symmetric or skewed. If a distribution has a
gap in the middle, like the one shown, two separate
clusters of data may result, forming a bimodal
distribution. How can the center and spread of a
bimodal distribution be described?
The birth months of the players do not display central
tendency.
SOLUTION: Sample answer: Since the distribution has two
clusters, an overall summary of center and spread
would give an inaccurate depiction of the data.
Instead, summarize the center and spread of each
cluster individually using its mean and standard
deviation.
ANSWER: Sample answer: Since the distribution has two
clusters, an overall summary of center and spread
would give an inaccurate depiction of the data.
Instead, summarize the center and spread of each
cluster individually using its mean and standard
deviation.
15. OPEN ENDED Find a real-world data set that
appears to represent a symmetric distribution and one
that does not. Describe each distribution. Create a
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visual representation of each set of data.
SOLUTION: ANSWER: Sample answer: The heights of the players on the
Pittsburgh Steelers roster appear to represent aPage 14
normal distribution.
11-2 Distributions of Data
ANSWER: Sample answer: The heights of the players on the
Pittsburgh Steelers roster appear to represent a
normal distribution.
The mean of the data is about 73.61 in. or 6 ft 1.61
in. The standard deviation is about 2.97 in.
The birth months of the players do not display central
tendency.
16. WRITING IN MATH Explain the difference
between positively skewed, negatively skewed, and
symmetric sets of data, and give an example of each.
SOLUTION: Sample answer: The distribution for a data set is
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positively
the majority of the data is on
the left of the mean and a tail appears to the right of
the mean. An example is when the data set includes
16. WRITING IN MATH Explain the difference
between positively skewed, negatively skewed, and
symmetric sets of data, and give an example of each.
SOLUTION: Sample answer: The distribution for a data set is
positively skewed when the majority of the data is on
the left of the mean and a tail appears to the right of
the mean. An example is when the data set includes
the height of everyone in an elementary school, most
of the data will be on the left side (the students),
while a comparatively small amount will be on the
right (the teachers and staff). The distribution for a
data set is negatively skewed when the majority of
the data is on the right of the mean and a tail appears
to the left of the mean. An example is when the
batting averages of a baseball lineup are listed, most
of the data will be at a certain level while the pitcher
will typically be much lower. The distribution for a
data set is symmetric when the data are evenly
distributed on both sides of the mean. An example is
when test scores are calculated for an entire state,
most of the students will place in the middle, while
some will place above or below.
ANSWER: Sample answer: The distribution for a data set is
positively skewed when the majority of the data is on
the left of the mean and a tail appears to the right of
the mean. An example is when the data set includes
the height of everyone in an elementary school, most
of the data will be on the left side (the students),
while a comparatively small amount will be on the
right (the teachers and staff). The distribution for a
data set is negatively skewed when the majority of
the data is on the right of the mean and a tail appears
to the left of the mean. An example is when the
batting averages of a baseball lineup are listed, most
of the data will be at a certain level while the pitcher
will typically be much lower. The distribution for a
data set is symmetric when the data are evenly
distributed on both sides of the mean. An example is
when test scores are calculated for an entire state,
most of the students will place in the middle, while
some will place above or below.
17. DISTRIBUTIONS Which of the following is a
characteristic of a negatively skewed distribution?
A The majority of the data are on the left of the
mean.
B The mean and median are approximately equal.
C The mean is greater than the median.
D The mean is less than the median.
Page 15
SOLUTION: In a negatively skewed distribution, the data to the
left of the median are distributed over a wider range
data set is symmetric when the data are evenly
distributed on both sides of the mean. An example is
when test scores are calculated for an entire state,
of the students
will place in the middle, while
11-2most
Distributions
of Data
some will place above or below.
17. DISTRIBUTIONS Which of the following is a
characteristic of a negatively skewed distribution?
A The majority of the data are on the left of the
mean.
B The mean and median are approximately equal.
C The mean is greater than the median.
D The mean is less than the median.
SOLUTION: In a negatively skewed distribution, the data to the
left of the median are distributed over a wider range
than the data to the right and the mean is less than
the median. Also, the majority of the data are on the
right of the mean. Therefore, the correct choice is D.
ANSWER: 19. SAT/ACT What is the multiplicative inverse of 2i?
F –2i
G –2
H
J
K
SOLUTION: The multiplicative inverse of 2i is
.
ANSWER: D
18. SHORT RESPONSE The average of the test
scores of a class of c students is 80, and the average
test scores of a class of d students is 85. When the
scores of both classes are combined, the average
score is 82. What is the value of
Therefore, option H is the correct answer.
?
SOLUTION: The total marks of c and d students are 80c and 85d.
The combined average score is 82.
ANSWER: H
20. Which equation best represents the graph?
A y = 4x
2
B y = x + 4
C y = 4–x D y = –4x
SOLUTION: –x
The equation of the given graph is y = 4 .
Therefore, option C is the correct answer.
ANSWER: 19. SAT/ACT What is the multiplicative inverse of 2i?
F –2i
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H
ANSWER: C
Determine whether each survey question is
biased. Explain your reasoning.
21. What toppings do you prefer on your pizza?
SOLUTION: The question is not confusing, does not cause a
Page 16
–x
The equation of the given graph is y = 4 .
Therefore, option C is the correct answer.
11-2ANSWER: Distributions of Data
C
formula is correct for n = 1.
Step 2: Assume that after k guests have arrived, a
handshakes have taken place, for total of
Determine whether each survey question is
biased. Explain your reasoning.
21. What toppings do you prefer on your pizza?
SOLUTION: The question is not confusing, does not cause a
strong reaction, does not cause a certain response,
and addresses only one issue. Therefore, it is
unbiased.
some positive integer k .
Step 3: When the (k + 1)st guest arrives, he or she
shakes hands with the k guests already there, so the
total number of handshakes that have then taken
place is
.
ANSWER: unbiased
22. What is your favorite class, and what teacher gives
the easiest homework?
SOLUTION: Biased; sample answer: The question combines two
issues.
ANSWER: Biased; sample answer: The question combines two
issues.
23. Don’t you hate how high gas prices are?
SOLUTION: Biased; sample answer: The question encourages a
certain response. The phrase “Don’t you hate” encourages you to agree that gas prices are too high.
ANSWER: Biased; sample answer: The question encourages a
certain response. The phrase “Don’t you hate” encourages you to agree that gas prices are too high.
24. PARTIES Suppose each time a new guest arrives at
a party, he or she shakes hands with each person
already at the party. Prove that after n guests have
arrived, a total of
The last expression is the formula to be proved,
where n = k + 1. Thus, the formula is true for n = k
+ 1. Therefore, the total number of handshakes is
for all positive integers n.
ANSWER: Step 1: After the first guest has arrived, no
handshakes have taken place.
, so the
formula is correct for n = 1.
Step 2: Assume that after k guests have arrived, a
handshakes have taken place, for total of
some positive integer k .
Step 3: When the (k + 1)st guest arrives, he or she
shakes hands with the k guests already there, so the
total number of handshakes that have then taken
place is
.
handshakes have taken place.
SOLUTION: Step 1: After the first guest has arrived, no
handshakes have taken place.
, so the
formula is correct for n = 1.
Step 2: Assume that after k guests have arrived, a
total of
handshakes have taken place, for some positive integer k .
Step 3: When the (k + 1)st guest arrives, he or she
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shakes
hands
with the
k guests already there, so the
total number of handshakes that have then taken
place is
The last expression is the formula to be proved,
where n = k + 1. Thus, the formula is true for n = k
+ 1. Therefore, the total number of handshakes is
for all positive integers n.
Page 17
25. ASTRONOMY The orbit of Pluto can be modeled
by the equation
, where the units are
The last expression is the formula to be proved,
where n = k + 1. Thus, the formula is true for n = k
+ 1. Therefore, the total number of handshakes is
11-2 Distributions of Data
for all positive integers n.
25. ASTRONOMY The orbit of Pluto can be modeled
by the equation
, where the units are
astronomical units. Suppose a comet is following a
2
path modeled by the equation x = y + 20.
a. Find the point(s) of intersection of the orbits of
Pluto and the comet.
b. Will the comet necessarily hit Pluto? Explain.
2
2
c. Where do the graphs of y = 2x + 1 and 2x + y =
11 intersect?
d. What are the coordinates of the points that lie on
2
2
2
2
the graphs of both x + y = 25 and 2x + 3y = 66?
SOLUTION: d. Graph the equations x2 + y 2 = 25 and 2x2 + 3y 2 =
66.
The intersection points are (3, ±4) and (–3, ±4).
ANSWER: a. (39.2, ±4.4)
b. No; the comet and Pluto may not be at either point
of intersection at the same time.
c.
d. (3, ±4), (–3, ±4)
Determine whether each situation involves a
permutation or a combination. Then find the
number of possibilities.
26. the winner of the first, second, and third runners-up
in a contest with 8 finalists
SOLUTION: Permutation.
2
a. Solve the second equation for y .
2
Substitute the x – 20 for y and solve for x.
ANSWER: permutation; 336
27. selecting two of eight employees to attend a business
seminar
Substitute 39.4 for x in the second equation and solve
for y.
SOLUTION: Combination
The points of intersection are
.
b. No; the comet and Pluto may not be at either point
of intersection at the same time.
2
2
c. Graph the equations y = 2x + 1 and 2x + y = 11.
The intersection points are
.
d. Graph the equations x2 + y 2 = 25 and 2x2 + 3y 2 =
66.
The intersection points are (3, ±4) and (–3, ±4).
ANSWER: a. (39.2, ±4.4)
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b. No; the comet and Pluto may not be at either point
of intersection at the same time.
ANSWER: combination; 28
28. an arrangement of the letters in the word MATH
SOLUTION: Permutation.
Page 18
11-2ANSWER: Distributions of Data
combination; 28
28. an arrangement of the letters in the word MATH
SOLUTION: Permutation.
ANSWER: permutation; 24
29. placing an algebra book, a geometry book, a
chemistry book, an English book, and a health book
on a shelf
SOLUTION: Permutation.
ANSWER: permutation; 120
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Page 19
The random variable X is the number of cars passing
through an intersection. The cars are countable, so X
is discrete.
11-3 Probability Distributions
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
1. the number of pages linked to a Web page
SOLUTION: The random variable X is the number of pages linked
to a Web page. The pages are countable, so X is
discrete.
ANSWER: The random variable X is the number of pages linked
to a Web page. The pages are countable, so X is
discrete.
2. the number of stations in a cable package
SOLUTION: The random variable X is the number of stations in a
cable package. The stations are countable, so X is
discrete.
ANSWER: The random variable X is the number of stations in a
cable package. The stations are countable, so X is
discrete.
3. the amount of precipitation in a city per month
SOLUTION: The random variable X is the amount of precipitation
in a city per month. Precipitation can be anywhere
within a certain range. Therefore, X is continuous.
5. X represents the sum of the values of two spins of
the wheel.
a. Construct a relative-frequency table showing the
theoretical probabilities.
b. Graph the theoretical probability distribution.
c. Construct a relative-frequency table for 100 trials.
d. Graph the experimental probability distribution.
e. Find the expected value for the sum of two spins
of the wheel.
f. Find the standard deviation for the sum of two
spins of the wheel.
SOLUTION: a.
There are 64 total outcomes possible for two spins
of the wheel. List each possible sum in the first
column. List the number of times the sum can
occur in the second column. For example, a sum of
4 can only occur when a 2 is spun both times and a
sum of 6 can occur when a 2 and 4 are spun or
when a 4 and 2 are spun. To determine the relative
frequency, or theoretical probability, of each
outcome, divide the frequency by 64.
ANSWER: The random variable X is the amount of precipitation
in a city per month. Precipitation can be anywhere
within a certain range. Therefore, X is continuous.
4. the number of cars passing through an intersection in
a given time interval
SOLUTION: The random variable X is the number of cars passing
through an intersection. The cars are countable, so X
is discrete.
ANSWER: The random variable X is the number of cars passing
through an intersection. The cars are countable, so X
is discrete.
5. X represents the sum of the values of two spins of
the wheel.
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Page 1
4 can only occur when a 2 is spun both times and a
sum of 6 can occur when a 2 and 4 are spun or
when a 4 and 2 are spun. To determine the relative
or Distributions
theoretical probability, of each
11-3frequency,
Probability
outcome, divide the frequency by 64.
c. Create a spinner equivalent to the one in the book
or use a random number generator to complete the
simulation and create a simulation tally sheet. Then
determine the frequency and relative frequency.
b. The graph shows the probability distribution for the
sum of two spins X. The bars are separated on the
graph because the distribution is discrete (no other
values of X are possible).
d.
c. Create a spinner equivalent to the one in the book
or use a random number generator to complete the
simulation and create a simulation tally sheet. Then
determine the frequency and relative frequency.
e. Using the theoretical probability distribution,
multiply each sum by the corresponding relative
frequency. Then find the sum of those values. Note
the last column is the sums converted to a common
denominator.
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Page 2
e. Using the theoretical probability distribution,
multiply each sum by the corresponding relative
frequency. Then find the sum of those values. Note
last column
is the sums converted to a common
11-3the
Probability
Distributions
denominator.
subtract the expected value from each outcome.
Square each difference. Then multiply by the each
corresponding probability. Finally, find the sum of
these values. The values in the last column are
rounded to the nearest ten-thousandth.
The expected value is 13.5.
f. Using the table from part e and E(X) = 13.5,
subtract the expected value from each outcome.
Square each difference. Then multiply by the each
corresponding probability. Finally, find the sum of
these values. The values in the last column are
rounded to the nearest ten-thousandth.
The standard deviation is about 4.29.
ANSWER: a.
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Page 3
The standard deviation is about 4.29.
11-3 Probability Distributions
ANSWER: a.
c.
b.
c.
d. e. 13.5
f. 4.29
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Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
Page 4
d. 11-3e.Probability Distributions
13.5
f. 4.29
10. CCSS PERSEVERANCE A contestant has won a
prize on a game show. The frequency table at the
right shows the number of winners for 3200
hypothetical players.
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
6. the number of texts received per week
SOLUTION: The random variable X is the number of texts per
week. The texts are countable, so X is discrete.
ANSWER: The random variable X is the number of texts per
week. The texts are countable, so X is discrete.
7. the number of diggs (or “likes”) for a Web page
SOLUTION: The random variable X is the number of diggs for a
web page. The diggs are countable, so X is discrete.
ANSWER: The random variable X is the number of diggs for a
web page. The diggs are countable, so X is discrete.
8. the height of a plant after a specific amount of time
SOLUTION: The random variable X is the height of a plant.
Height can be anywhere within a certain range.
Therefore, X is continuous.
a. Construct a relative-frequency table showing the
theoretical probability.
b. Graph the theoretical probability distribution.
c. Construct a relative-frequency table for 50 trials.
d. Graph the experimental probability distribution.
e. Find the expected value.
f. Find the standard deviation.
SOLUTION: a.
There are 3200 total outcomes. List each possible
prize in the first column. The relative frequency is the
number of winners divided by 3200. This is also the
associated probability P(X). List this in the second
column.
ANSWER: The random variable X is the height of a plant.
Height can be anywhere within a certain range.
Therefore, X is continuous.
9. the number of files infected by a computer virus
SOLUTION: The random variable X is the number of files infected
by a computer virus. The files are countable, so X is
discrete.
ANSWER: The random variable X is the number of files infected
by a computer virus. The files are countable, so X is
discrete.
10. CCSS PERSEVERANCE A contestant has won a
prize on a game show. The frequency table at the
right shows the number of winners for 3200
hypothetical players.
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b. The graph shows the probability distribution for the
prize amounts X. The bars are separated on the
graph because the distribution is discrete (no other
values of X are possible).
Page 5
The graph shows
the probability distribution for the
11-3b.
Probability
Distributions
prize amounts X. The bars are separated on the
graph because the distribution is discrete (no other
values of X are possible).
e . Using the theoretical probability distribution,
multiply each prize value by the corresponding
relative frequency P(X). Then find the sum of those
values.
c. Use a random number generator to complete the
simulation and create a simulation tally sheet. Let 135 represent $100, 36-60 represent $250, 61-75
represent $500, and so on. On your graphing
calculator, select randInt(1, 100, 50) for 50 trials.
Then determine the frequency and relative
frequency. The relative frequency is the frequency
divided by the number of trials, 50.
The expected value is $922.50.
f. Using the table from part e and E(X) = 922.5,
subtract the expected value from each outcome.
Square each difference. Then multiply by each
corresponding probability. Finally, find the sum of
these values. The values in the last column are
rounded to the nearest hundredth.
The standard deviation is about 1711.91.
ANSWER: a. d.
e . Using the theoretical probability distribution,
corresponding
relative frequency P(X). Then find the sum of those
values.
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multiply
each
prize value
by the
Page 6
The standard deviation is about 1711.91.
11-3 Probability Distributions
ANSWER: a. d. e. $922.50
f. 1711.91
11. SNOW DAYS The following probability distribution
lists the probable number of snow days per school
year at North High School. Use this information to
determine the expected number of snow days per
year.
b. SOLUTION: Find the sum of the weighted values of each variable.
c. E(X) = Σ[X · P(X)] = 3.34
ANSWER: 3.34
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12. CARDS In a standard deck of 52 cards, there are 4
different suits.
a. If jacks = 11, queens = 12, kings = 13, and aces =
1, what is the expected value of a card that is drawn
from a standard deck?
b. If you are dealt 7 cards with replacement, what is
the expected number of spades?
Page 7
SOLUTION: a. The relative frequency P(X) of each card value is
12. CARDS In a standard deck of 52 cards, there are 4
different suits.
a. If jacks = 11, queens = 12, kings = 13, and aces =
what is the Distributions
expected value of a card that is drawn
11-31,Probability
from a standard deck?
b. If you are dealt 7 cards with replacement, what is
the expected number of spades?
SOLUTION: a. The relative frequency P(X) of each card value is
or . Multiply each card value by the
corresponding relative frequency P(X). Then find the
sum of those values.
b. 1.75
13. RAFFLES The table shows the probability distributi
for a raffle if 100 tickets are sold for $1 each. There
1 prize for $20, 5 prizes for $10, and 10 prizes for $5.
a. Graph the theoretical probability distribution.
b. Find the expected value.
c. Interpret the results you found in part b. What can
you conclude about the raffle?
SOLUTION: a. Use the given probabilities to create the distributio
The graph shows the probability distribution for the
prize amounts X. The bars are separated on the grap
because the distribution is discrete (no other values o
X are possible).
.
b. List each prize value X along with the correspondi
relative frequency P(X). Find X ∙ P(X). Then find the
sum of those values.
The expected value is
or 7.
b. Each time you get a card, there is a probability of
that it will be a spade. Since the card is being
replaced each time, this probability will be the same
each time you get a card. Therefore, if you are given
7 cards, then the expected number of spades is 7 ∙ or 1.75.
ANSWER: a. 7
b. 1.75
13. RAFFLES The table shows the probability distributi
for a raffle if 100 tickets are sold for $1 each. There
1 prize for $20, 5 prizes for $10, and 10 prizes for $5.
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X
0
20
10
5
P(X)
0.84
0.01
0.05
0.1
Total
XP(X)
0
0.2
0.5
0.5
1.20
The expected value of winnings is $1.20.
c. Sample answer: The expected value is positive, so
person buying a ticket can expect to win $0.20 even
after the cost of the ticket is considered. Thus, a pers
would want to participate in this raffle. On the other
hand, this raffle is guaranteed to lose money for the
organizers and they should change the distribution of
prizes or not do the raffle.
ANSWER: a.
Page 8
after the cost of the ticket is considered. Thus, a pers
would want to participate in this raffle. On the other
hand, this raffle is guaranteed to lose money for the
11-3organizers
Probability
andDistributions
they should change the distribution of
prizes or not do the raffle.
SOLUTION: a. List each value X along with the corresponding
relative frequency P(X). Find X ∙ P(X). Then find the
sum of those values.
ANSWER: a.
The expected number is 4.2, so we can expect there
to be 4 students running. We cannot have 0.2 people,
so we round to the nearest whole number.
b. $1.20
c. Sample answer: The expected value is positive, so
person buying a ticket can expect to win $0.20 even
after the cost of the ticket is considered. Thus, a pers
would want to participate in this raffle. On the other
hand, this raffle is guaranteed to lose money for the
organizers and they should change the distribution of
prizes or not do the raffle.
b. Use a random number generator to complete the
simulation and create a simulation tally sheet. Let 1-5
represent 1 student, 6-20 represent 2 students, 21-30
represent 3 students, 31-45 represent 4 students, 4680 represent 5 students, and 81-100 represent 6
students. Then determine the frequency and relative
frequency.
14. CCSS TOOLS Based on previous data, the
probability distribution of the number of students
running for class president is shown.
c. Use the given probabilities to create the
distribution. The graph shows the probability
distribution for the number of students X. The bars
are separated on the graph because the distribution is
discrete (no other values of X are possible).
a. Determine the expected number of students who
will run. Interpret your results.
b. Construct a relative-frequency table for 50 trials.
c. Graph the experimental probability distribution.
SOLUTION: a. List each value X along with the corresponding
relative frequency P(X). Find X ∙ P(X). Then find the
sum of those values.
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The expected number is 4.2, so we can expect there
ANSWER: a. 4.2; Sample answer: The expected number is 4.2,
so we can expect there to be 4 students running. We
cannot have 0.2 people, so we round to the nearest
Page 9
whole number.
b. Sample answer:
11-3ANSWER: Probability Distributions
a. 4.2; Sample answer: The expected number is 4.2,
so we can expect there to be 4 students running. We
cannot have 0.2 people, so we round to the nearest
whole number.
b. Sample answer:
SOLUTION: a. Multiply the value of X by the corresponding
relative frequency P(X). Then find the sum of those
values. Note the last column is the sums converted to
a common denominator.
c. Sample answer:
The expected number of upsets is
or about 4.34.
15. BASKETBALL The distribution below lists the
probability of the number of major upsets in the first
round of a basketball tournament each year.
b. Using the table from part a and E(X) =
,
subtract the expected value from each outcome.
Square each difference. Then multiply by each
corresponding probability. Finally, find the sum of
these values. The values in the third and column are
rounded to the nearest ten-thousandth. You can
avoid calculation errors by using a graphing
calculator. Enter the first column as L1 and the
second column as L2. Set L3 equal to [L1 –
2
a. Determine the expected number of upsets.
Interpret your results.
b. Find the standard deviation.
c. Construct a relative-frequency table for 50 trials.
d. Graph the experimental probability distribution.
(139/32)] . Set L4 equal to L3 ∙ L2. Then find the
sum of L4.
SOLUTION: a. Multiply the value of X by the corresponding
relative frequency P(X). Then find the sum of those
values. Note the last column is the sums converted to
a common denominator.
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0
Page 10
c. Use a random number generator to complete the
simulation and create a simulation tally sheet. One
11-3 Probability Distributions
d. Use the given probabilities to create the
distribution. The graph shows the probability
distribution for the number of upsets X. The bars are
separated on the graph because the distribution is
discrete (no other values of X are possible).
0
c. Use a random number generator to complete the
simulation and create a simulation tally sheet. One
challenge is identifying what numbers to use. First,
convert P(X) to a common denominator for each
value of X. Then we can let the numerators dictate
the values that represent each number of upsets.
ANSWER: a. 4.34; Sample answer: The expected number is
4.34, so we can expect there to be 4 upsets. We
cannot have 0.34 upsets, so we round to the nearest
whole number.
b. 1.90
c. On your graphing calculator, select randInt(1, 32,
50) for 50 trials. Then determine the frequency and
relative frequency.
d. d. Use the given probabilities to create the
distribution. The graph shows the probability
distribution for the number of upsets X. The bars are
separated on the graph because the distribution is
discrete (no other values of X are possible).
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16. RAFFLES The French Club sold 500 raffle tickets
for $1 each. The first prize ticket will win $100, 2
second prize tickets will each win $10, and 5 third
prize tickets each win $5.
a. What is the expected value of a single ticket?
b. Calculate the standard deviation of the probability
distribution.
c. DECISION MAKING The Glee Club is offering
a raffle with a similar expected value and a standard
deviation of 2.2. In which raffle should you
Page 11
participate? Explain your reasoning.
SOLUTION: prize tickets each win $5.
a. What is the expected value of a single ticket?
b. Calculate the standard deviation of the probability
11-3distribution.
Probability Distributions
c. DECISION MAKING The Glee Club is offering
a raffle with a similar expected value and a standard
deviation of 2.2. In which raffle should you
participate? Explain your reasoning.
SOLUTION: a. List each prize value X along with the
corresponding relative frequency P(X). Note that
each raffle ticket costs $1, so the actual prize values
are the listed values minus $1. The probability of not
winning a prize is one minus the sum of the
probabilities of all of the other prizes. Find X ∙ P(X).
Then find the sum of those values.
expected values, the riskier raffle will also have the
potential to win more, so both raffles have good and
bad qualities. It is up to the individual participant to
decide which one to choose.
ANSWER: a. -0.71
b. 4.53
c. Sample answer: The standard deviation of the
probability distribution for the Glee Club raffle is
about half the standard deviation for the French Club
raffle, so the Glee Club raffle is less risky. Since they
have similar expected values, the riskier raffle will
also have the potential to win more, so both raffles
have good and bad qualities. It is up to the individual
participant to decide which one to choose.
17. DECISION MAKING Carmen is thinking about
investing $10,000 in two different investment funds.
The expected rates of return and the corresponding
probabilities for each fund are listed below. Compare
the two investments using the expected value and
standard deviation. Which investment would you
advise Carmen to choose, and why?
The expected value is –$0.71.
b. Using the table from part a and E(X) = –0.71,
subtract the expected value from each outcome.
Square each difference. Then multiply by each
corresponding probability. Finally, find the sum of
these values. The values in the last column are
rounded to the nearest hundredth.
c. The standard deviation of the probability
distribution for the Glee Club raffle is about half the
standard deviation for the French Club raffle, so the
Glee Club raffle is less risky. Since they have similar
expected values, the riskier raffle will also have the
potential to win more, so both raffles have good and
bad qualities. It is up to the individual participant to
decide which one to choose.
ANSWER: a. -0.71
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b. 4.53
c. Sample answer: The standard deviation of the
probability distribution for the Glee Club raffle is
SOLUTION: Find the expected value of each investment. Multiply
the possible value of each fund by the associated
probability. A profit is a positive value while a loss is
a negative value.
Fund A:
Fund B:
Page 12
11-3 Probability Distributions
Fund B:
The expected values of Funds A and B are $595 and
$540, respectively. Calculate each standard
deviation.
Sample answer: The expected value of Funds A and
B is $595 and $540, respectively. The standard
deviation for Fund A is about 951.6, while the
standard deviation for Fund B is about 941.5. Since
the standard deviations are about the same, the funds
have about the same amount of risk. Therefore, with
a higher expected value, Fund A is the better
investment.
18. MULTIPLE REPRESENTATIONS In this
problem, you will investigate geometric probability.
a. Tabular The spinner shown has a radius of 2.5
inches. Copy and complete the table below.
Fund A:
Fund B:
The expected value of Funds A and B is $595 and
$540, respectively. The standard deviation for Fund
A is about 951.6, while the standard deviation for
Fund B is about 941.5. Since the standard deviations
are about the same, the funds have about the same
amount of risk. Therefore, with a higher expected
value, Fund A is the better investment.
ANSWER: Sample answer: The expected value of Funds A and
B is $595 and $540, respectively. The standard
deviation for Fund A is about 951.6, while the
standard deviation for Fund B is about 941.5. Since
the standard deviations are about the same, the funds
have about the same amount of risk. Therefore, with
a higher expected value, Fund A is the better
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investment.
18. MULTIPLE REPRESENTATIONS In this
b. Verbal Make a conjecture about the relationship
between the ratio of the area of the sector to the
total area and the probability of the spinner landing on
each color.
c. Analytical Consider the dartboard shown.
Predict the probability of a dart landing in each area
of the board. Assume that any dart thrown will land
on the board and is equally likely to land at any point
on the board.
d. Tabular Construct a relative-frequency table for
throwing 100 darts.
e. Graphical Graph the experimental probability
distribution.
SOLUTION: a. The green and blue sectors take up
of the circle
each while the other colors take up each. These
values represent the probabilities of the spinner
landing on each color. The area of the circle is π(2.5)
2
2
Pageto
13
or about 19.63 in . Multiply this by the probability
find each sector area.
on the outer ring.
SOLUTION: a. The green and blue sectors take up
of the circle
while theDistributions
other colors take up each. These
11-3each
Probability
values represent the probabilities of the spinner
landing on each color. The area of the circle is π(2.5)
2
2
or about 19.63 in . Multiply this by the probability to
find each sector area.
On your graphing calculator, select randInt(1, 9,
100) for 100 trials. Then determine the frequency
and relative frequency.
b. The second column is equal to the fifth column, so
the probability is equal to the ratio of the sector area
to the total area.
e. Use the given probabilities to create the
distribution. The graph shows the probability
distribution for the number of landings X. The bars
are separated on the graph because the distribution is
discrete (no other values of X are possible).
c. The area of the entire circle is
area of the green circle is
or 81π. The
or 9π. The
probability of the dart landing on the green circle is
the area of the green circle divided by the total area:
or
.
The area of the yellow ring is the area of the circle
formed by the yellow and green sections minus the
area of the green circle. The area of the bigger circle
is
or 36π. So, the area of the yellow ring is
36π – 9π or 27π. The probability of the dart landing
on the yellow ring is the area of the yellow ring
divided by the total area:
or .
ANSWER: a.
The probability of the dart landing on the blue ring is
1 minus the sum of the probabilities of it landing on
the other colors.
d. Use a random number generator to complete the
simulation and create a simulation tally sheet. Let 1
represent landing on the green circle, 2-4 represent
landing on the yellow ring, and 5-9 represent landing
on the outer ring.
b. Sample answer: The probability is equal to the
ratio of the sector area to the total area. c. green: ; yellow: ; blue:
d.
On your graphing calculator, select randInt(1, 9,
100) for 100 trials. Then determine the frequency
and relative frequency.
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e.
Page 14
11-3 Probability Distributions
Sample answer: Liana; Shannon didn’t consider
every scenario in determining the total probability.
For example, in calculating the probability of a sum of
5, she considered spinning a 3 then a 2, but not a 2
then a 3.
e.
20. REASONING Determine whether the following
statement is true or false. Explain.
If you roll a die 10 times, you will roll the
expected value at least twice.
SOLUTION: Sample answer: False; the expected value is 3.5
which is not a possible outcome of a single roll.
19. CCSS CRITIQUE Liana and Shannon each
created a probability distribution for the sum of two
spins on the spinner. Is either of them correct?
Explain your reasoning.
ANSWER: Sample answer: False; the expected value is 3.5
which is not a possible outcome of a single roll.
21. OPEN ENDED Create a discrete probability
distribution that shows five different outcomes and
their associated probabilities.
SOLUTION: Ensure that each outcome is independent of the
others and that the sum of the probabilities of the
outcomes is one. Sample answer: A spinner with 5
equal-sided areas shaded red, blue, yellow, green,
and brown.
ANSWER: Sample answer: A spinner with 5 equal-sided areas
shaded red, blue, yellow, green, and brown.
SOLUTION: Sample answer: Liana; Shannon didn’t consider
every scenario in determining the total probability.
For example, in calculating the probability of a sum of
5, she considered spinning a 3 then a 2, but not a 2
then a 3.
ANSWER: Sample answer: Liana; Shannon didn’t consider
every scenario in determining the total probability.
For example, in calculating the probability of a sum of
5, she considered spinning a 3 then a 2, but not a 2
then a 3.
20. REASONING Determine whether the following
statement is true or false. Explain.
If you roll a die 10 times, you will roll the
expected value at least twice.
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Sample answer: False; the expected value is 3.5
which is not a possible outcome of a single roll.
22. REASONING Determine whether the following
statement is true or false. Explain.
Random variables that can take on an infinite
number of values are continuous.
SOLUTION: Sample answer: False; a random variable X
representing the number of Web sites on the Internet
at any given time is infinite and countable. However,
there cannot be 1.3 Web sites, so it is also discrete.
ANSWER: Sample answer: False; a random variable X
representing the number of Web sites on the Internet
at any given time is infinite and countable. However,
there cannot be 1.3 Web sites, so it is also discrete.
23. OPEN ENDED Provide examples of a discrete
probability distribution and a continuous probability
distribution. Describe the differences between them.
Page 15
SOLUTION: Sample answer: A discrete probability distribution
ANSWER: Sample answer: False; a random variable X
representing the number of Web sites on the Internet
any given time
is infinite and countable. However,
11-3atProbability
Distributions
there cannot be 1.3 Web sites, so it is also discrete.
23. OPEN ENDED Provide examples of a discrete
probability distribution and a continuous probability
distribution. Describe the differences between them.
SOLUTION: Sample answer: A discrete probability distribution
can be the uniform distribution of the roll of a die. In
this type of distribution, there are only a finite number
of possibilities. A continuous probability distribution
can be the distribution of the lives of 400 batteries. In
this distribution, there are an infinite number of
possibilities.
ANSWER: Sample answer: A discrete probability distribution
can be the uniform distribution of the roll of a die. In
this type of distribution, there are only a finite number
of possibilities. A continuous probability distribution
can be the distribution of the lives of 400 batteries. In
this distribution, there are an infinite number of
possibilities.
24. WRITING IN MATH Compare and contrast two
investments that have identical expected values and
significantly different standard deviations.
SOLUTION: Sample answer: Since the investments have identical
expected values, an investor would expect to earn
the same amount of money on each investment.
However, since they have significantly different
standard deviations, the investment with the higher
standard deviation is much riskier than the other
investment. The greater standard deviation indicates
a greater variability, so the riskier investment will
have an opportunity to earn more money than the
other investment, but it will also have an opportunity
to lose more as well.
ANSWER: Sample answer: Since the investments have identical
expected values, an investor would expect to earn
the same amount of money on each investment.
However, since they have significantly different
standard deviations, the investment with the higher
standard deviation is much riskier than the other
investment. The greater standard deviation indicates
a greater variability, so the riskier investment will
have an opportunity to earn more money than the
other investment, but it will also have an opportunity
to lose more as well.
25. GRIDDED RESPONSE The height f (x) of a
bouncing ball after x bounces is represented by f (x) =
eSolutions Manual - Powered by Cognero
x
140(0.8) . How many times higher is the first bounce
than the fifth bounce?
investment. The greater standard deviation indicates
a greater variability, so the riskier investment will
have an opportunity to earn more money than the
other investment, but it will also have an opportunity
to lose more as well.
25. GRIDDED RESPONSE The height f (x) of a
bouncing ball after x bounces is represented by f (x) =
x
140(0.8) . How many times higher is the first bounce
than the fifth bounce?
SOLUTION: Substitute 1 and 5 for x and evaluate.
Therefore, the first bounce is about 2.4 times higher
than the fifth bounce.
ANSWER: 2.4
26. PROBABILITY Andres has a bag that contains 4
red, 6 yellow, 2 blue, and 4 green marbles. If he
reaches into the bag and removes a marble without
looking, what is the probability that it will not be
yellow?
A
B
C
D
SOLUTION: The probability that it will not be yellow = 1 – The
probability that it will be yellow
The probability that it will be yellow is
The probability that it will not be yellow =
.
.
Option D is the correct answer.
ANSWER: D
27. GEOMETRY Find the area of the shaded portion of
Page 16
the figure to the nearest square inch.
The probability that it will not be yellow =
.
Option D is the correct answer.
11-3ANSWER: Probability Distributions
D
27. GEOMETRY Find the area of the shaded portion of
the figure to the nearest square inch.
F 79
G 94
H 589
J 707
E
29. ARTICLES Peter and Paul each write articles for
an online magazine. Their employer keeps track of
the number of likes received by each article.
a. Use a graphing calculator to create a histogram
for each data set. Then describe the shape of each
distribution.
b. Compare the distributions using either the means
and standard deviations or the five-number
summaries. Justify your choice.
SOLUTION: The central angle of the shaded region is 300º.
The area of a sector is
.
The area of the shaded region is
2
in .
Option H is the correct answer.
ANSWER: H
SOLUTION: a.
Peter’s Articles:
First, press STAT ENTER and enter each data
value. Then, press 2ND [STAT PLOT] ENTER
ENTER and choose „ . Finally, adjust the window to
the dimensions shown.
28. SAT/ACT If x and y are positive integers, which of
the following expressions is equivalent to
A
B
C
D
E
SOLUTION: The majority of the data are on the left of the mean,
so the distribution is positively skewed.
Option E is the correct answer.
Paul’s Articles:
First, press STAT ENTER and enter each data
value in L2. Then, press 2ND [STAT PLOT]
ENTER ENTER and choose „ . Finally, adjust the
window to the dimensions shown.
ANSWER: E
29. ARTICLES Peter and Paul each write articles for
an online magazine. Their employer keeps track of
the number of likes received by each article.
a. Use a graphing calculator to create a histogram
for each
data
set. Then
describe the shape of each
eSolutions
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distribution.
b. Compare the distributions using either the means
and standard deviations or the five-number
Page 17
The data are evenly distributed about the median, so
Paul’s Articles:
First, press STAT ENTER and enter each data
value in L2. Then, press 2ND [STAT PLOT]
ENTER
and choose „ . Finally, adjust the
11-3 ENTER
Probability
Distributions
window to the dimensions shown.
more likes (and are more popular) than 75% of
Peter’s articles. Therefore, we can conclude that
Paul’s articles are more popular overall.
ANSWER: a.
The data are evenly distributed about the median, so
the distribution is symmetric.
b. One distribution is skewed, so use the five-number
summaries.
For the first distribution, press STAT ► ENTER
ENTER and scroll down to view the five-number
summary. For the second distribution, press STAT
► ENTER ENTER L2 and scroll down to view the
five-number summary.
Peter’s Articles:
Paul’s Articles:
Peter’s articles, positively skewed; Paul’s articles,
symmetric
b. Sample answer: One of the distributions is
symmetric and the other is skewed, so use the fivenumber summaries. The range for Peter’s articles is
64, and the range for Paul’s articles is 53. However,
the upper quartile for Peter’s is 33, while the lower
quartile for Paul’s is 34. This means that 75% of
Paul’s articles have more likes (and are more
popular) than 75% of Peter’s articles. Therefore, we
can conclude that Paul’s articles are more popular
overall.
Determine whether the situation calls for a
survey, an observational study, or an experiment.
Explain your reasoning.
30. You want to test a drug that reverses male pattern
baldness.
The range for Peter’s articles is 64, and the range for
Paul’s articles is 53. However, the upper quartile for
Peter’s is 33, while the lower quartile for Paul’s is
34. This means that 75% of Paul’s articles have
more likes (and are more popular) than 75% of
Peter’s articles. Therefore, we can conclude that
Paul’s articles are more popular overall.
ANSWER: a.
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SOLUTION: This is an experiment because the sample will be
divided into groups in which one group will be given
blueberries while the other group will be given a
placebo. The participants are affected by the study.
ANSWER: Sample answer: This situation calls for an experiment
because the treatment will be tested on a sample
group, which means that the members of the sample
will be affected by the study.
31. You want to find voters’ opinions on recent
legislation.
SOLUTION: Page 18
The first five terms are 0.125, 0.1875, 0.28125,
0.421875 and 0.632813.
Sample answer: This situation calls for an experiment
because the treatment will be tested on a sample
group, which means that the members of the sample
11-3will
Probability
Distributions
be affected
by the study.
31. You want to find voters’ opinions on recent
legislation.
ANSWER: 0.125, 0.1875, 0.28125, 0.421875, 0.632813
33. = 0.5, r = 2.5
SOLUTION: SOLUTION: This is a survey because data are collected from
responses to the question. ANSWER: Sample answer: This situation calls for a survey
because the data will be collected from responses
from members of a sample of the population.
Find the first five terms of each geometric
sequence described.
32. = 0.125, r = 1.5
SOLUTION: The first five terms are 0.5, 1.25, 3.125, 7.8125 and
19.53125.
ANSWER: 0.5, 1.25, 3.125, 7.8125, 19.53125
34. = 4, r = 0.5
SOLUTION: The first five terms are 0.125, 0.1875, 0.28125,
0.421875 and 0.632813.
ANSWER: 0.125, 0.1875, 0.28125, 0.421875, 0.632813
33. = 0.5, r = 2.5
SOLUTION: The first five terms are 4, 2, 1, 0.5 and 0.25.
ANSWER: 4, 2, 1, 0.5, 0.25
35. = 12,
SOLUTION: eSolutions Manual - Powered by Cognero
The first five terms are 0.5, 1.25, 3.125, 7.8125 and
19.53125.
Page 19
The first five terms are 4, 2, 1, 0.5 and 0.25.
ANSWER: 11-3ANSWER: Probability Distributions
4, 2, 1, 0.5, 0.25
35. 36. = 12,
= 21,
SOLUTION: SOLUTION: The first five terms are 12, 4,
36. 12, 4,
.
The first five terms are 21, 14,
ANSWER: ANSWER: 12, 4,
21, 14,
= 21,
SOLUTION: eSolutions Manual - Powered by Cognero
37. .
= 80,
SOLUTION: Page 20
ANSWER: ANSWER: 11-3 Probability Distributions
21, 14,
37. 80, 100, 125,
38. COMMUNICATION A microphone is placed at
the focus of a parabolic reflector to collect sound for
the television broadcast of a football game. Write an
equation for the cross section, assuming that the
focus is at the origin, the focus is 6 inches from the
vertex, and the parabola opens to the right.
= 80,
SOLUTION: SOLUTION: Given:
The focus of the parabola is (0, 0).
That is:
The focus is 6 inches from the vertex and the
parabola opens to the right.
Therefore, the vertex of the parabola (h, k) is (–6,
0).
Find the value of a.
The first five terms are 80, 100, 125,
.
ANSWER: The standard form of a parabolic equation open to
80, 100, 125,
the right is
.
Substitute the corresponding values.
38. COMMUNICATION A microphone is placed at
the focus of a parabolic reflector to collect sound for
the television broadcast of a football game. Write an
equation for the cross section, assuming that the
focus is at the origin, the focus is 6 inches from the
vertex, and the parabola opens to the right.
SOLUTION: Given:
The focus of the parabola is (0, 0).
That is:
ANSWER: Solve each equation. Check your solutions.
x=
39. The focus is 6 inches from the vertex and the
parabola opens to the right.
Therefore, the vertex of the parabola (h, k) is (–6,
0).
eSolutions
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the value
of a. by Cognero
SOLUTION: Page 21
ANSWER: 11-3xProbability
Distributions
=
Solve each equation. Check your solutions.
39. ANSWER: 3
Expand each power.
3
42. (a – b)
SOLUTION: SOLUTION: ANSWER: 3
2
43. (m + n)
4
2
a – 3a b + 3ab – b
3
SOLUTION: ANSWER: 27
ANSWER: 4
3
2 2
3
m + 4m n + 6m n + 4mn + n
40. 44. (r + n)
SOLUTION: 4
8
SOLUTION: ANSWER: 8
7
6 2
5 3
4 4
3 5
r + 8r n + 28r n + 56r n + 70r n + 56r n
2 6
7
8
+ 28r n + 8rn + n
ANSWER: 1000
41. SOLUTION: ANSWER: 3
Expand each power.
3
42. (a – b)
SOLUTION: eSolutions Manual - Powered by Cognero
ANSWER: 3
2
2
a – 3a b + 3ab – b
3
Page 22
11-4 The Binomial Distribution
Determine whether each experiment is a
binomial experiment or can be reduced to a
binomial experiment. If so, describe a trial,
determine the random variable, and state n, p ,
and q.
1. A study finds that 58% of people have pets. You ask
100 people how many pets they have.
SOLUTION: People can have 0, 1, 2, or more pets. There are
more than two possible outcomes, so this cannot be
reduced to a binomial experiment. If the question
was changed to Do you own a pet? then we could
do a binomial experiment.
4. Conduct a binomial experiment to determine the
probability of drawing an ace or a king from a deck
of cards. Then compare the experimental and
theoretical probabilities of the experiment.
SOLUTION: Sample answer:
Step 1 A trial is drawing a card from a deck. The
simulation will consist of 20 trials.
Step 2 A success is drawing an ace or a king. The
probability of success is
and the probability of
failure is .
Step 3 The random variable X represents the number
of aces or kings drawn in 20 trials.
Step 4 Use a random number generator. Let 0−1
represent drawing an ace or a king. Let 2−12
represent drawing all other cards. Make a frequency
table and record the results as you run the generator.
ANSWER: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
2. You roll a die 15 times and find the sum of all of the
rolls.
SOLUTION: There are more than two possible sums, so this
cannot be reduced to a binomial experiment.
ANSWER: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
3. A poll found that 72% of students plan on going to
the homecoming dance. You ask 30 students if they
are going to the homecoming dance.
SOLUTION: This experiment can be reduced to a binomial
experiment. Success is yes, failure is no, a trial is
asking a student, and the random variable is the
number of yeses; n = 30, p = 0.72, q = 0.28.
ANSWER: This experiment can be reduced to a binomial
experiment. Success is yes, failure is no, a trial is
asking a student, and the random variable is the
number of yeses; n = 30, p = 0.72, q = 0.28.
4. Conduct a binomial experiment to determine the
probability of drawing an ace or a king from a deck
of cards. Then compare the experimental and
theoretical probabilities of the experiment.
SOLUTION: eSolutions
Manual
- Powered by Cognero
Sample
answer:
Step 1 A trial is drawing a card from a deck. The
simulation will consist of 20 trials.
The experimental probability is
or 30%. This is
greater than the theoretical probability of
15.4%.
or about
ANSWER: Sample answer:
Step 1 A trial is drawing a card from a deck. The
simulation will consist of 20 trials.
Step 2 A success is drawing an ace or a king. The
probability of success is
and the probability of
failure is .
Step 3 The random variable X represents the number
of aces or kings drawn in 20 trials.
Step 4 Use a random number generator. Let 0−1
represent drawing an ace or a king. Let 2−12
represent drawing all other cards. Make a frequency
table and record the results as you run the generator.
The experimental probability is
or 30%. This is
greater than the theoretical probability of
15.4%.
or about
5. GAMES Aiden has earned five spins of the wheel.
He will receive a prize each time the spinner lands on
Page 1
WIN. What is the probability that he receives three
prizes?
The experimental probability is
or 30%. This is
than the theoretical
probability of
11-4greater
The Binomial
Distribution
15.4%.
or about
5. GAMES Aiden has earned five spins of the wheel.
He will receive a prize each time the spinner lands on
WIN. What is the probability that he receives three
prizes?
n
q) .
or 0.25. So, q = 1 − 0.25 or 0.75 and n =
3
5
P(3) = 5C3(0.25) (0.75) − 3
≈ 0.0879
The probability of Aiden receiving three prizes is
about 0.088 or 8.8%. So, the correct answer is D.
ANSWER: D
6. CCSS PRECISION A poll at Steve’s high school
spending class money to expand the junior-senior par
from the population.
a. Determine the probabilities associated with the nu
favor of expanding the parking lot by calculating the
b. What is the probability that no more than 2 people
c. How many students should Steve expect to find w
SOLUTION: a. In this binomial experiment, n = 6, p = 0.85, and q
n Manual - Powered by Cognero
eSolutions
(p + q)
n
(p + q)
n
= 1p6 +
(0.15)
= 6
+
≈0.00001+
6p5 q
6(0.15)5
(0.85)
0.00039
+
+
+
15p4 q2
15(0.15)4
(0.85)2
0.00549
+
+
+
20p3 q3
20(0.15)3
(0.85)3
0.04145
X n
P(X) = nCXp q - X
6. CCSS PRECISION A poll at Steve’s high school
spending class money to expand the junior-senior par
from the population.
SOLUTION: a. In this binomial experiment, n = 6, p = 0.85, and q
SOLUTION: Find the probability that the spinner lands on WIN
three times out of five spins. Since the spinner is
divided into eight equal spaces and WIN is on two of
q) .
ANSWER: D
a. Determine the probabilities associated with the nu
favor of expanding the parking lot by calculating the
b. What is the probability that no more than 2 people
c. How many students should Steve expect to find w
A 4.2%
B 5.8%
C 7.1%
D 8.8%
them, p =
5.
The probability of Aiden receiving three prizes is
about 0.088 or 8.8%. So, the correct answer is D.
0 in favor, 0.00001 or 0.001%; 1 in favor, 0.00039 or
favor, 0.04145 4.145%; 4 in favor, 0.17618 or 17.618
0.37715 or 37.715%
b. P(no more than 2) = P(0) + P(1) + P(2) = 0.00001
c. The greatest probability is 0.39933 which correspo
ANSWER: a. 0 in favor, 0.00001 or 0.001%;
1 in favor, 0.00039 or 0.039%;
2 in favor, 0.00549 or 0.549%;
3 in favor, 0.04145 or 4.145%;
4 in favor, 0.17618 or 17.618%;
5 in favor, 0.39933 or 39.933%;
6 in favor, 0.37715 or 37.715%
b. 0.00589 or 0.589%
c. 5
Determine whether each experiment is a
binomial experiment or can be reduced to a
binomial experiment. If so, describe a trial,
determine the random variable, and state n, p ,
and q.
7. There is a 35% chance that it rains each day in a
given month. You record the number of days thatPage
it 2
rains for that month.
SOLUTION: 5 in favor, 0.39933 or 39.933%;
6 in favor, 0.37715 or 37.715%
0.00589
or 0.589%
11-4b.
The
Binomial
Distribution
c. 5
Determine whether each experiment is a
binomial experiment or can be reduced to a
binomial experiment. If so, describe a trial,
determine the random variable, and state n, p ,
and q.
7. There is a 35% chance that it rains each day in a
given month. You record the number of days that it
rains for that month.
SOLUTION: This experiment can be reduced to a binomial
experiment. Success is a day that it rains, failure is a
day it does not rain, a trial is a day, and the random
variable X is the number of days it rains; n = the
number of days in the month, p = 0.35, q = 0.65.
ANSWER: This experiment can be reduced to a binomial
experiment. Success is a day that it rains, failure is a
day it does not rain, a trial is a day, and the random
variable X is the number of days it rains; n = the
number of days in the month, p = 0.35, q = 0.65.
8. A survey found that on a scale of 1 to 10, a movie
received a 7.8 rating. A movie theater employee asks
200 patrons to rate the movie on a scale of 1 to 10.
SOLUTION: There are more than two possible outcomes when
rating something from 1–10, so this cannot be
reduced to a binomial experiment.
ANSWER: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
ANSWER: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
9. A ball is hidden under one of the hats shown below.
A hat is chosen, one at a time, until the ball is found.
SOLUTION: This experiment cannot be reduced to a binomial
experiment because the events are not independent.
The probability of choosing the hat that covers the
ball changes after each selection.
ANSWER: This experiment cannot be reduced to a binomial
experiment because the events are not independent.
The probability of choosing the hat that covers the
ball changes after each selection.
10. DICE Conduct a binomial experiment to determine
the probability of rolling a 7 with two dice. Then
compare the experimental and theoretical
probabilities of the experiment.
SOLUTION: Sample answer:
Step 1 A trial is rolling two dice. The simulation will
consist of 25 trials.
Step 2 A success is rolling a 7. The probability of
success is and the probability of failure is .
Step 3 The random variable X represents the number
of times a 7 is rolled in 25 trials.
Step 4 Use a random number generator. Let 0
represent rolling a 7. Let 1−5 represent all other
outcomes. Make a frequency table and record the
results as you run the generator.
9. A ball is hidden under one of the hats shown below.
A hat is chosen, one at a time, until the ball is found.
The experimental probability is
SOLUTION: This experiment cannot be reduced to a binomial
experiment because the events are not independent.
The probability of choosing the hat that covers the
ball changes after each selection.
ANSWER: This experiment cannot be reduced to a binomial
experiment because the events are not independent.
The probability of choosing the hat that covers the
ball Manual
changes
after each
selection.
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10. DICE Conduct a binomial experiment to determine
the probability of rolling a 7 with two dice. Then
or 16%. This is
approximately equal to the theoretical probability of
or about 16.7%.
ANSWER: Sample answer:
Step 1 A trial is rolling two dice. The simulation will
consist of 25 trials.
Step 2 A success is rolling a 7. The probability of
success is and the probability of failure is .
Step 3 The random variable X represents the number
Page 3
of times a 7 is rolled in 25 trials.
Step 4 Use a random number generator. Let 0
ANSWER: This experiment cannot be reduced to a binomial
experiment because the events are not independent.
11-4The
Theprobability
Binomial of
Distribution
choosing the hat that covers the
ball changes after each selection.
10. DICE Conduct a binomial experiment to determine
the probability of rolling a 7 with two dice. Then
compare the experimental and theoretical
probabilities of the experiment.
SOLUTION: Sample answer:
Step 1 A trial is rolling two dice. The simulation will
consist of 25 trials.
Step 2 A success is rolling a 7. The probability of
success is and the probability of failure is .
Step 3 The random variable X represents the number
of times a 7 is rolled in 25 trials.
Step 4 Use a random number generator. Let 0
represent rolling a 7. Let 1−5 represent all other
outcomes. Make a frequency table and record the
results as you run the generator.
The experimental probability is
or 16%. This is
approximately equal to the theoretical probability of
or about 16.7%.
ANSWER: Sample answer:
Step 1 A trial is rolling two dice. The simulation will
consist of 25 trials.
Step 2 A success is rolling a 7. The probability of
success is and the probability of failure is .
Step 3 The random variable X represents the number
of times a 7 is rolled in 25 trials.
Step 4 Use a random number generator. Let 0
represent rolling a 7. Let 1−5 represent all other
outcomes. Make a frequency table and record the
results as you run the generator.
The experimental probability is
or 16%. This is
approximately equal to the theoretical probability of
or about 16.7%.
11. MARBLES Conduct a binomial experiment to
determine the probability of pulling a red marble from
the bag. Then compare the experimental and
theoretical probabilities of the experiment.
eSolutions Manual - Powered by Cognero
The experimental probability is
or 16%. This is
approximately equal to the theoretical probability of
or about 16.7%.
11. MARBLES Conduct a binomial experiment to
determine the probability of pulling a red marble from
the bag. Then compare the experimental and
theoretical probabilities of the experiment.
SOLUTION: Sample answer:
Step 1 A trial is pulling out a marble. The simulation
will consist of 20 trials.
Step 2 A success is pulling out a red marble. The
probability of success is
and the probability of
failure is .
Step 3 The random variable X represents the number
of red marbles pulled out in 20 trials.
Step 4 Use a random number generator. Let 0−4
represent pulling out a red marble. Let 5−11
represent all other outcomes. Make a frequency
table and record the results as you run the generator.
The experimental probability is
or 50%. This is
greater than the theoretical probability of
41.7%.
or about
ANSWER: Sample answer:
Step 1 A trial is pulling out a marble. The simulation
will consist of 20 trials.
Step 2 A success is pulling out a red marble. The
probability of success is
and the probability of
failure is .
Step 3 The random variable X represents the number
of red marbles pulled out in 20 trials.
Step 4 Use a random number generator. Let 0−4
represent pulling out a red marble. Let 5−11
represent all other outcomes. Make a frequency
table and record the results as you run the generator.
Page 4
failure is .
Step 3 The random variable X represents the number
of red marbles pulled out in 20 trials.
Use a random
number generator. Let 0−4
11-4Step
The 4Binomial
Distribution
represent pulling out a red marble. Let 5−11
represent all other outcomes. Make a frequency
table and record the results as you run the generator.
The experimental probability is
or 50%. This is
greater than the theoretical probability of
41.7%.
or about
12. SPINNER Conduct a binomial experiment to
determine the probability of the spinner stopping on an even number. Then compare the experimental and
theoretical probabilities of the experiment.
The experimental probability is
slightly greater than the theoretical probability of
40%.
probability of failure is .
Step 3 The random variable X represents the number
of times the spinner stops on an even number in 25
trials.
Step 4 Use a random number generator. Let 0−1
represent the spinner stopping on an even number.
Let 2−4 represent all other outcomes. Make a
frequency table and record the results as you run the
generator.
The experimental probability is
or
ANSWER: Sample answer:
Step 1 A trial is spinning the spinner. The simulation
will consist of 25 trials.
Step 2 A success is the spinner landing on an even
number. The probability of success is and the
probability of failure is .
Step 3 The random variable X represents the number
of times the spinner stops on an even number in 25
trials.
Step 4 Use a random number generator. Let 0−1
represent the spinner stopping on an even number.
Let 2−4 represent all other outcomes. Make a
frequency table and record the results as you run the
generator.
The experimental probability is
SOLUTION: Sample answer:
Step 1 A trial is spinning the spinner. The simulation
will consist of 25 trials.
Step 2 A success is the spinner landing on an even
number. The probability of success is and the
or 44%. This is
or 44%. This is
slightly greater than the theoretical probability of
40%.
or
13. CARDS Conduct a binomial experiment to
determine the probability of drawing a face card out
of a standard deck of cards. Then compare the
experimental and theoretical probabilities of the
experiment.
SOLUTION: Sample answer:
Step 1 A trial is drawing a card from a deck. The
simulation will consist of 20 trials.
Step 2 A success is drawing a face card. The
probability of success is
and the probability of
failure is .
Step 3 The random variable X represents the number
of face cards drawn in 20 trials.
Step 4 Use a random number generator. Let 0−2
represent drawing a face card. Let 3−12 represent
all other outcomes. Make a frequency table and
record the results as you run the generator.
or 44%. This is
slightly greater than the theoretical probability of
40%.
or
ANSWER: eSolutions
Manual - Powered by Cognero
Sample answer:
Step 1 A trial is spinning the spinner. The simulation
The experimental probability is
Page 5
or 10%. This is
less than the theoretical probability of
or about
≈ 0.04
Simplify.
ANSWER: 0.04 or 4%
11-4 The Binomial Distribution
The experimental probability is
or 10%. This is
less than the theoretical probability of
23.1%
or about
15. CARS According to a recent survey, 92% of high
school seniors drive their own car. What is the
probability that 10 out of 12 random high school
students drive their own car?
SOLUTION: ANSWER: Sample answer:
Step 1 A trial is drawing a card from a deck. The
simulation will consist of 20 trials.
Step 2 A success is drawing a face card. The
probability of success is
and the probability of
failure is .
Step 3 The random variable X represents the number
of face cards drawn in 20 trials.
Step 4 Use a random number generator. Let 0−2
represent drawing a face card. Let 3−12 represent
all other outcomes. Make a frequency table and
record the results as you run the generator.
Binomial Probability
Formula
X n –X
P(X) = nCxp q
10
P(10)
= 12C10(0.92)
12 – 10
(0.08)
≈ 0.183
n = 12, X = 10, p =
0.92, and q = 0.08
Simplify.
ANSWER: 0.183 or 18.3%
16. SENIOR PROM According to a recent survey,
25% of high school upperclassmen think that the
junior-senior prom is the most important event of the
school year. What is the probability that 3 out of 15
random high school upperclassmen think this way?
SOLUTION: Binomial Probability
Formula
X n –X
P(X) = nCxp q
The experimental probability is
3
or 10%. This is
less than the theoretical probability of
23.1%
SOLUTION: X n –X
6
P(6)
= 10C6(0.85)
10 – 6
(0.15)
≈ 0.04
Binomial Probability
Formula
= 15C3(0.25)
(0.75)
or about
14. PERSONAL MEDIA PLAYERS According to a
recent survey, 85% of high school students own a
personal media player. What is the probability that 6
out of 10 random high school students own a
personal media player?
P(X) = nCxp q
P(3)
n = 15, X = 3, p =
0.25, and q = 0.75
15 – 3
≈ 0.225
Simplify.
ANSWER: 0.225 or 22.5%
17. FOOTBALL A certain football team has won
75.7% of their games. Find the probability that they
win 7 of their next 12 games.
SOLUTION: Binomial Probability
Formula
X n –X
P(X) = nCxp q
n = 10, X = 6, p =
0.85, and q = 0.15
7
P(7)
Simplify.
= 12C7(0.757)
(0.243)
12 – 7
≈ 0.096
n = 12, X = 7, p =
0.757, and q = 0.243
Simplify.
ANSWER: 0.04 or 4%
15. CARS According to a recent survey, 92% of high
school seniors drive their own car. What is the
probability that 10 out of 12 random high school
students drive their own car?
eSolutions Manual - Powered by Cognero
SOLUTION: X n –X
P(X) = nCxp q
Binomial Probability
Formula
ANSWER: 0.096 or 9.6%
18. GARDENING Peter is planting 24 irises in his front
yard. The flowers he bought were a combination of
two varieties, blue and white. The flowers are not
Page 6
blooming yet, but Peter knows that the probability
of
having a blue flower is 75%. What is the probability
that 20 of the flowers will be blue?
≈ 0.096
Simplify.
11-4ANSWER: The Binomial Distribution
0.096 or 9.6%
The probability of the kicker making 7 of his next 10
kicks from within 35 yards is about 0.25 or 25%.
ANSWER: 0.25 or 25%
18. GARDENING Peter is planting 24 irises in his front
yard. The flowers he bought were a combination of
two varieties, blue and white. The flowers are not
blooming yet, but Peter knows that the probability of
having a blue flower is 75%. What is the probability
that 20 of the flowers will be blue?
20. BABIES Mr. and Mrs. Davis are planning to have 3
children. The probability of each child being a boy is
50%. What is the probability that they will have 2
boys?
SOLUTION: X n –X
P(X) = nCxp q
SOLUTION: 20
P(20)
2
Binomial Probability
Formula
X n –X
P(X) = nCxp q
= 24C20(0.75)
24 – 20
(0.25)
≈ 0.132
n = 24, X = 20, p =
0.75, and q = 0.25
P(2)
= 3C2(0.50)
3–2
(0.50)
≈ 0.375
Binomial Probability
Formula
n = 3, X = 2, p = 0.50,
and q = 0.50
Simplify.
Simplify.
ANSWER: 0.375 or 37.5%
ANSWER: 0.132 or 13.2%
19. FOOTBALL A field goal kicker is accurate 75% of
the time from within 35 yards. What is the probability
that he makes exactly 7 of his next 10 kicks from
within 35 yards?
21. CCSS SENSE-MAKING According to a recent
survey, 52% of high school students own a laptop.
Ten random students are chosen.
a. Determine the probabilities associated with the
number of students that own a laptop by calculating
the probability distribution.
b. What is the probability that at least 8 of the 10
students own a laptop?
c. How many students should you expect to own a
laptop?
SOLUTION: a.
SOLUTION: Find the probability that the kicker makes 7 of his
next 10 kicks from within 35 yards. A success is
making a field goal, so p = 0.75, q = 1 – 0.75 or 0.25,
and n = 10.
X n
P(X) = nCXp q – X
7
10
P(7) = 10C7(0.75) (0.25) – 7
≈ 0.25
The probability of the kicker making 7 of his next 10
kicks from within 35 yards is about 0.25 or 25%.
ANSWER: 0.25 or 25%
20. BABIES Mr. and Mrs. Davis are planning to have 3
children. The probability of each child being a boy is
50%. What is the probability that they will have 2
boys?
SOLUTION: eSolutions Manual - Powered by Cognero
X n –X
P(X) = nCxp q
2
= C (0.50)
Binomial Probability
Formula
b. P(at least 8) = P(8) + P(9) + P(10) = 0.0014 +
0.0133 + 0.0554 = 0.0701
The probability is about 7.01%.
c. The greatest individual probability (0.2441) is with
an X value of 5. Therefore, you can expect 5 in 10
students to own a laptop.
ANSWER: a. 0 own a laptop, 0.0006 or 0.06%; 1 owns a laptop,
0.007 or 0.7%, 2 own a laptop, 0.0343 or 3.43%; 3
own a laptop, 0.0991 or 9.91%; 4 own a laptop,
0.1878 or 18.78%; 5 own a laptop, 0.2441 or 24.41%;
6 own a laptop, 0.2204 or 22.04%; 7 own a laptop,
0.1364 or 13.64%; 8 own a laptop, 0.0554 or 5.54%;
Page 7
9 own a laptop, 0.0133 or 1.33%; 10 own a laptop,
0.0014 or 0.14%
b. about 7%
≈ 0.375
Simplify.
11-4ANSWER: The Binomial Distribution
0.375 or 37.5%
21. CCSS SENSE-MAKING According to a recent
survey, 52% of high school students own a laptop.
Ten random students are chosen.
a. Determine the probabilities associated with the
number of students that own a laptop by calculating
the probability distribution.
b. What is the probability that at least 8 of the 10
students own a laptop?
c. How many students should you expect to own a
laptop?
SOLUTION: a.
b. P(at least 8) = P(8) + P(9) + P(10) = 0.0014 +
0.0133 + 0.0554 = 0.0701
The probability is about 7.01%.
0.1364 or 13.64%; 8 own a laptop, 0.0554 or 5.54%;
9 own a laptop, 0.0133 or 1.33%; 10 own a laptop,
0.0014 or 0.14%
b. about 7%
c. 5
22. ATHLETICS A survey was taken to see the
percent of students that participate in sports for their
school. Six random students are chosen.
a. Determine the probabilities associated with the
number of students playing in at least one sport by
calculating the probability distribution.
b. What is the probability that no more than 2 of the
students participated in a sport?
c. How many students should you expect to have
participated in at least one sport?
SOLUTION: a. The probability of a student playing in at least one
sport is P(1) + P(2) + P(3+) = 55% + 20% + 5% or
80%.
c. The greatest individual probability (0.2441) is with
an X value of 5. Therefore, you can expect 5 in 10
students to own a laptop.
ANSWER: a. 0 own a laptop, 0.0006 or 0.06%; 1 owns a laptop,
0.007 or 0.7%, 2 own a laptop, 0.0343 or 3.43%; 3
own a laptop, 0.0991 or 9.91%; 4 own a laptop,
0.1878 or 18.78%; 5 own a laptop, 0.2441 or 24.41%;
6 own a laptop, 0.2204 or 22.04%; 7 own a laptop,
0.1364 or 13.64%; 8 own a laptop, 0.0554 or 5.54%;
9 own a laptop, 0.0133 or 1.33%; 10 own a laptop,
0.0014 or 0.14%
b. about 7%
c. 5
22. ATHLETICS A survey was taken to see the
percent of students that participate in sports for their
school. Six random students are chosen.
eSolutions
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a. Determine the probabilities associated with the
number of students playing in at least one sport by
b. P(no more than 2) = P(0) + P(1) + P(2) = 0.00006
+ 0.00154 + 0.01536 = 0.01696
c. The greatest individual probability (0.39322) is with
an X value of 5. Therefore, you can expect 5 in 10
students to participate in at least 1 sport.
ANSWER: a. 0 play at least 1 sport, 0.00006 or 0.006%; 1 plays
at least 1 sport, 0.00154 or 0.154%; 2 play at least 1
sport, 0.01536 or 1.536%; 3 play at least 1 sport,
0.08192 or 8.192%; 4 play at least 1 sport, 0.24576 or
24.576%; 5 play at least 1 sport, 0.39322 or
39.322%; 6 play at least 1 sport, 0.26214 or 26.214%
b. 0.01696 or 1.696%
c. 5
23. CCSS MODELING An online poll showed that
57% of adults still own vinyl records. Moe surveyed
8 random adults from the population.
a. Determine the probabilities associated with the
number of adults that still own vinyl records by Page 8
calculating the probability distribution.
b. What is the probability that no less than 6 of the
people surveyed still own vinyl records?
0.08192 or 8.192%; 4 play at least 1 sport, 0.24576 or
24.576%; 5 play at least 1 sport, 0.39322 or
39.322%; 6 play at least 1 sport, 0.26214 or 26.214%
0.01696
or 1.696%
11-4b.
The
Binomial
Distribution
c. 5
23. CCSS MODELING An online poll showed that
57% of adults still own vinyl records. Moe surveyed
8 random adults from the population.
a. Determine the probabilities associated with the
number of adults that still own vinyl records by
calculating the probability distribution.
b. What is the probability that no less than 6 of the
people surveyed still own vinyl records?
c. How many people should Moe expect to still own
vinyl records?
SOLUTION: a.
0.268 or 26.8%; 6 own vinyl records, 0.178 or
17.8%; 7 own vinyl records, 0.067 or 6.7%; 8 own
vinyl records, 0.011 or 1.1%
b. 0.256 or 25.6%
c. 5
A binomial distribution has a 60% rate of
success. There are 18 trials.
24. What is the probability that there will be at least 12
successes?
SOLUTION: The probability of a failure is 1 – 0.6 or 0.4.
The probability that there will be at least 12
successes is
or 0.37 or 37%.
ANSWER: 0.37 or 37%
25. What is the probability that there will be 12 failures?
b. P(no less than 6) = P(6) + P(7) + P(8)] = 0.1776 + 0.0672 + 0.0111 = 0.2559 or about 25.6%
c. The greatest individual probability (0.2679) is with
an X value of 5. Therefore, you can expect 5 in 8
adults to own vinyl records.
ANSWER: a. 0 own vinyl records, 0.001 or 0.1%; 1 owns vinyl records, 0.012 or 1.2%; 2 own vinyl records, 0.058 or
5.8%; 3 own vinyl records, 0.152 or 15.2%; 4 own
vinyl records, 0.253 or 25.3%; 5 own vinyl records,
0.268 or 26.8%; 6 own vinyl records, 0.178 or
17.8%; 7 own vinyl records, 0.067 or 6.7%; 8 own
vinyl records, 0.011 or 1.1%
b. 0.256 or 25.6%
c. 5
A binomial distribution has a 60% rate of
success. There are 18 trials.
24. What is the probability that there will be at least 12
successes?
SOLUTION: The probability of a failure is 1 – 0.6 or 0.4.
The probability that there will be at least 12
successes is
or 0.37 or 37%.
ANSWER: 0.37 or 37%
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SOLUTION: The probability of a failure is 1 – 0.6 or 0.4.
The probability that there will be 12 failures is equal
to the probability that there will be exactly 6
successes.
That is
or 0.0145 or 1.45%.
ANSWER: 0.0145 or 1.45%
26. What is the expected number of successes?
SOLUTION: The expected number of success is
or 10.8
ANSWER: 10.8
27. DECISION MAKING Six roommates randomly
select someone to wash the dishes each day.
a. What is the probability that the same person has to
wash the dishes 3 times in a given week?
b. What method can the roommates use to select
who washes the dishes each day?
SOLUTION: a. Since the dishwasher is randomly selected, each
roommate has an equal chance of being selected, so
the probability of one being selected (a success) Page
is 1 9
out of 6 or about 0.167. The probability of a failure is
1 – 0.167 or 0.833. There are 7 days in the week, so
n = 7.
of 6 chances.
or 10.8
11-4ANSWER: The Binomial Distribution
10.8
27. DECISION MAKING Six roommates randomly
select someone to wash the dishes each day.
a. What is the probability that the same person has to
wash the dishes 3 times in a given week?
b. What method can the roommates use to select
who washes the dishes each day?
SOLUTION: a. Since the dishwasher is randomly selected, each
roommate has an equal chance of being selected, so
the probability of one being selected (a success) is 1
out of 6 or about 0.167. The probability of a failure is
1 – 0.167 or 0.833. There are 7 days in the week, so
n = 7.
3
4
P(3) = 7C3(0.167) (0.833) ≈ 0.078
The probability is about 7.8% that the same person
will wash the dishes 3 times in a given week.
ANSWER: a. about 7.8%
b. Sample answer: They can roll a six-sided die.
28. DECISION MAKING A committee of five people
randomly selects someone to take the notes of each
meeting.
a. What is the probability that a person takes notes
less than twice in 10 meetings?
b. What method can the committee use to select the
notetaker each meeting?
c. If the method described in part b results in the
same person being notetaker for nine straight
meetings, would this result cause you to question the
method
SOLUTION: a. Since the notetaker is randomly selected, each
person has an equal chance of being selected, so the
probability of one being selected (a success) is 1 out
of 5 or 0.20. The probability of a failure is 1 – 0.20 or
0.80. There are 10 meetings, so n = 10.
P(less than twice) = P(1) + P(0)
b. A six-sided die can be used to represent the 1 out
of 6 chances.
P(1) = 10C1(0.20) (0.80) ≈ 0.2684
ANSWER: a. about 7.8%
b. Sample answer: They can roll a six-sided die.
28. DECISION MAKING A committee of five people
randomly selects someone to take the notes of each
meeting.
a. What is the probability that a person takes notes
less than twice in 10 meetings?
b. What method can the committee use to select the
notetaker each meeting?
c. If the method described in part b results in the
same person being notetaker for nine straight
meetings, would this result cause you to question the
method
SOLUTION: a. Since the notetaker is randomly selected, each
person has an equal chance of being selected, so the
probability of one being selected (a success) is 1 out
of 5 or 0.20. The probability of a failure is 1 – 0.20 or
0.80. There are 10 meetings, so n = 10.
P(less than twice) = P(1) + P(0)
1
9
0
10
P(1) = 10C1(0.20) (0.80) ≈ 0.2684
P(0) = 10C0(0.20) (0.80) ≈ 0.1074
1
9
0
10
P(0) = 10C0(0.20) (0.80) ≈ 0.1074
0.2684 + 0.1074 = 0.3758
The probability is about 37.6% that a person takes
notes less than twice in 10 meetings.
b. The can roll a six-sided die. If a six is rolled, they
can roll again.
c. Sample answer: While it is possible for the same
person to be chosen nine straight times, it is rather
unlikely. The fairness of the die should be questioned.
ANSWER: a. about 37.6%
b. Sample answer: They can roll a six-sided die. If a
six is rolled, they roll again.
c. Sample answer: While it is possible for the same
person to be chosen nine straight times, it is rather
unlikely. The fairness of the die should be questioned.
Each binomial distribution has n trials and p
probability of success. Determine the most
likely number of successes.
29. n = 8, p = 0.6
SOLUTION: 0.2684 + 0.1074 = 0.3758
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The probability is about 37.6% that a person takes
notes less than twice in 10 meetings.
Page 10
b. Sample answer: They can roll a six-sided die. If a
six is rolled, they roll again.
c. Sample answer: While it is possible for the same
11-4person
The Binomial
Distribution
to be chosen
nine straight times, it is rather
unlikely. The fairness of the die should be questioned.
Each binomial distribution has n trials and p
probability of success. Determine the most
likely number of successes.
29. n = 8, p = 0.6
The most likely number of success is 5.
ANSWER: 5
30. n = 10, p = 0.4
SOLUTION: SOLUTION: The most likely number of success is 5.
ANSWER: 5
30. n = 10, p = 0.4
SOLUTION: The most likely number of success is 4.
ANSWER: 4
31. n = 6, p = 0.8
SOLUTION: eSolutions Manual - Powered by Cognero
Page 11
The most likely number of success is 4.
The most likely number of success is 5.
11-4ANSWER: The Binomial Distribution
4
31. n = 6, p = 0.8
SOLUTION: ANSWER: 5
32. n = 12, p = 0.55
SOLUTION: The most likely number of success is 5.
ANSWER: 5
32. n = 12, p = 0.55
SOLUTION: The most likely number of success is 7.
ANSWER: 7
33. n = 9, p = 0.75
SOLUTION: eSolutions Manual - Powered by Cognero
Page 12
The most likely number of success is 7.
The most likely number of success is 7.
11-4ANSWER: The Binomial Distribution
7
33. n = 9, p = 0.75
SOLUTION: ANSWER: 7
34. n = 11, p = 0.35
SOLUTION: The most likely number of success is 7.
ANSWER: 7
34. n = 11, p = 0.35
SOLUTION: The most likely number of success is 4.
ANSWER: 4
35. SWEEPSTAKES A beverage company is having a
sweepstakes. The probability of winning selected
prizes is shown below. If Ernesto purchases 8
beverages, what is the probability that he wins at
least one prize?
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Page 13
The most likely number of success is 4.
11-4ANSWER: The Binomial Distribution
4
Therefore the probability of winning a prize is 1 – P
(0) ≈ 0.603 or 60.3%
ANSWER: 0.603 or 60.3%
35. SWEEPSTAKES A beverage company is having a
sweepstakes. The probability of winning selected
prizes is shown below. If Ernesto purchases 8
beverages, what is the probability that he wins at
least one prize?
Each binomial distribution has n trials and p
probability of success. Determine the
probability of s successes
36. n = 8, p = 0.3, s 2
SOLUTION: The probability of a success is 0.3.
The probability of a failure is 1 – 0.3 or 0.7.
ANSWER: 0.744 or 74.4%
SOLUTION: The probability of winning is
37. n = 10, p = 0.2, s > 2
SOLUTION: The probability of a success is 0.2.
The probability of a failure is 1 – 0.2 or 0.8.
The probability of not winning is (1–.10904004)=.89094996
With 8 beverage purchase, the probability of not
winning a prize is
8
0
P(0) = (0.89094996 )(1 – 0.89094996) ≈ 0.397
ANSWER: 0.322 or 32.2%
Therefore the probability of winning a prize is 1 – P
(0) ≈ 0.603 or 60.3%
ANSWER: 0.603 or 60.3%
38. n = 6, p = 0.6, s
4
SOLUTION: The probability of a success is 0.6.
The probability of a failure is 1 – 0.6 or 0.4.
Each binomial distribution has n trials and p
probability of success. Determine the
probability of s successes
36. n = 8, p = 0.3, s 2
SOLUTION: The probability of a success is 0.3.
The probability of a failure is 1 – 0.3 or 0.7.
ANSWER: 0.767 or 76.7%
39. n = 9, p = 0.25, s
ANSWER: 0.744 or 74.4%
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37. n = 10, p = 0.2, s > 2
SOLUTION: 5
SOLUTION: The probability of a success is 0.25.
The probability of a failure is 1 – 0.25 or 0.75.
Page 14
11-4ANSWER: The Binomial Distribution
0.767 or 76.7%
39. n = 9, p = 0.25, s
5
SOLUTION: The probability of a success is 0.25.
The probability of a failure is 1 – 0.25 or 0.75.
ANSWER: 0.99 or 99%
ANSWER: 0.889 or 88.9%
42. CHALLENGE A poll of students determined that
88% wanted to go to college. Eight random students
are chosen. The probability that at least x students
want to go to college is about 0.752 or 75.2%. Solve
for x.
SOLUTION: Create the probability distribution and keep track of
the cumulative probability with each new value of X.
Find the cumulative probability by taking the sum of
all of the P(X) values up to that point.
40. n = 10, p = 0.75, s ≥ 8
SOLUTION: The probability of a success is 0.75.
The probability of a failure is 1 – 0.75 or 0.25.
The cumulative probability of 0.7519 is about 75.2%,
so P(7) + P(8), or at least 7 students want to go to
college. Therefore, x = 7.
ANSWER: 0.526 or 52.6%
41. n = 12, p = 0.1, s < 3
SOLUTION: The probability of a success is 0.1.
The probability of a failure is 1 – 0.1 or 0.9.
ANSWER: 0.889 or 88.9%
42. CHALLENGE A poll of students determined that
88% wanted to go to college. Eight random students
are chosen. The probability that at least x students
want to go to college is about 0.752 or 75.2%. Solve
for x.
SOLUTION: Create the probability distribution and keep track of
the cumulative probability with each new value of X.
Find the cumulative probability by taking the sum of
all of the P(X) values up to that point.
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ANSWER: 7
43. WRITING IN MATH What should you consider
when using a binomial distribution to make a
decision?
SOLUTION: Sample answer: You should consider the type of
situation for which the binomial distribution is being
used. For example, if a binomial distribution is being
used to predict outcomes regarding an athletic event,
the probabilities of success and failure could change
due to other variables such as weather conditions or
player health. So, binomial distributions should be
used cautiously when making decisions involving
events that are not completely random.
ANSWER: Sample answer: You should consider the type of
situation for which the binomial distribution is being
used. For example, if a binomial distribution is being
used to predict outcomes regarding an athletic event,
the probabilities of success and failure could change
due to other variables such as weather conditions or
player health. So, binomial distributions should be
used cautiously when making decisions involving
events that are not completely random.
44. OPEN ENDED Describe a real-world setting within
Page 15
your school or community activities that seems to fit
a binomial distribution. Identify the key components
the probabilities of success and failure could change
due to other variables such as weather conditions or
player health. So, binomial distributions should be
11-4used
Thecautiously
Binomialwhen
Distribution
making decisions involving
events that are not completely random.
44. OPEN ENDED Describe a real-world setting within
your school or community activities that seems to fit
a binomial distribution. Identify the key components
of your setting that connect to binomial distributions.
SOLUTION: Sample answer: During May and June, lunches are
held outside, weather permitting. Also during this
time, there has historically been a 15% chance of
rain. So, to determine the probability of not having
rain for at least 24 of these 28 days, the binomial
distribution would use p = 0.85, q = 0.15, and n = 28.
ANSWER: Sample answer: During May and June, lunches are
held outside, weather permitting. Also during this
time, there has historically been a 15% chance of
rain. So, to determine the probability of not having
rain for at least 24 of these 28 days, the binomial
distribution would use p = 0.85, q = 0.15, and n = 28.
45. WRITING IN MATH Describe how binomial
distributions are connected to Pascal’s triangle.
SOLUTION: Sample answer: A full binomial distribution can be
determined by expanding the binomial, which itself
utilizes Pascal’s triangle.
ANSWER: Sample answer: A full binomial distribution can be
determined by expanding the binomial, which itself
utilizes Pascal’s triangle.
46. WRITING IN MATH Explain the relationship
between a binomial experiment and a binomial
distribution.
SOLUTION: Sample answer: A binomial distribution shows the
probabilities of the outcomes of a binomial
experiment.
ANSWER: Sample answer: A binomial distribution shows the
probabilities of the outcomes of a binomial
experiment.
47. EXTENDED RESPONSE Carly is taking a 10question multiple-choice test in which each question
has four choices. If she guesses on each question,
what is the probability that she will get
a. 7 questions correct?
b. 9 questions correct?
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c. 0 questions correct?
d. 3 questions correct?
ANSWER: Sample answer: A binomial distribution shows the
probabilities of the outcomes of a binomial
experiment.
47. EXTENDED RESPONSE Carly is taking a 10question multiple-choice test in which each question
has four choices. If she guesses on each question,
what is the probability that she will get
a. 7 questions correct?
b. 9 questions correct?
c. 0 questions correct?
d. 3 questions correct?
SOLUTION: The probability of a success is 0.25.
The probability of a failure is 1 – 0.25 or 0.75.
a. the probability that she will get 7 question corrects
is
or 0.003 or 0.3%.
b. the probability that she will get 9 question corrects
is
or 0.00003 or 0.003%.
c. the probability that she will get 0 question corrects
is
or 0.0563 or 5.6%.
d. the probability that she will get 3 question corrects
is
or 0.2503 or 25%.
ANSWER: a. 0.003 or 0.3%
b. 0.00003 or 0.003%
c. 0.056 or 5.6%
d. 0.25 or 25%
48. What is the maximum point of the graph of the
2
equation y = –2x + 16x + 5?
A (–4, –59)
B (–4, –91)
C (4, 37)
D (4, 101)
SOLUTION: 2
Compare the equation y = –2x + 16x + 5 with the
2
equation y = ax + bx + c.
Here a = –2, b = 16 and c = 5.
Since a < 0, the maximum point of the graph is its
vertex
.
Now find f (4).
Page 16
2
2x + 16x + 5 is (4, 37).
Option C is the correct answer.
a. 0.003 or 0.3%
b. 0.00003 or 0.003%
0.056
or 5.6%Distribution
11-4c.The
Binomial
d. 0.25 or 25%
ANSWER: C
48. What is the maximum point of the graph of the
2
equation y = –2x + 16x + 5?
A (–4, –59)
B (–4, –91)
C (4, 37)
D (4, 101)
of the way from X to Y. What is the coordinate of
SOLUTION: 2
Compare the equation y = –2x + 16x + 5 with the
2
equation y = ax + bx + c.
Here a = –2, b = 16 and c = 5.
Since a < 0, the maximum point of the graph is its
vertex
49. GEOMETRY On a number line, point X has
coordinate –8 and point Y has coordinate 4. Point P is
P?
F –4
G –2
H0
J2
SOLUTION: The distance between the points X and Y is 12.
.
The coordinate of P is –8 + 8 or 0.
Therefore, option H is the correct answer.
Now find f (4).
ANSWER: H
50. SAT/ACT The cost of 4 CDs is d dollars. At this
rate, what is the cost, in dollars, of 36 CDs?
A 9d
B 144d
C
Therefore, the maximum point of the graph of y = –
2
2x + 16x + 5 is (4, 37).
Option C is the correct answer.
ANSWER: C
49. GEOMETRY On a number line, point X has
coordinate –8 and point Y has coordinate 4. Point P is
of the way from X to Y. What is the coordinate of
P?
F –4
G –2
H0
J2
SOLUTION: The distance between the points X and Y is 12.
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The coordinate of P is –8 + 8 or 0.
Therefore, option H is the correct answer.
D
E
SOLUTION: The cost of 36 CDs is 9d.
Option A is the correct answer.
ANSWER: A
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
51. the number of customers at an amusement park
SOLUTION: The random variable X is the number of customers at
an amusement park. The customers are finite and
countable, so X is discrete.
ANSWER: The random variable X is the number of customers at
Page 17
an amusement park. The customers are finite and
countable, so X is discrete.
The cost of 36 CDs is 9d.
Option A is the correct answer.
11-4ANSWER: The Binomial Distribution
A
cities. The distance can be anywhere within a certain
range, so X is continuous.
55. FINANCIAL LITERACY The prices of entrees
offered by a restaurant are shown.
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
51. the number of customers at an amusement park
SOLUTION: The random variable X is the number of customers at
an amusement park. The customers are finite and
countable, so X is discrete.
ANSWER: The random variable X is the number of customers at
an amusement park. The customers are finite and
countable, so X is discrete.
52. the running time of a movie
SOLUTION: The random variable X is the running time of a
movie. The time can be anywhere within a certain
range, so X is continuous.
a. Use a graphing calculator to create a box-andwhisker plot. Then describe the shape of the
distribution.
b. Describe the center and spread of the data using
either the mean and standard deviation or the fivenumber summary. Justify your choice.
SOLUTION: a. Enter the data as L1. Press 2ND [STAT PLOT]
ENTER ENTER and choose fl. Adjust the window
to the dimensions shown.
ANSWER: The random variable X is the running time of a
movie. The time can be anywhere within a certain
range, so X is continuous.
53. the number of hot dogs sold at a sporting event
SOLUTION: The random variable X is the number of hot dogs sold
at a sporting event. The hot dogs are finite and
countable, so X is discrete.
ANSWER: The random variable X is the number of hot dogs sold
at a sporting event. The hot dogs are finite and
countable, so X is discrete.
54. the distance between two cities
The data are equally distributed to the left and right
of the median. Therefore, the distribution is
symmetric.
b. The distribution is symmetric, so use the mean and
standard deviation. Press STAT ► ENTER
ENTER and scroll down to view the mean and
standard deviation.
SOLUTION: The random variable X is the distance between two
cities. The distance can be anywhere within a certain
range, so X is continuous.
ANSWER: The random variable X is the distance between two
cities. The distance can be anywhere within a certain
range, so X is continuous.
55. FINANCIAL LITERACY The prices of entrees
offered by a restaurant are shown.
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The mean is about $16.02 with a standard deviation
of about $4.52.
ANSWER: a.
Page 18
The mean is about $16.02 with a standard deviation
about
$4.52. Distribution
11-4of
The
Binomial
ANSWER: 11
ANSWER: a.
57. = –34,
= 44, d = ?
SOLUTION: There are 22 – 9 + 1 or 14 terms between a 9 and
a 22.
Find d.
symmetric
b. Sample answer: The distribution is symmetric, so
use the mean and standard deviation. The mean is
about $16.02 with standard deviation of about $4.52.
Find the missing value for each arithmetic
sequence.
56. = 12,
= 133, d = ?
ANSWER: 6
58. = 18,
= 95, d = 7, n = ?
SOLUTION: There are 16 – 5 + 1 or 12 terms between a 5 and
SOLUTION: There are n – 4 + 1 or n – 3 terms between a 4 and
a 16.
Find d.
a n.
Find n.
ANSWER: 11
ANSWER: 15
57. = –34,
= 44, d = ?
59. = 31, d = 8
SOLUTION: There are 19 – 8 + 1 or 12 terms between a 8 and
SOLUTION: There are 22 – 9 + 1 or 14 terms between a 9 and
a 19.
a 22.
Find d.
Find a 8.
ANSWER: –57
ANSWER: eSolutions
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6
= ?,
60. = ?,
= 64, d = 7
Page 19
11-4ANSWER: The Binomial Distribution
15
59. = ?,
= 31, d = 8
ANSWER: –34
61. = –28,
= 76, d = 8, n = ?
SOLUTION: There are 19 – 8 + 1 or 12 terms between a 8 and
SOLUTION: There are n – 7 + 1 or n – 6 terms between a 7 and
a 19.
a n.
Find n.
Find a 8.
ANSWER: –57
60. = ?,
= 64, d = 7
SOLUTION: There are 20 – 6 + 1 or 15 terms between a 6 and
a 20.
ANSWER: 20
62. ASTRONOMY The table shows the closest and
farthest distances of Venus and Jupiter from the
center of the Sun in millions of miles.
Find a 6.
ANSWER: –34
61. = –28,
= 76, d = 8, n = ?
SOLUTION: There are n – 7 + 1 or n – 6 terms between a 7 and
a n.
Find n.
a. Write an equation for the orbit of each planet.
Assuming that the center of the orbit is the origin and
the center of the Sun is a focus that lies on the xaxis.
b. Which planet has an orbit that is closer to a circle?
SOLUTION: a. Venus: The value of a is one half the length of
the major axis.
The value of c is the distance from the center of the
ellipse to the focus.
Use the values of a and c to find the value of b.
ANSWER: 20
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62. ASTRONOMY The table shows the closest and
farthest distances of Venus and Jupiter from the
Page 20
b. Venus
11-4 The Binomial Distribution
Use the values of a and c to find the value of b.
ANSWER: a.
Venus:
Jupiter:
b. Venus
Write an equivalent exponential or logarithmic
function.
The equation for the orbit of Venus is
63. or
SOLUTION: .
ANSWER: –x = ln 5
Jupiter:
The value of a is one half the length of the major
axis.
64. SOLUTION: The value of c is the distance from the center of the
ellipse to the focus.
ANSWER: 2 = ln 6x
65. ln e = 1
Use the values of a and c to find the value of b.
SOLUTION: ANSWER: 1
e =e
66. ln 5.2 = x
SOLUTION: The equation for the orbit of Jupiter
is
or
ANSWER: x
e = 5.2
.
b. Venus
67. e
x+1
=9
SOLUTION: ANSWER: a.
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Venus:
Jupiter:
ANSWER: x + 1 = ln 9
Page 21
ANSWER: 11-4 The Binomial Distribution
x
e = 5.2
67. e
x+1
=9
SOLUTION: ANSWER: x + 1 = ln 9
68. e
–1
2
=x
SOLUTION: ANSWER: 3
e =e
x
71. MUSIC Tina owns 11 pop, 6 country, 16 rock, and 7
rap CDs. Find each probability if she randomly
selects 4 CDs.
a. P(2 rock)
b. P(1 rap)
c. P(1 rock and 2 country)
SOLUTION: a. We are choosing 4 songs out of 40 which gives a
total of 40C4 different choices. The total number of
ways of choosing exactly 2 rock songs and exactly 2
non-rock songs is 16C2·24C2. We can calculate the
probability of getting 2 rock songs as follows:
ANSWER: –1 = ln x
69. ln
2
= 2x
SOLUTION: b. We are choosing 4 songs out of 40 which gives a
total of 40C4 different choices. The total number of
ways of choosing exactly 1 rap song and exactly 3
non-rock songs is 7C1·33C3. We can calculate the
probability of getting 2 rock songs as follows:
ANSWER: e
2x
=
x
70. ln e = 3
SOLUTION: c. We are choosing 4 songs out of 40 which gives a
total of 40C4 different choices. The total number of
ways of choosing exactly 1 rock song, 2 country
songs, and exactly 1 non-rock or country song is
16C1·6C2·18C1. We can calculate the probability of
getting 1 rock song and 2 country songs as follows:
ANSWER: 3
e =e
x
71. MUSIC Tina owns 11 pop, 6 country, 16 rock, and 7
rap CDs. Find each probability if she randomly
selects 4 CDs.
a. P(2 rock)
b. P(1 rap)
c. P(1 rock and 2 country)
SOLUTION: a. We are choosing 4 songs out of 40 which gives a
total of 40C4 different choices. The total number of
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ways of choosing exactly 2 rock songs and exactly 2
non-rock songs is 16C2·24C2. We can calculate the
ANSWER: a. 0.36
b. 0.42
c. 0.05
Page 22
are only concerned with the lower tail, so 16% of the
data will be less than 361.
11-5 The Normal Distribution
A normal distribution has a mean of 416 and a
standard deviation of 55.
1. Find the range of values that represent the middle
99.7% of the distribution.
SOLUTION: The middle 99.7% of data in a normal distribution is
the range from µ – 3σ to µ + 3σ. The standard
deviation is 55, so 3σ = 3 ∙ 55 or 165.
416 + 165 = 581 and 416 – 165 = 251
Therefore, the range of values in the middle 99.7% is
251 < X < 581.
ANSWER: 251 < X < 581
ANSWER: 16%
3. CCSS TOOLS The number of texts sent per day
by a sample of 811 teens is normally distributed with
a mean of 38 and a standard deviation of 7.
a. About how many teens sent between 24 and 38
texts?
b. What is the probability that a teen selected at
random sent less than 818 texts?
SOLUTION: a. 24 is 2σ away from the mean, so the range of
values are between µ – 2σ and µ. This represents
half of 95% of the distribution, or 47.5%.
47.5% of 811 = 385.225
Therefore, about 386 teens sent between 24 and 38
texts.
2. What percent of the data will be less than 361?
SOLUTION: 361 is 55 less than 416, so the µ – σ represents the
data value of 361. 68% of the data fall within µ – σ
and µ + σ, so the remaining data values represented
by the two tails covers 32% of the distribution. We
are only concerned with the lower tail, so 16% of the
data will be less than 361.
ANSWER: 16%
3. CCSS TOOLS The number of texts sent per day
by a sample of 811 teens is normally distributed with
a mean of 38 and a standard deviation of 7.
a. About how many teens sent between 24 and 38
texts?
b. What is the probability that a teen selected at
random sent less than 818 texts?
SOLUTION: a. 24 is 2σ away from the mean, so the range of
values are between µ – 2σ and µ. This represents
half of 95% of the distribution, or 47.5%.
47.5% of 811 = 385.225
Therefore, about 386 teens sent between 24 and 38
texts.
b. 45 is σ away from the mean, so the range of
values are less than µ + σ. The range of values from
µ – σ to µ + σ represent 68% of the data, so the
range from µ to µ + σ represents half of these
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values, or 34% of the data. 50% of the data are less
than µ, so the percentage of data values less than µ
+ σ is 50% + 34% or 84%. Therefore, the probability
b. 45 is σ away from the mean, so the range of
values are less than µ + σ. The range of values from
µ – σ to µ + σ represent 68% of the data, so the
range from µ to µ + σ represents half of these
values, or 34% of the data. 50% of the data are less
than µ, so the percentage of data values less than µ
+ σ is 50% + 34% or 84%. Therefore, the probability
that a teen selected at random sent less than 45 texts
is 84%.
ANSWER: a. about 386 teens
b. 84%
Find the missing variable. Indicate the position
of X in the distribution.
4. z if μ = 89, X = 81, and σ = 11.5
SOLUTION: z =
Formula for z-values
z =
≈ –0.70
X = 81, µ = 89, and σ =
11.5
Simplify.
The z-value that corresponds to X = 81 is
approximately 0.70 standard deviations less than the
mean.
ANSWER: −0.70; 0.70 standard deviations less than the mean
5. z if μ = 13.3, X = 17.2, and σ = 1.9
SOLUTION: Page 1
z =
z
Formula for z-values
X = 17.2, µ = 13.3, and
approximately 0.70 standard deviations less than the
mean.
ANSWER: 11-5−0.70;
The Normal
Distribution
0.70 standard
deviations less than the mean
5. z if μ = 13.3, X = 17.2, and σ = 1.9
SOLUTION: approximately 1.38 standard deviations less than the
mean.
ANSWER: 59.8; 1.38 standard deviations less than the mean
7. σ if μ = 21.1, X = 13.7, and z =-2.40
SOLUTION: z =
Formula for z-values
z =
≈ 2.05
X = 17.2, µ = 13.3, and
σ = 1.9
Simplify.
z =
–2.40 =
–2.40 =
–2.4σ = –7.4
The z-value that corresponds to X = 17.2 is
approximately 2.05 standard deviations more than the
mean.
ANSWER: 2.05; 2.05 standard deviations greater than the mean
6. X if z =-1.38, μ = 68.9, and σ = 6.6
SOLUTION: z =
–1.38 =
–9.108 = X – 68.9
59.792 = X
Formula for z-values
z = –1.38, µ = 68.9,
and σ = 6.6
Multiply each side by
6.6.
Add 68.9 to each side.
The z-value that corresponds to X = 59.8 is
approximately 1.38 standard deviations less than the
mean.
σ =
σ ≈ 3.08
σ =
σ ≈ 3.08
Multiply each side by σ.
Divide each side by –
2.4.
Simplify.
ANSWER: 3.08; 2.40 standard deviations less than the mean
8. CONCERTS The number of concerts attended per
year by a sample of 925 teens is normally distributed
with a mean of 1.8 and a standard deviation of 0.5.
Find each probability. Then use a graphing calculator
to sketch the area under each curve.
a. P(X < 2)
b. P(1 < X < 3)
SOLUTION: a.
z =
=
= 0.4
Formula for z-values
X = 2, σ = 0.5, and µ =
1.8
Simplify.
SOLUTION: –2.40 =
–2.4σ = –7.4
Subtract.
The z-value that corresponds to X = 13.7 is
approximately 2.4 standard deviations less than the
mean.
7. σ if μ = 21.1, X = 13.7, and z =-2.40
–2.40 =
z = –2.40, X = 13.7,
and µ = 21.1
ANSWER: 59.8; 1.38 standard deviations less than the mean
z =
Formula for z-values
Formula for z-values
z = –2.40, X = 13.7,
and µ = 21.1
Subtract.
Multiply each side by σ.
Divide each side by –
2.4.
Simplify.
P(X < 2) corresponds to a z-value of 0.4, so find the
area between z = –4 and z = 0.4.
Select ` ú. Then, under the DRAW menu, select
ShadeNorm(lower z value, upper z value).
The area between z = –4 and z = 0.44 is about 0.655
as shown in the graph.
The z-value that corresponds to X = 13.7 is
approximately 2.4 standard deviations less than the
mean.
ANSWER: 3.08;Manual
2.40 standard
deviations
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8. CONCERTS The number of concerts attended per
year by a sample of 925 teens is normally distributed
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
b. Find the z-values for X = 1 and X = 3.
Page 2
11-5 The Normal Distribution
b. 93.7%
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
b. Find the z-values for X = 1 and X = 3.
z =
=
= –1.6
Formula for z-values
X = 1, σ = 0.5, and µ =
1.8
Simplify.
z =
=
= 2.4
Formula for z-values
X = 3, σ = 0.5, and µ =
1.8
Simplify.
P(1 < X < 3) corresponds to z-values of –1.6 and 2.4,
so find the area between z = –1.6 and z = 2.4.
Select ` ú. Then, under the DRAWmenu, select
ShadeNorm(lower z value, upper z value).
The area between z = –1.6 and z = 2.4 is about 0.937
as shown in the graph.
A normal distribution has a mean of 29.3 and a
standard deviation of 6.7.
9. Find the range of values that represent the outside
5% of the distribution.
SOLUTION: The outside 5% of data in a normal distribution is
equal to 1 minus the middle 95%.
The middle 95% is the range from µ – 2σ to µ + 2σ.
The standard deviation is 6.7, so σ = 2 ∙ 6.7 or 13.4
29.3 – 13.4 = 15.9 and 29.3 + 13.4 = 42.7
The range of values in the middle 95% is 15.9 < X <
42.7. Therefore, the outside 5% is represented by X
< 15.9 or X > 42.7.
ANSWER: X < 15.9 or X > 42.7
10. What percent of the data will be between 22.6 and
42.7?
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
ANSWER: a. 65.5%
SOLUTION: 22.6 is 6.7 less than 29.3 and 42.7 is 13.4 more than
29.3, so the µ – σ to µ + 2σ represents the data that
are between 22.6 and 42.7.
For u – σ to u:
68% of the data fall within µ – σ and µ + σ. We are
only concerned with the lower section, so 34% of the
data will be between µ – σ and µ.
b. 93.7%
For u to u + 2σ:
95% of the data fall within µ – 2σ and µ + 2σ. We
are only concerned with the upper section, so 47.5%
of the data will be between µ and µ + 2σ.
Therefore, 34% + 47.5% or 81.5% of the data are
between µ – σ and µ + 2σ.
So, 81.5% of the data will be between 22.6 and 42.7.
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ANSWER: 81.5%
Page 3
42.7. Therefore, the outside 5% is represented by X
< 15.9 or X > 42.7.
ANSWER: 11-5XThe
Normal
< 15.9
or X >Distribution
42.7
10. What percent of the data will be between 22.6 and
42.7?
SOLUTION: 22.6 is 6.7 less than 29.3 and 42.7 is 13.4 more than
29.3, so the µ – σ to µ + 2σ represents the data that
are between 22.6 and 42.7.
For u – σ to u:
68% of the data fall within µ – σ and µ + σ. We are
only concerned with the lower section, so 34% of the
data will be between µ – σ and µ.
For u to u + 2σ:
95% of the data fall within µ – 2σ and µ + 2σ. We
are only concerned with the upper section, so 47.5%
of the data will be between µ and µ + 2σ.
So, 81.5% of the data will be between 22.6 and 42.7.
ANSWER: 81.5%
11. GYMS The number of visits to a gym per year by a
sample of 522 members is normally distributed with a
mean of 88 and a standard deviation of 19.
a. About how many members went to the gym at
least 50 times?
b. What is the probability that a member selected at
random went to the gym more than 145 times?
SOLUTION: a. 50 is 2σ away from the mean of 88, so the range
of values are greater than µ – 2σ. µ – 2σ to µ
represents half of 95% of the distribution, or 47.5%.
50% of the distribution is greater than µ, so
everything greater than µ – 2σ is 47.5% + 50% or
97.5%.
97.5% of 522 = 508.95
Therefore, 34% + 47.5% or 81.5% of the data are
between µ – σ and µ + 2σ.
Therefore, about 509 members went to the gym at
least 50 times.
So, 81.5% of the data will be between 22.6 and 42.7.
b. 145 is 3σ away from the mean, so the range of
values are more than µ + 3σ. The range of values
from µ – 3σ to µ + 3σ represent 99.7% of the data,
so the range from µ to µ + 3σ represents half of
these values, or 49.85% of the data. The outer tail of
the distribution is 50% – 49.85% or 0.15% of the
data. Therefore, the probability that a member went
to the gym more than 145 times is 0.15%.
ANSWER: 81.5%
11. GYMS The number of visits to a gym per year by a
sample of 522 members is normally distributed with a
mean of 88 and a standard deviation of 19.
a. About how many members went to the gym at
least 50 times?
b. What is the probability that a member selected at
random went to the gym more than 145 times?
SOLUTION: a. 50 is 2σ away from the mean of 88, so the range
of values are greater than µ – 2σ. µ – 2σ to µ
represents half of 95% of the distribution, or 47.5%.
50% of the distribution is greater than µ, so
everything greater than µ – 2σ is 47.5% + 50% or
97.5%.
ANSWER: a. about 509 members
b. 0.15%
Find the missing variable. Indicate the position
of X in the distribution.
12. z if μ = 3.3, X = 3.8, and σ = 0.2
SOLUTION: 97.5% of 522 = 508.95
Therefore, about 509 members went to the gym at
least 50 times.
b. 145 is 3σ away from the mean, so the range of
values are more than µ + 3σ. The range of values
from µ – 3σ to µ + 3σ represent 99.7% of the data,
so the range from µ to µ + 3σ represents half of
these values, or 49.85% of the data. The outer tail of
the distribution is 50% – 49.85% or 0.15% of the
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data. Therefore, the probability that a member went
to the gym more than 145 times is 0.15%.
z =
Formula for z-values
z =
= 2.5
X = 3.8, µ = 3.3, and σ =
0.2
Simplify.
The z-value that corresponds to X = 3.8 is
approximately 2.5 standard deviations more than the
mean.
ANSWER: 2.5; 2.5 standard deviations greater than the mean
13. z if μ = 19.9, X = 18.7, and σ = 0.9
SOLUTION: Page 4
approximately 2.5 standard deviations more than the
mean.
ANSWER: 11-52.5;
The2.5
Normal
Distribution
standard
deviations greater than the mean
13. z if μ = 19.9, X = 18.7, and σ = 0.9
SOLUTION: z =
Formula for z-values
z =
≈ –1.33
X = 18.7, µ = 19.9, and
σ = 0.9
Simplify.
ANSWER: 177.7; 1.73 standard deviations greater than the mea
16. VENDING A vending machine dispenses about 8.2
ounces of coffee. The amount varies and is normally
distributed with a standard deviation of 0.3 ounce.
Find each probability. Then use a graphing calculator
to sketch the corresponding area under the curve.
a. P(X < 8)
b. P(X > 7.5)
SOLUTION: a. Find the z-value that corresponds to X = 8.
z =
The z-value that corresponds to X = 18 is
approximately 1.33 standard deviations less than the
mean.
ANSWER: −1.33; 1.33 standard deviations less than the mean
14. μ if z = −0.92, X = 44.2, and σ = 8.3
SOLUTION: =
= –0.667
Formula for z-values
X = 8, µ = 8.2, and σ =
0.3
Simplify.
P(X < 8) corresponds to a z-value of –0.667, so find
the area between z = –4 and z = –0.667.
z =
Formula for z-val
Select ` ú. Then, under the DRAWmenu, select
ShadeNorm(lower z value, upper z value).
The area between z = –4 and z = –0.667 is about
–0.92 =
–7.636 = 44.2 – µ
µ – 7.636 = 44.2
µ = 51.836
X = 44.2, z = –0.9
0.252 as shown in the graph.
Multiply each side
Add µ to each sid
Add 7.636 to eac
The z-value that corresponds to X = 44.28 is approxi
standard deviations less than the mean.
ANSWER: 51.8; 0.92 standard deviations less than the mean
15. X if μ = 138.8, σ = 22.5, and z = 1.73
1.73 =
38.925 = X – 138.8
177.725 = X
b. Find the z-value that corresponds to X = 7.5.
SOLUTION: z =
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
z =
Formula for z-val
=
= –2.33
µ = 138.8, z = 1.7
Formula for z-values
X = 7.5, µ = 8.2, and σ =
0.3
Simplify.
Multiply each sid
Add 138.8 to eac
P(X > 7.5) corresponds to a z-value of –2.33, so find
the area between z = –2.33 and z = 4.
The z-value that corresponds to X = 177.725 is appro
standard deviations more than the mean.
ANSWER: 177.7; 1.73 standard deviations greater than the mea
16. VENDING A vending machine dispenses about 8.2
ounces of coffee. The amount varies and is normally
distributed with a standard deviation of 0.3 ounce.
Find each probability. Then use a graphing calculator
eSolutions Manual - Powered by Cognero
to sketch the corresponding area under the curve.
a. P(X < 8)
b. P(X > 7.5)
Select ` ú. Then, under the DRAWmenu, select
ShadeNorm(lower z value, upper z value).
The area between z = –2.33 and z = 4 is about 0.99
as shown in the graph
Page 5
Select ` ú. Then, under the DRAWmenu, select
ShadeNorm(lower z value, upper z value).
The area between z = –2.33 and z = 4 is about 0.99
11-5 The Normal Distribution
as shown in the graph
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
Use a graphing calculator to find the area between
the z-values.
ANSWER: a. 25.2%
b. 99.0%
17. CAR BATTERIES The useful life of a certain car
battery is normally distributed with a mean of 113,627
miles and a standard deviation of 14,266 miles. The
company makes 20,000 batteries a month.
a. About how many batteries will last between
90,000 and 110,000 miles?
b. About how many batteries will last more than
125,000 miles?
c. What is the probability that if you buy a car battery
at random, it will last less than 100,000 miles?
SOLUTION: a. Find the z-values associated with 90,000 and
110,000. The mean μ is 113,627 and the standard
deviation σ is 14,266.
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The value 0.35 is the percentage of batteries that will
last between 90,000 and 110,000 miles. The total
number of batteries in this group is 0.35 × 20,000 or 7000. b. Find the z-value associated with 125,000.
We are looking for values greater than 125,000, so
we can use a graphing calculator to find the area
between z = 0.797 and z = 4.
The value 0.21 is the percentage of batteries that will
last between more than 125,000 miles. The total
number of batteries in this group is 0.21 × 20,000 or 4200.
c. Find the z-value associated with 100,000.
Page 6
We are looking for values less than 100,000, so we
can use a graphing calculator to find the area
last between more than 125,000 miles. The total
number of batteries in this group is 0.21 × 20,000 or 4200.
11-5 The Normal Distribution
c. Find the z-value associated with 100,000.
z =
=
≈ 1.13
Formula for z-values
X = 200, σ = 23.6, and µ
= 173.3
Simplify.
P(150 < X < 200) corresponds to z-values of –0.99
and 1.13, so find the area between z = –0.99 and z =
1.13.
We are looking for values less than 100,000, so we
can use a graphing calculator to find the area
between z = −4 and z = −0.955.
The probability that if you buy a car battery at
random, it will last less than 100,000 miles is about
17.0%.
Select ` ú. Then, under the DRAWmenu, select
ShadeNorm(lower z value, upper z value).
The area between z = –0.99 and z = 1.13 is about
0.710 as shown in the graph.
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
About 71.0% of the product will last between 150
and 200 days.
ANSWER: a. about 7000 batteries
b. about 4200 batteries
c. 17.0%
b. Find the z-value for X = 225.
z =
18. FOOD The shelf life of a particular snack chip is
normally distributed with a mean of 173.3 days and a
standard deviation of 23.6 days.
a. About what percent of the product lasts between
150 and 200 days?
b. About what percent of the product lasts more than
225 days?
c. What range of values represents the outside 5% of
the distribution?
SOLUTION: a. Find the z-values for X = 150 and X = 200.
=
≈ 2.19
Formula for z-values
X = 225, σ = 23.6, and µ
= 173.3
Simplify.
P(X > 225) corresponds to a z-value of about 2.19, so
find the area between z = 2.19 and z = 4.
Select ` ú. Then, under the DRAWmenu, select
ShadeNorm(lower z value, upper z value).
The area between z = 2.19 and z = 4 is about 0.014
as shown in the graph.
z =
=
≈ –0.99
Formula for z-values
X = 150, σ = 23.6, and µ
= 173.3
Simplify.
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
z =
=
≈ 1.13
Formula for z-values
X = 200, σ = 23.6, and µ
= 173.3
Simplify.
About 1.4% of the product will last more than 225
days.
P(150 < X < 200) corresponds to z-values of –0.99
and 1.13, so find the area between z = –0.99 and z =
1.13.
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c. The outside 5% of the distribution is everything
outside of the middle 95% of the distribution. The
middle 95% of the distribution is represented by Page
the 7
range of values from µ – 2σ to µ + 2σ. The standard
deviation is 23.6, so µ – 2σ = 173.3 – 2(23.6) or
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
About 1.4% of the product will last more than 225
11-5days.
The Normal Distribution
c. The outside 5% of the distribution is everything
outside of the middle 95% of the distribution. The
middle 95% of the distribution is represented by the
range of values from µ – 2σ to µ + 2σ. The standard
deviation is 23.6, so µ – 2σ = 173.3 – 2(23.6) or
126.1. µ + 2σ = 173.3 + 2(23.6) or 220.5.
We are looking for values more than 1000, so we can
use a graphing calculator to find the area between z
= 1.487 and z = 4.
Therefore, the outside 5% of the data are
represented by X < 126.1 or X > 220.5.
ANSWER: a. 71.0%
b. 1.4%
c. X > 220.5 or X < 126.1
19. FINANCIAL LITERACY The insurance industry
uses various factors including age, type of car driven,
and driving record to determine an individual’s
insurance rate. Suppose insurance rates for a sample
population are normally distributed.
a. If the mean annual cost per person is $829 and the
standard deviation is $115, what is the range of rates
you would expect the middle 68% of the population
to pay annually?
b. If 900 people were sampled, how many would you
expect to pay more than $1000 annually?
c. Where on the distribution would you expect a
person with several traffic citations to lie? Explain
your reasoning.
d. How do you think auto insurance companies use
each factor to calculate an individual’s insurance
rate?
SOLUTION: a. The middle 68% represents all data values within
one standard deviation of the mean. Add ±$115 to
$829. The range of rates is $714 to $944.
b. Find the z-value associated with 1000.
We are looking for values more than 1000, so we can
use a graphing calculator to find the area between z
= 1.487 and z = 4.
The probability that a customer selected at random
will pay more than $1000 is about 6.8%. Out of 900
people, about 0.06847575 · 900 or 62 people will pay more than $1000.
c. Sample answer: I would expect people with
several traffic citations to lie to the far right of the
distribution where insurance costs are highest,
because I think insurance companies would charge
them more.
d. Sample answer: I think auto insurance companies
would charge younger people more than older people
because they have not been driving as long. I think
they would charge more for expensive cars and
sports cars and less for cars that have good safety
ratings. I think they would charge a person less if
they have a good driving record and more if they
have had tickets and accidents.
ANSWER: a. between $714 and $944
b. 62
c. Sample answer: I would expect people with
several traffic citations to lie to the far right of the
distribution where insurance costs are highest,
because I think insurance companies would charge
them more.
d. Sample answer: As the probability of an accident
occurring increases, the more an auto insurance
company is going to charge. I think auto insurance
companies would charge younger people more than
older people because they have not been driving as
long. I think they would charge more for expensive
cars and sports cars and less for cars that have good
safety ratings. I think they would charge a person
less if they have a good driving record and more if
they have had tickets and accidents.
20. STANDARDIZED TESTS Nikki took three
national standardized tests and scored an 86 on all
three. The table shows the mean and standard
deviation of each test.
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Page 8
20. STANDARDIZED TESTS Nikki took three
national standardized tests and scored an 86 on all
The tableDistribution
shows the mean and standard
11-5three.
The Normal
deviation of each test.
a. Calculate the z-values that correspond to her score
on each test.
b. What is the probability of a student scoring an 86
or lower on each test?
c. On which test was Nikki’s standardized score the
highest? Explain your reasoning.
SOLUTION: a.
Math:
z =
=
≈ 1.03
P(X < 86) Science:
Select ` ú. Then, select normalcdf(lower z
value, upper z value). P(X < 86) ≈ 79.1%
P(X < 86) Social Studies:
Select ` ú. Then, select normalcdf(lower z
value, upper z value). P(X < 86) ≈ 88.7%
Formula for z-values
X = 86, µ = 76 and σ =
9.7
Simplify.
Science:
z =
=
≈ 0.81
Formula for z-values
X = 86, µ = 81 and σ =
6.2
Simplify.
c. Social Studies; Sample answer: When the
distributions are standardized, Nikki’s relative scores
on each test are 84.8%, 79.1%, and 88.7%. These
standardized scores are Nikki’s scores in relation to
the population of scores for each individual test.
Therefore, her 86% on the Social Studies test was
better than 88.7% of the other test takers’ scores on
that test. This is the highest percentage of the three
tests, so Nikki performed the best in Social Studies.
Social Studies:
z =
=
≈ 1.21
Formula for z-values
X = 86, µ = 72 and σ =
11.6
Simplify.
b. P(X < 86) Math:
Select ` ú. Then, select normalcdf(lower z
value, upper z value). P(X < 86) ≈ 84.8%
P(X < 86) Science:
Select ` ú. Then, select normalcdf(lower z
value, upper z value). P(X < 86) ≈ 79.1%
ANSWER: a. Math: 1.03; Science: 0.81; Social Studies: 1.21
b. Math: 84.8%; Science: 79.1%; Social Studies:
88.7%
c. Social Studies; Sample answer: When the
distributions are standardized, Nikki’s relative scores
on each test are 84.8%, 79.1%, and 88.7%. These
standardized scores are Nikki’s scores in relation to
the population of scores for each individual test.
Therefore, her 86% on the Social Studies test was
better than 88.7% of the other test takers’ scores on
that test. This is the highest percentage of the three
tests, so Nikki performed the best in Social Studies.
21. CCSS CRITIQUE A set of normally distributed
tree diameters have mean 11.5 cm, standard
deviation 2.5, and range 3.6 to 19.8. Monica and
Hiroko are to find the range that represents the
middle 68% of the data. Is either of them correct?
Explain.
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Page 9
the population of scores for each individual test.
Therefore, her 86% on the Social Studies test was
better than 88.7% of the other test takers’ scores on
This is
the highest percentage of the three
11-5that
Thetest.
Normal
Distribution
tests, so Nikki performed the best in Social Studies.
21. CCSS CRITIQUE A set of normally distributed
tree diameters have mean 11.5 cm, standard
deviation 2.5, and range 3.6 to 19.8. Monica and
Hiroko are to find the range that represents the
middle 68% of the data. Is either of them correct?
Explain.
Use a graphing calculator to calculate the area
between z = 1.5714 and z = 6.
The probability of having 8 cases with a battery life
higher than 9.3 (having a battery life higher than
1.571 standard deviations above the mean) is
0.058091.
Let N be the total number of cases:
SOLUTION: Hiroko; Monica’s solution would work with a uniform
distribution.
ANSWER: Hiroko; Monica’s solution would work with a uniform
distribution.
22. CHALLENGE A case of digital audio players has
an average battery life of 8.2 hours with a standard
deviation of 0.7 hour. Eight of the players have a
battery life greater than 9.3 hours. If the sample is
normally distributed, how many players are in the
case?
SOLUTION: Given
There are about 138 players in the case.
ANSWER: 138
23. REASONING The term six sigma process comes
from the notion that if one has six standard deviations
between the mean of a process and the nearest
specification limit, there will be practically no items
that fail to meet the specifications. Is this a true
assumption? Explain.
SOLUTION: Sample answer: True; according to the Empirical
Rule, 99% of the data lie within 3 standard deviations
of the mean. Therefore, only 1% will fall outside of
three sigma. An infinitesimally small amount willPage
fall10
outside of six-sigma.
.
Use a graphing calculator to calculate the area
between
z -=Powered
1.5714byand
z = 6.
eSolutions
Manual
Cognero
ANSWER: 11-5ANSWER: The Normal Distribution
138
23. REASONING The term six sigma process comes
from the notion that if one has six standard deviations
between the mean of a process and the nearest
specification limit, there will be practically no items
that fail to meet the specifications. Is this a true
assumption? Explain.
SOLUTION: Sample answer: True; according to the Empirical
Rule, 99% of the data lie within 3 standard deviations
of the mean. Therefore, only 1% will fall outside of
three sigma. An infinitesimally small amount will fall
outside of six-sigma.
ANSWER: Sample answer: True; according to the Empirical
Rule, 99% of the data lie within 3 standard deviations
of the mean. Therefore, only 1% will fall outside of
three sigma. An infinitesimally small amount will fall
outside of six-sigma.
24. REASONING True or false: According to the
Empirical Rule, in a normal distribution, most of the
data will fall within one standard deviation of the
mean. Explain.
SOLUTION: Sample answer: True; according to the Empirical
Rule, 68% of the data lie within 1 standard deviation
of the mean.
ANSWER: Sample answer: True; according to the Empirical
Rule, 68% of the data lie within 1 standard deviation
of the mean.
25. OPEN ENDED Find a set of real-world data that
appears to be normally distributed. Calculate the
range of values that represent the middle 68%, the
middle 95%, and the middle 99.7% of the distribution.
SOLUTION: Sample answer: The scores per team in each game
of the first round of the 2010 NBA playoffs. The
mean is 96.56 and the standard deviation is 11.06.
The middle 68% of the distribution is 85.50 < X <
107.62. The middle 95% is 74.44 < X < 118.68. The
middle 99.7% is 63.38 < X < 129.74.
ANSWER: Sample answer: The scores per team in each game
of the first round of the 2010 NBA playoffs. The
mean is 96.56 and the standard deviation is 11.06.
The middle 68% of the distribution is 85.50 < X <
eSolutions
Manual
Cognero
107.62.
The- Powered
middle by
95%
is 74.44 < X < 118.68. The
middle 99.7% is 63.38 < X < 129.74.
ANSWER: Sample answer: True; according to the Empirical
Rule, 68% of the data lie within 1 standard deviation
of the mean.
25. OPEN ENDED Find a set of real-world data that
appears to be normally distributed. Calculate the
range of values that represent the middle 68%, the
middle 95%, and the middle 99.7% of the distribution.
SOLUTION: Sample answer: The scores per team in each game
of the first round of the 2010 NBA playoffs. The
mean is 96.56 and the standard deviation is 11.06.
The middle 68% of the distribution is 85.50 < X <
107.62. The middle 95% is 74.44 < X < 118.68. The
middle 99.7% is 63.38 < X < 129.74.
ANSWER: Sample answer: The scores per team in each game
of the first round of the 2010 NBA playoffs. The
mean is 96.56 and the standard deviation is 11.06.
The middle 68% of the distribution is 85.50 < X <
107.62. The middle 95% is 74.44 < X < 118.68. The
middle 99.7% is 63.38 < X < 129.74.
26. WRITING IN MATH Describe the relationship
between the z-value, the position of an interval of X
in the normal distribution, the area under the normal
curve, and the probability of the interval occurring.
Use an example to explain your reasoning.
SOLUTION: Sample answer: The z-value represents the position
of a value X in a normal distribution. A z-value of
1.43 means that the corresponding data value X is
1.43 standard deviations to the right of the mean in
the distribution. An interval of all of the values
greater than X in the distribution will be represented
by the area under the curve from z = 1.43 to z = 4.
This area is equivalent to the probability of the
interval occurring (a random data value falling within
the interval).
ANSWER: Sample answer: The z-value represents the position
of a value X in a normal distribution. A z-value of
1.43 means that the corresponding data value X is
1.43 standard deviations to the right of the mean in
the distribution. An interval of all of the values
greater than X in the distribution will be represented
by the area under the curve from z = 1.43 to z = 4.
This area is equivalent to the probability of the
interval occurring (a random data value falling within
the interval).
27. The lifetimes of 10,000 light bulbs are normally
distributed. The mean lifetime is 300 days, and the
standard deviation is 40 days. How many light bulbs
Page 11
will last between 260 and 340 days?
A 2500
B 3400
greater than X in the distribution will be represented
by the area under the curve from z = 1.43 to z = 4.
This area is equivalent to the probability of the
occurring
(a random data value falling within
11-5interval
The Normal
Distribution
the interval).
middle and right, the data are positively skewed.
Option J is the correct answer.
27. The lifetimes of 10,000 light bulbs are normally
distributed. The mean lifetime is 300 days, and the
standard deviation is 40 days. How many light bulbs
will last between 260 and 340 days?
A 2500
B 3400
C 5000
D 6800
29. SHORT RESPONSE In the figure below, RT = TS
and QR = QT. What is the value of x?
SOLUTION: Given
The probability that a randomly selected value in the
distribution is between μ – σ and μ + σ, that is,
between 300 – 40 or 260 and 300 + 40 or 340.
Option D is the correct answer.
ANSWER: J
SOLUTION: Given RT = TS and QR = QT.
Since RT = TS,
.
Since QR = QT,
.
ANSWER: D
28. Which description best represents the graph?
ANSWER: 32.5
F negatively skewed
G no correlation
H normal distribution
J positively skewed
SOLUTION: Since the graph is high on the left and low in the
middle and right, the data are positively skewed.
Option J is the correct answer.
ANSWER: J
29. SHORT RESPONSE In the figure below, RT = TS
and QR = QT. What is the value of x?
30. SAT/ACT The integer 99 can be expressed as a
sum of n consecutive positive integers. The value of
n could be which of the following?
I. 2
II. 3
III. 6
A I only
B II only
C III only
D I and II only
E I, II, and III
SOLUTION: 49 + 50 = 99
32 + 33 + 34 = 99
14 + 15 + 16 + 17 + 18 + 19 = 99
Option E is the correct answer.
ANSWER: E
SOLUTION: Given RT = TS and QR = QT.
Since RT = TS,
.
Since
QR =- Powered
QT, by Cognero
eSolutions
Manual
31. SNOW There is a 25% chance that it snows each da
the probability that it snows 3 out of the next 7 days.
SOLUTION: .
X n –X
P(X) = nCxp q
3
PagePr
12
Binomial
7– 3
P(3) = 7C3(0.25) (0.75)
n = 7, X = 3
ANSWER: The random variable X is the amount of precipitation
in a city per month. Precipitation can be anywhere
within a certain range. Therefore, X is continuous.
Option E is the correct answer.
11-5ANSWER: The Normal Distribution
E
31. SNOW There is a 25% chance that it snows each da
the probability that it snows 3 out of the next 7 days.
34. BRIDGES The Sydney Harbour Bridge connects
the Sydney central business district to northern
metropolitan Sydney. It has an arch in the shape of
a parabola that opens downward. Write an
equation of a parabola to model the arch, assuming
that the origin is at the surface of the water, beneath
the vertex of the arch.
SOLUTION: X n –X
P(X) = nCxp q
3
Binomial Pr
7– 3
P(3) = 7C3(0.25) (0.75)
≈ 0.173
n = 7, X = 3
Simplify.
Refer to Page 778.
The probability that it will snow in 3 of the next 7 day
ANSWER: 17.3%
SOLUTION: The coordinates of the top of the bridge (vertex) is
(0, 440).
The bridge touches the water level at (825, 0) and (–
8825, 0).
Substitute the values in the standard equation of a
parabola.
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
32. the number of pages in a newspaper
SOLUTION: The number of pages in a newspaper is countable,
and therefore discrete.
ANSWER: The random variable X is the number of stations in a
cable package. The stations are finite and countable,
so X is discrete.
Substitute the point (825, 0) in the equation and solve
for a.
33. the amount of precipitation in a city per month
SOLUTION: The amount of precipitation in a city is continuous
because it can take on any value.
ANSWER: The random variable X is the amount of precipitation
in a city per month. Precipitation can be anywhere
within a certain range. Therefore, X is continuous.
ANSWER: The equation of the parabola is about y = –0.00065x
+ 440.
2
2
about y = –0.00065x + 440
Identify the type of function represented by
each graph.
34. BRIDGES The Sydney Harbour Bridge connects
the Sydney central business district to northern
metropolitan Sydney. It has an arch in the shape of
a parabola that opens downward. Write an
equation of a parabola to model the arch, assuming
that the origin is at the surface of the water, beneath
the vertex of the arch.
Refer to Page 778.
SOLUTION: The coordinates of the top of the bridge (vertex) is
(0, 440).
The bridge touches the water level at (825, 0) and (–
eSolutions Manual - Powered by Cognero
8825, 0).
Substitute the values in the standard equation of a
parabola.
35. SOLUTION: greatest integer
ANSWER: greatest integer
Page 13
The equation of the parabola is about y = –0.00065x
+ 440.
ANSWER: 11-5 The Normal Distribution
2
about y = –0.00065x + 440
ANSWER: constant
38. Calculate the standard deviation of the population of
data.
Identify the type of function represented by
each graph.
SOLUTION: The sum of the values is 293.
Find the mean.
Find the square of the differences.
35. 2
SOLUTION: greatest integer
(13 – 14.65) = 2.7225
2
(18 – 14.65) = 11.2225
ANSWER: greatest integer
(17 – 14.65) = 5.5225
2
(21 – 14.65) = 40.3225
2
2
(16 – 14.65) = 1.8225
2
(9 – 14.65) = 31.9225
2
(11 – 14.65) = 13.3225
2
(28 – 14.65) = 178.2225
2
(8 – 14.65) = 44.2225
2
(10 – 14.65) = 21.6225
2
(7 – 14.65) = 58.5225
2
(19 – 14.65) = 18.9225
36. 2
SOLUTION: inverse variation or rational
(16 – 14.65) = 1.8225
2
(16 – 14.65) = 1.8225
ANSWER: inverse variation or rational
(12 – 14.65) = 7.0225
2
(19– 14.65) = 18.9225
2
2
(21 14.65) = 40.3225
2
(11– 14.65) = 13.3225
2
(8 – 14.65) = 44.2225
2
(13 – 14.65) = 2.7225
The sum of the squares is 558.55.
Find the variance.
37. SOLUTION: constant
Find the standard deviation.
ANSWER: constant
38. Calculate the standard deviation of the population of
data.
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SOLUTION: ANSWER: 5.28
Page 14
The 99% confidence interval is 1.52 ≤ µ ≤ 1.68.
11-6 Confidence Intervals and Hypothesis Testing
1. LUNCH A sample of 145 high school seniors was
asked how many times they go out for lunch per
week. The mean number of times was 2.4 with a
standard deviation of 0.7. Use a 90% confidence
level to calculate the maximum error of estimate.
SOLUTION: E
≈ 0.096
Maximum Error of
Estimate
z = 1.645, s = 0.7,
and n = 145
Simplify.
The maximum error of estimate is approximately
0.096.
ANSWER: 0.096
Identify the null and alternative hypotheses for
each statement. Then identify the statement
that represents the claim.
3. Lori thinks it takes a fast-food restaurant less than 2
minutes to serve her meal after she orders it.
SOLUTION: less than 2 minutes: μ < 2,
not less than 2 minutes: μ ≥ 2
The claim is μ < 2, and it is the alternative hypothesis
because it does not include equality. The null
hypothesis is μ ≥ 2, which is the complement of μ <
2.
H0: μ ≥ 2
Ha: μ < 2 (claim)
2. PRACTICE A poll of 233 randomly chosen high
school athletes showed that they spend an average of
1.6 hours practicing their sport during the off-season.
The standard deviation is 0.5 hour. Determine a 99%
confidence interval for the population mean.
SOLUTION: Find the maximum error of estimate.
Maximum Error of
E
Estimate
z = 2.576, s = 0.5, and
n = 233
≈ 0.08
Simplify.
Use the maximum error of estimate to find the
confidence interval CI.
CI =
ANSWER: ±E
= 1.6 ± 0.08
Confidence Interval for
Population Mean
= 1.6 and E = 0.08
The 99% confidence interval is 1.52 ≤ µ ≤ 1.68.
ANSWER: Identify the null and alternative hypotheses for
each statement. Then identify the statement
that represents the claim.
3. Lori thinks it takes a fast-food restaurant less than 2
minutes to serve her meal after she orders it.
SOLUTION: less than 2 minutes: μ < 2,
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not less than 2 minutes: μ ≥ 2
The claim is μ < 2, and it is the alternative hypothesis
ANSWER: H0: μ ≥ 2; Ha: μ < 2 (claim)
4. A snack label states that one serving contains one
gram of fat.
SOLUTION: contains 1 gram: μ = 1
does not contain 1 gram: μ ≠ 1
The claim is μ = 1, and it is the null hypothesis
because it does include equality. The alternative
hypothesis is μ ≠ 1, which is the complement of μ =
1.
H0: μ = 1 (claim)
Ha: μ ≠ 1
ANSWER: H0: μ = 1 (claim); Ha: μ ≠ 1
5. Mrs. Hart’s review game takes at least 20 minutes to
complete.
SOLUTION: at least 20 minutes: μ ≥ 20
less than 20 minutes: μ < 20
The claim is μ ≥ 20, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ < 20, which is the complement of μ ≥ 20.
H0: μ ≥ 20 (claim)
Ha: μ < 20
Page 1
Ha: μ ≠ 1
ANSWER: 11-6 Confidence Intervals and Hypothesis Testing
H0: μ = 1 (claim); Ha: μ ≠ 1
5. Mrs. Hart’s review game takes at least 20 minutes to
complete.
SOLUTION: at least 20 minutes: μ ≥ 20
less than 20 minutes: μ < 20
The claim is μ ≥ 20, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ < 20, which is the complement of μ ≥ 20.
per disc with a standard deviation of 1.9 minutes.
Test the hypothesis at 5% significance.
SOLUTION: State the claim and hypothesis.
H0: μ ≥ 84 (claim)
Ha: μ < 84
Determine the critical region.
The alternative hypothesis is μ < 84, so this is a lefttailed test. We are testing at 5% significance, so we
need to identify the z-value that corresponds with the
lower 5% of the distribution. Keystrokes: ` ú
InvNorm .05 e
H0: μ ≥ 20 (claim)
Ha: μ < 20
ANSWER: H0: μ ≥ 20 (claim); Ha: μ < 20
6. The tellers at a bank can complete no more than 18
transactions per hour.
SOLUTION: no more than 18 transactions: μ ≤ 18
more than 18 transactions: μ > 18
The claim is μ ≤ 18, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ > 18, which is the complement of μ ≤ 18.
H0: μ ≤ 18 (claim)
Ha: μ > 18
ANSWER: H0: μ ≤ 18 (claim); Ha: μ > 18
CCSS REASONING Identify the hypotheses and claim, decide whether to reject the null
hypothesis, and make a conclusion about the
claim.
7. COMPACT DISCS A manufacturer of blank
compact discs claims that each disc can hold at least
84 minutes of music. Using a sample of 219 compact
discs, Cayla calculated a mean time of 84.1 minutes
per disc with a standard deviation of 1.9 minutes.
Test the hypothesis at 5% significance.
SOLUTION: State the claim and hypothesis.
H0: μ ≥ 84 (claim)
The critical region is z < –1.645.
Calculate the z-statistic for the sample data.
z=
Formula for z-statistic
=
= 84.1, µ = 84, s = 1.9,
and n = 219
≈ 0.78
Decide whether to reject the null hypothesis.
H0 is not rejected because the z-statistic for the
sample does not fall within the critical region.
Make a conclusion about the claim.
There is not enough evidence to reject the claim that
the compact discs can hold at least 84 minutes of
music.
ANSWER: H0: μ ≥ 84 (claim); Ha: μ < 84; Do not reject H0;
The manufacturer’s claim that the discs can hold at
least 84 minutes cannot be rejected.
8. GOLF TEES A company claims that each golf tee
they produce is 5 centimeters in length. Using a sample of 168 tees, Angelene calculated a mean of
5.1 centimeters with a standard deviation of 0.3. Test
the hypothesis at 10% significance.
Ha: μ < 84
SOLUTION: State the claim and hypothesis.
Determine the critical region.
The alternative hypothesis is μ < 84, so this is a left-
H0: μ = 5 (claim)
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Page 2
they produce is 5 centimeters in length. Using a sample of 168 tees, Angelene calculated a mean of
5.1 centimeters with a standard deviation of 0.3. Test
11-6the
Confidence
and Hypothesis Testing
hypothesisIntervals
at 10% significance.
SOLUTION: State the claim and hypothesis.
H0: μ = 5 (claim)
Ha: μ ≠ 5
Determine the critical region.
The alternative hypothesis is μ ≠ 5, so this is a twotailed test. We are testing at 10% significance, so we
need to identify the z-value that corresponds with the
upper and lower 5% of the distribution. Keystrokes: `
ú InvNorm .05 e and ` ú InvNorm .95 e.
Make a conclusion about the claim.
There is enough evidence to reject the claim that
each golf tee is 5 centimeters in length.
ANSWER: H0: μ = 5 (claim); Ha: μ ≠ 5; Reject H0; The
company’s claim that each tee is 5 centimeters is
rejected.
9. MUSIC A sample of 76 albums had a mean run
time of 61.3 minutes with a standard deviation of 5.2
minutes. Use a 95% confidence level to calculate the
maximum error of estimate.
SOLUTION: The sample size n is 76 and the standard deviation s
is 5.2. The mean is not needed to calculate the
maximum error of estimate. The z-values which
correspond to 95% significance are ±1.96.
The maximum error of estimate is about 1.17.
ANSWER: 1.17
The critical region is –1.645 > z or z > 1.645.
Calculate the z-statistic for the sample data.
10. COLLEGE A poll of 218 students at a university
showed that they spend 11.8 hours per week
studying. The standard deviation is 3.7 hours.
Determine a 90% confidence interval for the
population mean.
SOLUTION: Find the maximum error of estimate.
z=
Formula for z-statistic
E
=
= 5.1, µ = 5, s = 0.3,
and n = 168
≈ 4.32
Decide whether to reject the null hypothesis.
H0 is rejected because the z-statistic for the sample
≈ 0.41
Use the maximum error of estimate to find the
confidence interval CI.
falls within the critical region.
CI = ± E
= 11.8 ±
0.41
Make a conclusion about the claim.
There is enough evidence to reject the claim that
each golf tee is 5 centimeters in length.
ANSWER: H0: μManual
= 5 (claim);
Ha:Cognero
μ ≠ 5; Reject H0; The
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company’s claim that each tee is 5 centimeters is
rejected.
Maximum Error of
Estimate
z = 1.645, s = 3.7, and
n = 218
Simplify.
Confidence Interval for
Population Mean
= 11.8 and E = 0.41
The 90% confidence interval is 11.39 ≤ µ ≤ 12.21.
ANSWER: 11.39 ≤ µ ≤ 12.21
Page 3
Identify the null and alternative hypotheses for
Ha: μ < 6
The maximum error of estimate is about 1.17.
11-6ANSWER: Confidence Intervals and Hypothesis Testing
1.17
10. COLLEGE A poll of 218 students at a university
showed that they spend 11.8 hours per week
studying. The standard deviation is 3.7 hours.
Determine a 90% confidence interval for the
population mean.
SOLUTION: Find the maximum error of estimate.
Maximum Error of
Estimate
z = 1.645, s = 3.7, and
n = 218
Simplify.
E
≈ 0.41
Use the maximum error of estimate to find the
confidence interval CI.
CI = ± E
= 11.8 ±
0.41
Confidence Interval for
Population Mean
= 11.8 and E = 0.41
The 90% confidence interval is 11.39 ≤ µ ≤ 12.21.
ANSWER: 11.39 ≤ µ ≤ 12.21
Identify the null and alternative hypotheses for
each statement. Then identify the statement
that represents the claim.
11. Julian sends at least six text messages to his best
friend every day.
SOLUTION: at least 6 texts: μ ≥ 6
less than 6 texts: μ < 6
The claim is μ ≥ 6, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ < 6, which is the complement of μ ≥ 6.
H0: μ ≥ 6 (claim)
Ha: μ < 6
ANSWER: H0: μ ≥ 6 (claim); Ha: μ < 6
12. A car company states that one of their vehicles gets
27 miles per gallon.
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SOLUTION: gets 27 miles per gallon: μ = 27
does not get 27 miles per gallon: μ ≠ 27
ANSWER: H0: μ ≥ 6 (claim); Ha: μ < 6
12. A car company states that one of their vehicles gets
27 miles per gallon.
SOLUTION: gets 27 miles per gallon: μ = 27
does not get 27 miles per gallon: μ ≠ 27
The claim is μ = 27, and it is the null hypothesis
because it does include equality. The alternative
hypothesis is μ ≠ 27, which is the complement of μ =
27.
H0: μ = 27 (claim)
Ha: μ ≠ 27
ANSWER: H0: μ = 27 (claim); Ha: μ ≠ 27
13. A company advertisement states that it takes no
more than 2 hours to paint a 200-square-foot room.
SOLUTION: no more than 2 hours: μ ≤ 2
more than 2 hours: μ > 2
The claim is μ ≤ 2, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ > 2, which is the complement of μ ≤ 2.
H0: μ ≤ 2 (claim)
Ha: μ > 2
ANSWER: H0: μ ≤ 2 (claim); Ha: μ > 2
14. A singer plays at least 18 songs at every concert.
SOLUTION: at least 18 songs: μ ≥ 18
less than 18 songs: μ < 18
The claim is μ ≥ 18, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ < 18, which is the complement of μ ≥ 18.
H0: μ ≥ 18 (claim)
Ha: μ < 18
ANSWER: H0: μ ≥ 18 (claim); Ha: μ < 18
Identify the hypotheses and claim, decide
Page 4
Ha: μ > 2
ANSWER: 11-6 Confidence Intervals and Hypothesis Testing
H0: μ ≤ 2 (claim); Ha: μ > 2
14. A singer plays at least 18 songs at every concert.
SOLUTION: at least 18 songs: μ ≥ 18
less than 18 songs: μ < 18
The claim is μ ≥ 18, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ < 18, which is the complement of μ ≥ 18.
The critical region is z < –2.326.
Calculate the z-statistic for the sample data.
z=
Formula for z-statistic
=
= 29.6, µ = 30, s = 3.9,
and n = 38
≈ –0.63
H0: μ ≥ 18 (claim)
Decide whether to reject the null hypothesis.
H0 is not rejected because the z-statistic for the
Ha: μ < 18
sample does not fall within the critical region.
ANSWER: H0: μ ≥ 18 (claim); Ha: μ < 18
Identify the hypotheses and claim, decide
whether to reject the null hypothesis, and make
a conclusion about the claim.
15. PIZZA A pizza chain promises a delivery time of
less than 30 minutes. Using a sample of 38 deliveries,
Chelsea calculated a mean delivery time of 29.6
minutes with a standard deviation of 3.9 minutes.
Test the hypothesis at 1% significance.
SOLUTION: State the claim and hypothesis.
H0: μ ≥ 30
Make a conclusion about the claim.
There is not enough evidence to support the claim of
a delivery time of less than 30 minutes.
ANSWER: H0: μ ≥ 30; Ha: μ < 30 (claim); Do not reject H0;
There is not enough evidence to support the pizza
chain’s claim of a delivery time of less than 30
minutes cannot be rejected.
16. CHEESE A company claims that each package of
cheese contains exactly 24 slices. Using a sample of
93 packages, Mr. Matthews calculated a mean of
24.1 slices with a standard deviation of 0.5. Test the
hypothesis at 5% significance.
Ha: μ < 30 (claim)
SOLUTION: State the claim and hypothesis.
Determine the critical region.
The alternative hypothesis is μ < 30, so this is a lefttailed test. We are testing at 1% significance, so we
need to identify the z-value that corresponds with the
lower 1% of the distribution. Keystrokes: ` ú
InvNorm .01 e
H0: μ = 24 (claim)
Ha: μ ≠ 24
Determine the critical region.
The alternative hypothesis is μ ≠ 24, so this is a twotailed test. We are testing at 5% significance, so we
need to identify the z-value that corresponds with the
upper and lower 2.5% of the distribution. Keystrokes:
` ú InvNorm .025 e and ` ú InvNorm .975 e.
The critical region is z < –2.326.
Calculate the z-statistic for the sample data.
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z=
Formula for z-statistic
The critical region is –1.965 > z or z > 1.96.
Page 5
11-6 Confidence Intervals and Hypothesis Testing
The critical region is –1.965 > z or z > 1.96.
The sample mean and standard deviation are 12.9
and 1.08, respectively.
Calculate the z-statistic for the sample data.
State the claim and hypothesis.
H0: μ = 12 (claim)
z=
Formula for z-statistic
Ha: μ ≠ 12
=
= 24.1, µ = 24, s = 0.5,
and n = 93
≈ 1.93
Decide whether to reject the null hypothesis.
H0 is not rejected because the z-statistic for the
Determine the critical region.
The alternative hypothesis is μ ≠ 12, so this is a twotailed test. Test at 10% significance. Identify the zvalue that corresponds with the upper and lower 5%
of the distribution. Keystrokes: ` ú InvNorm .05 e
and ` ú InvNorm .95 e.
sample does not fall within the critical region.
Make a conclusion about the claim.
There is not enough evidence to reject the claim that
each package contains exactly 24 slices.
ANSWER: H0: μ = 24 (claim); Ha: μ ≠ 24; Do not reject H0;
The company’s claim that each package contains 24
slices is not rejected.
17. DECISION MAKING The number of peaches in
40 random cans is shown below. Should the
manufacturer place a label on the can promising
exactly 12 peaches in every can? Explain your
reasoning.
13, 14, 13, 14, 12, 12, 12, 11, 15, 12, 13, 13, 14, 13, 14,
12, 15, 11, 11, 14,
13, 14, 14, 13, 12, 12, 12, 12, 13, 13, 11, 14, 14, 13, 14,
13, 13, 14, 12, 12
SOLUTION: Find the mean and standard deviation of the sample
data.
Enter the data as L1 in a graphing calculator and
then select STAT CALC 1 to obtain the statistics.
The critical region is –1.645 > z or z > 1.645.
Calculate the z-statistic for the sample data.
z=
Formula for z-statistic
=
= 12.9, µ = 12, s =
1.08, and n = 40
≈ 5.27
Decide whether to reject the null hypothesis.
H0 is rejected because the z-statistic for the sample
falls within the critical region. It would also be
rejected at 5% and 1% significance.
Make a conclusion about the claim.
There is enough evidence to reject the claim that
each can contains exactly 12 peaches. Therefore, the
company should not put the claim on the label.
The sample mean and standard deviation are 12.9
and 1.08, respectively.
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ANSWER: H0 : μ = 12; Ha : μ ≠ 12; The mean of the sample data is 12.9 with a standard deviation of about 1.08.
The z-statistic is about 5.27, which falls in the critical
region at 1% significance. Therefore, the null
hypothesis is rejected and the company should not
Page 6
make the claim on the label.
State the claim and hypothesis.
18. CCSS ARGUMENTS The number of chocolate
ANSWER: H0 : μ = 12; Ha : μ ≠ 12; The mean of the sample is 12.9 with
a standard
deviation
of about
1.08.
11-6data
Confidence
Intervals
and
Hypothesis
Testing
The z-statistic is about 5.27, which falls in the critical
region at 1% significance. Therefore, the null
hypothesis is rejected and the company should not
make the claim on the label.
The critical region is –1.645 > z or z > 1.645.
18. CCSS ARGUMENTS The number of chocolate
chips in 40 random cookies is shown below. Should
the manufacturer place a label on the package
promising exactly 20 chips on every cookie? Explain
your reasoning.
21, 19, 20, 20, 19, 19, 18, 21, 19, 17, 19, 18, 18, 20, 20,
19, 18, 20, 19, 20,
21, 21, 19, 17, 17, 18, 19, 19, 20, 17, 22, 21, 21, 20, 19,
18, 19, 17, 17, 21
SOLUTION: Find the mean and standard deviation of the sample
data.
Enter the data as L1 in a graphing calculator and
then select STAT CALC 1 to obtain the statistics.
Calculate the z-statistic for the sample data.
z=
Formula for z-statistic
=
= 19.175, µ = 20, s =
1.375, and n = 40
≈ –3.79
Decide whether to reject the null hypothesis.
H0 is rejected because the z-statistic for the sample
falls within the critical region. It would also be
rejected at 5% and 1% significance.
Make a conclusion about the claim.
There is enough evidence to reject the claim that
each cookies has exactly 20 chocolate chips.
Therefore, the company should not put the claim on
the label.
The sample mean and standard deviation are 19.175
and 1.375, respectively.
State the claim and hypothesis.
H0: μ = 20 (claim)
Ha: μ ≠ 20
Determine the critical region.
The alternative hypothesis is μ ≠ 20, so this is a twotailed test. Test at 10% significance. Identify the zvalue that corresponds with the upper and lower 5%
of the distribution. Keystrokes: ` ú InvNorm .05 e
and ` ú InvNorm .95 e.
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eSolutions
The critical region is –1.645 > z or z > 1.645.
ANSWER: H0 : μ = 20; Ha : μ ≠ 20; The mean of the sample data is 19.175 with a standard deviation of 1.375.
The z-statistic is about −3.79, which falls in the
critical region at 1% significance. Therefore, the null
hypothesis is rejected and the company should not
make the claim on the label.
19. MULTIPLE REPRESENTATIONS In this
problem, you will explore how the confidence interval
is affected by the sample size and the confidence
level. Consider a sample of data where
and s
=3.
a. GRAPHICAL Graph the 90% confidence
interval for n = 50, 100, and 250 on a number line.
b. ANALYTICAL How does the sample size affect
the confidence interval?
c. GRAPHICAL Graph the 90%, 95%, and 99%
confidence intervals for n = 150.
d. ANALYTICAL How does the confidence level
affect the confidence interval?
e. ANALYTICAL How does decreasing the size of
the confidence interval affect the accuracy of the
confidence interval?
SOLUTION: a. n = 50:
Find the maximum error of estimate.
Page 7
Maximum Error of
affect the confidence interval?
e. ANALYTICAL How does decreasing the size of
the confidence interval affect the accuracy of the
11-6confidence
Confidence
Intervals and Hypothesis Testing
interval?
SOLUTION: a. n = 50:
Find the maximum error of estimate.
Maximum Error of
Estimate
z = 1.645, s = 3, and n
= 50
Simplify.
E
≈ 0.70
Use the maximum error of estimate to find the
confidence interval CI.
CI =
±E
= 25 ± 0.31
Confidence Interval for
Population Mean
= 25 and E = 0.31
The 90% confidence interval is 24.69 ≤ µ ≤ 25.31.
Use the maximum error of estimate to find the
confidence interval CI.
CI =
±E
= 25 ± 0.70
Confidence Interval for
Population Mean
= 25 and E = 0.70
The 90% confidence interval is 24.30 ≤ µ ≤ 25.70.
n = 100:
Find the maximum error of estimate.
b. Sample answer: With everything else held
constant, increasing the same size will decrease the
size of the confidence interval.
c. 90%:
Find the maximum error of estimate.
E
Maximum Error of
Estimate
z = 1.645, s = 3, and
n = 100
Simplify.
E
≈ 0.49
Use the maximum error of estimate to find the
confidence interval CI.
≈ 0.40
Use the maximum error of estimate to find the
confidence interval CI.
CI =
CI =
±E
= 25 ± 0.49
= 25 and E = 0.49
The 90% confidence interval is 24.51 ≤ µ ≤ 25.49.
n = 250:
Find the maximum error of estimate.
Maximum Error of
Estimate
z = 1.645, s = 3, and
n = 250
Simplify.
≈ 0.31
Use the maximum error of estimate to find the
confidence interval CI.
CI =
±E
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= 25 ± 0.31
= 25 and E = 0.31
The 90% confidence interval is 24.69 ≤ µ ≤ 25.31.
= 25 and E = 0.40
95%:
Find the maximum error of estimate.
E
≈ 0.48
Maximum Error of
Estimate
z = 1.96, s = 3, and n
= 150
Simplify.
Use the maximum error of estimate to find the
confidence interval CI.
CI =
Confidence Interval for
Population Mean
Confidence Interval for
Population Mean
The 90% confidence interval is 24.60 ≤ µ ≤ 25.40.
E
±E
= 25 ± 0.40
Confidence Interval for
Population Mean
Maximum Error of
Estimate
z = 1.645, s = 3, and
n = 150
Simplify.
±E
= 25 ± 0.48
Confidence Interval for
Population Mean
= 25 and E = 0.48
The 95% confidence interval is 24.52 ≤ µ ≤ 25.48.
Page 8
99%:
Confidence Interval for
CI = ± E
Population Mean
11-6 Confidence Intervals and Hypothesis Testing
= 25 ± 0.48
= 25 and E = 0.48
b. Sample answer: With everything else held
constant, increasing the same size will decrease the
size of the confidence interval.
c. The 95% confidence interval is 24.52 ≤ µ ≤ 25.48.
99%:
Find the maximum error of estimate.
Maximum Error of
Estimate
z = 2.576, s = 3, and
n = 150
Simplify.
E
≈ 0.63
Use the maximum error of estimate to find the
confidence interval CI.
CI =
±E
= 25 ± 0.63
Confidence Interval for
Population Mean
= 25 and E = 0.63
The 99% confidence interval is 24.37 ≤ µ ≤ 25.63.
d. Sample answer: With everything else held
constant, increasing the confidence level will increase
the size of the confidence interval.
e. Sample answer: Expanding the confidence interval
reduces the accuracy of the estimate. So decreasing
the size of the confidence interval increases the
accuracy of the estimate. 20. ERROR ANALYSIS Tim and Judie want to test
whether a delivery service meets their promised time
of 45 minutes or less. Their hypotheses are shown
below. Is either of them correct?
d. Sample answer: With everything else held
constant, increasing the confidence level will
decrease the size of the confidence interval.
e. Expanding the confidence interval reduces the
accuracy of the estimate. So decreasing the size of
the confidence interval increases the accuracy of the
estimate.
ANSWER: a.
SOLUTION: Neither; Sample answer: The null hypothesis is a
statement of equality, so Tim is incorrect. The claim
is that the delivery time is less than or equal to 45, so
Judie is also incorrect.
ANSWER: Neither; Sample answer: The null hypothesis is a
statement of equality, so Tim is incorrect. The claim
is that the delivery time is less than or equal to 45, so
Judie is also incorrect.
b. Sample answer: With everything else held
constant, increasing the same size will decrease the
size of the confidence interval.
c. eSolutions Manual - Powered by Cognero
21. CHALLENGE A 95% confidence interval for the
mean weight of a 20-ounce box of cereal was 19.932
≤ μ ≤ 20.008 with a sample standard deviation of 0.128 ounces. Determine the sample size that led to
this interval.
SOLUTION: The range of values for the mean weight is 19.932 to
20.008. Half of this range is the maximum error of
Page 9
estimate.
ANSWER: Neither; Sample answer: The null hypothesis is a
statement of equality, so Tim is incorrect. The claim
11-6isConfidence
Intervals
that the delivery
time isand
lessHypothesis
than or equalTesting
to 45, so
Judie is also incorrect.
21. CHALLENGE A 95% confidence interval for the
mean weight of a 20-ounce box of cereal was 19.932
≤ μ ≤ 20.008 with a sample standard deviation of 0.128 ounces. Determine the sample size that led to
this interval.
SOLUTION: The range of values for the mean weight is 19.932 to
20.008. Half of this range is the maximum error of
estimate.
Use the formula for calculating the maximum error
of estimate to determine the same size. The z-values
which correspond to 95% significance are ±1.96.
ANSWER: 44
22. REASONING Determine whether the following
statement is sometimes, always, or never true.
Explain your reasoning.
If a confidence interval contains the H0 value of
μ, then it is not rejected.
SOLUTION: Sample answer: Always; if the null hypothesis value
of μ falls within the confidence interval, then it is not
rejected.
ANSWER: Sample answer: Always; if the null hypothesis value
of μ falls within the confidence interval, then it is not
rejected.
23. WRITING IN MATH How can a statistical test be
used in a decision-making process?
SOLUTION: Sample answer: You can use a statistical test to help
you to determine the strength of your decision.
ANSWER: Sample answer: You can use a statistical test to help
you to determine the strength of your decision.
24. OPEN ENDED Design and conduct your own
research study, and draw conclusions based on the
results of a hypothesis test. Write a brief summary of
your findings.
ANSWER: 44
22. REASONING Determine whether the following
statement is sometimes, always, or never true.
Explain your reasoning.
If a confidence interval contains the H0 value of
μ, then it is not rejected.
SOLUTION: Sample answer: Always; if the null hypothesis value
of μ falls within the confidence interval, then it is not
rejected.
ANSWER: Sample answer: Always; if the null hypothesis value
of μ falls within the confidence interval, then it is not
rejected.
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23. WRITING IN MATH How can a statistical test be
used in a decision-making process?
SOLUTION: Sample answer: A pizza company claims to put 100
pepperonis on its large pizza. H0 : μ = 100;
H1 : μ > 100; data collected: 100, 102, 101, 101, 100, 99, 99, 103,
102, 103, 103, 101, 102, 99, 105, 103, 102, 100, 101,
104;
, s = 1.70; a 95% confidence interval is
. H0 does not fall within the confidence interval, so we
reject the null hypothesis and accept the alternative
hypothesis.
ANSWER: Sample answer: A pizza company claims to put 100
pepperonis on its large pizza.
H0 : μ = 100;
H1 : μ > 100; Page 10
data collected: 100, 102, 101, 101, 100, 99, 99, 103,
102, 103, 103, 101, 102, 99, 105, 103, 102, 100, 101,
you to determine the strength of your decision.
ANSWER: answer:
You canand
use aHypothesis
statistical test
to help
11-6Sample
Confidence
Intervals
Testing
you to determine the strength of your decision.
24. OPEN ENDED Design and conduct your own
research study, and draw conclusions based on the
results of a hypothesis test. Write a brief summary of
your findings.
a 95% confidence interval is
. H0 does not fall within the confidence interval, so we
reject the null hypothesis and accept the alternative
hypothesis.
25. GEOMETRY In the graph below, line l passes
through the origin. What is the value of
?
SOLUTION: Sample answer: A pizza company claims to put 100
pepperonis on its large pizza. H0 : μ = 100;
H1 : μ > 100; data collected: 100, 102, 101, 101, 100, 99, 99, 103,
102, 103, 103, 101, 102, 99, 105, 103, 102, 100, 101,
104;
, s = 1.70; a 95% confidence interval is
. H0 does not fall within the confidence interval, so we
reject the null hypothesis and accept the alternative
hypothesis.
A –4
B
C
D4
ANSWER: Sample answer: A pizza company claims to put 100
pepperonis on its large pizza.
H0 : μ = 100;
SOLUTION: The coordinates of the point O is (0, 0).
Find the slope of the line using the points (0, 0) and
(1, 4).
H1 : μ > 100; data collected: 100, 102, 101, 101, 100, 99, 99, 103,
102, 103, 103, 101, 102, 99, 105, 103, 102, 100, 101,
104;
, s = 1.70; a 95% confidence interval is
. H0 does not fall within the confidence interval, so we
The slope of the line is
reject the null hypothesis and accept the alternative
hypothesis.
But
.
Find the slope of the line using the points (0, 0) and
(a, b).
25. GEOMETRY In the graph below, line l passes
Therefore,
through the origin. What is the value of
?
.
Option C is the correct answer.
ANSWER: C
2
A –4
B
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26. SAT/ACT If 5 + i and 5 – i are the roots of x –
10x + c = 0, what is the value of c?
F –26
G –25
H 25
J 26
SOLUTION: C
Given
D4
Page 11
.
Therefore, c = 26.
Option J is the correct answer.
Option C is the correct answer.
11-6ANSWER: Confidence Intervals and Hypothesis Testing
C
2
26. SAT/ACT If 5 + i and 5 – i are the roots of x –
10x + c = 0, what is the value of c?
F –26
G –25
H 25
J 26
ANSWER: J
27. The Service Club at Jake’s school was founded 8
years ago. The number of members of the club by
year is shown in the table. Which linear equation best
models the data?
SOLUTION: Given
.
Therefore, c = 26.
Option J is the correct answer.
ANSWER: J
27. The Service Club at Jake’s school was founded 8
years ago. The number of members of the club by
year is shown in the table. Which linear equation best
models the data?
A y
B y
C y
D y
= 1.4x
= 1.4x + 10.4
= 1.6x
= 1.6x + 11.1
SOLUTION: Keystrokes:
Press STAT ENTER 0 ENTER 2 ENTER 4 ENTER 6 ENTER 8 ENTER ► 1 1 ENTER 1 3 ENTER 1 5 ENTER 1 9 ENTER 2 2 ENTER STAT ► 4 ENTER .
A y
B y
C y
D y
= 1.4x
= 1.4x + 10.4
= 1.6x
= 1.6x + 11.1
SOLUTION: Keystrokes:
Press STAT ENTER 0 ENTER 2 ENTER 4 ENTER 6 ENTER 8 ENTER ► 1 1 ENTER 1 3 ENTER 1 5 ENTER 1 9 ENTER 2 2 ENTER STAT ► 4 ENTER .
Option B is the correct answer.
ANSWER: B
28. SHORT RESPONSE Solve for x:
.
SOLUTION: ANSWER: 14
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High12
29. HEALTH The heights of students at Madison Page
School are normally distributed with a mean of 66
inches and a standard deviation of 2 inches. Of the
1080 students in the school, how many would you
P(at least 4) = P(4) + P(5) = 0.312 + 0.116 = 0.428
or 42.8%
11-6ANSWER: Confidence Intervals and Hypothesis Testing
14
29. HEALTH The heights of students at Madison High
School are normally distributed with a mean of 66
inches and a standard deviation of 2 inches. Of the
1080 students in the school, how many would you
expect to be less than 62 inches tall?
ANSWER: 42.8%
Find a n for each geometric sequence.
31. =
, r = 3, n = 8
SOLUTION: SOLUTION: Given
.
The probability that a randomly selected value in the
distribution is less than μ – 2σ, that is, 66 – 2(2) or
62.
ANSWER: 729
P(x < 62) = 2% + 0.5% = 2.5%
So, 27 students expected to be less than 62 inches
tall.
32. , r = 4, n = 9
SOLUTION: ANSWER: 27 students
30. CAR WASH The Spanish Club is washing cars to
raise money. They have determined that 65% of the
customers donate more than the minimum amount for
the car wash. What is the probability that at least 4
of the next 5 customers will donate more than the
minimum?
X n –X
4
= 5C4(0.65)
5–4
(0.35)
≈ 0.312
= 16, r = 0.5, n = 8
SOLUTION: P(X) = nCxp q
ANSWER: 1024
33. SOLUTION: P(at least 4) = P(4) + P(5)
P(4)
=
Binomial Probability
Formula
n = 5, X = 4, p = 0.65,
and q = 0.35
Find a 1.
Simplify.
X n –X
P(X) = nCxp q
5
P(5)
= 5C5(0.65)
5–5
(0.35)
≈ 0.116
Binomial Probability
Formula
n = 5, X = 5, p = 0.65,
and q = 0.35
Simplify.
Find a 8.
P(at least 4) = P(4) + P(5) = 0.312 + 0.116 = 0.428
or 42.8%
ANSWER: 42.8%
Find a n for each geometric sequence.
eSolutions
, r =- Powered
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31. =Manual
SOLUTION: ANSWER: 1
Write each equation in standard form. State
whether the graph of the equation is a parabola,
Page 13
circle, ellipse, or hyperbola. Then graph the
equation.
2
2
11-6ANSWER: Confidence Intervals and Hypothesis Testing
1
Write each equation in standard form. State
whether the graph of the equation is a parabola,
circle, ellipse, or hyperbola. Then graph the
equation.
2
2
34. 4x + 2y = 8
2
35. x = 8y
SOLUTION: SOLUTION: The equation is in standard form of a parabola.
Therefore, it represents a parabola.
Graph the equation.
The equation is in standard form of an ellipse.
Therefore, it represents an ellipse.
Graph the equation.
ANSWER: parabola ANSWER: ellipse
y=
x
2
2
2
36. (x – 1) – 9(y – 4) = 36
SOLUTION: 2
35. x = 8y
SOLUTION: eSolutions Manual - Powered by Cognero
The equation is in standard form of a hyperbola.
Page 14
Therefore, it represents a hyperbola.
Graph the equation.
11-6 Confidence Intervals and Hypothesis Testing
2
2
Write an equation in slope-intercept form for
each graph.
36. (x – 1) – 9(y – 4) = 36
SOLUTION: The equation is in standard form of a hyperbola.
Therefore, it represents a hyperbola.
Graph the equation.
37. SOLUTION: The line passes trough the origin. Therefore, b = 0.
Find the slope of the line using the points (0, 0) and
(2.5, 2).
The equation of the line is
ANSWER: hyperbola
.
ANSWER: y = 0.8x
38. SOLUTION: Find the slope of the line using the points (4, 3) and
(7, –2).
Write an equation in slope-intercept form for
each graph.
Substitute 4, 3 and
for x1, y 1 and m in the point-
slope form and simplify.
37. SOLUTION: eSolutions
- Powered
by Cognero
The Manual
line passes
trough
the origin.
Therefore, b = 0.
Find the slope of the line using the points (0, 0) and
(2.5, 2).
Page 15
The equation of the line is
For all values of x the value of y is –4.
Therefore, the equation of the line is y = –4.
.
ANSWER: 11-6yConfidence
Intervals and Hypothesis Testing
= 0.8x
ANSWER: y=–4
Find each missing measure. Round to the
nearest tenth, if necessary.
40. 38. SOLUTION: Use the Pythagorean Theorem.
SOLUTION: Find the slope of the line using the points (4, 3) and
(7, –2).
2
2
2
a +b =c
2
2
2
9 + 15 = c
306 = c2
=c
17.5 ≈ c
Pythagorean Theorem
a = 9 and b = 15
Simplify.
Take the square root of
each side.
Evaluate the square
root.
Substitute 4, 3 and
for x1, y 1 and m in the point-
ANSWER: 17.5
slope form and simplify.
41. SOLUTION: Use the Pythagorean Theorem.
ANSWER: y=
+ a +b =c
2
2
2
Pythagorean Theorem
2
2
2
a = 7 and b = 24
Simplify.
Take the square root of
each side.
Evaluate the square
root.
7 + 24 = c
625 = c2
=c
25 = c
ANSWER: 25
39. SOLUTION: For all values of x the value of y is –4.
Therefore, the equation of the line is y = –4.
ANSWER: y=–4
Find each missing measure. Round to the
nearest tenth, if necessary.
42. eSolutions Manual - Powered by Cognero
Page 16
SOLUTION: Use the Pythagorean Theorem.
25 = c
root.
11-6ANSWER: Confidence Intervals and Hypothesis Testing
25
42. SOLUTION: Use the Pythagorean Theorem.
2
a +
2
2 =c
b
Pythagorean Theorem
2
10 +
2
2 = 17
b
100 +
2 = 289
b
2
b = 189
b
b ≈ 13.7
a = 10 and c = 17
Simplify.
Subtract 100 from
each side.
Take the square root
of each side.
Evaluate the square
root.
ANSWER: 13.7
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Page 17
Practice Test - Chapter 11
[59, 75] scl:1 by [0, 5] scl: 1
1. BUTTERFLIES Students in a biology class are
learning about the monarch butterfly’s life cycle.
Each student is given a caterpillar. When a caterpillar
turns into chrysalis, it is placed in a glass enclosure
with food and a heat lamp and examined.
a. Determine whether the situation describes a
survey, an experiment or an observational study.
b. Identify the sample, and suggest a population from
which it was selected.
SOLUTION: a. observational study
b. sample: the monarch butterflies observed in the
study; population: all monarch butterflies
ANSWER: a. observational study
b. sample: the monarch butterflies observed in the
study; population: all monarch butterflies
2. HEIGHTS The heights of Ms. Joy’s dance students
are shown.
a. Use a graphing calculator to create a box-andwhisker plot. Then describe the shape of the
distribution.
b. Describe the center and spread of the data using
either the mean and standard deviation or the fivenumber summary. Justify your choice.
SOLUTION: a. Enter the data as L1. Press 2ND [STAT PLOT]
ENTER ENTER and choose fl. Adjust the window
to the dimensions shown.
The data to the right of the median are distributed
over a wider range than the data to the left. There is
a longer tail to the right. Therefore, the distribution is
positively skewed.
b. The distribution is skewed, so use the five-number
summary. Press STAT ► ENTER ENTER and
scroll down to view the five-number summary.
The range is 60 to 74. The median is 64, and half of
the data are between 63 and 67.
ANSWER: a.
positively skewed
b. The distribution is skewed, so use the five-number
summary. The values range from 60 to 74. The
median is 64, and half of the data are between 63
and 67.
3. A binomial distribution has a 65% rate of success. Th
a. What is the probability that there will be exactly 12
b. What is the probability that there will be at least 10
SOLUTION: a.
X n –X
P(X) = nCxp q
12
P(12)
[59, 75] scl:1 by [0, 5] scl: 1
The data to the right of the median are distributed
over a wider range than the data to the left. There is
12
≈ 0.111
eSolutions Manual - Powered by Cognero
15 –
= 15C12(0.65) (0.35)
Binomial Probabilit
Formula
n = 15
X = 12
p = 0.65
q = 0.35
Page 1
Simplify.
b. P(at least 10) = P(10) + P(11) + P(12) + P(13) +
P(X) = nCxp q
12
15 –
= 15C12(0.65)
(0.35)
Practice Test
- Chapter
11
P(12)
12
≈ 0.111
Probabilit
Formula
n = 15
X = 12
p = 0.65
q = 0.35
Simplify.
X n –X
P(X) = nCxp q
15
b. P(at least 10) = P(10) + P(11) + P(12) + P(13) +
Binomial X n –X
P(X) = nCxp q
Probabilit
Formula
n = 15
10
15 –
= 15C10(0.65) (0.35)
X = 10
P(10)
p = 0.65
10
q = 0.35
≈ 0.212
Simplify.
15 –
Binomial
Probabilit
Formula
n = 15
X = 11
p = 0.65
q = 0.35
Simplify.
15 –
Binomial
Probabilit
Formula
n = 15
X = 12
p = 0.65
q = 0.35
Simplify.
X n –X
P(X) = nCxp q
11
P(11)
= 15C11(0.65) (0.35)
11
≈ 0.179
X n –X
P(X) = nCxp q
12
P(12)
= 15C12(0.65) (0.35)
12
≈ 0.111
X n –X
P(X) = nCxp q
13
P(13)
15 –
= 15C13(0.65) (0.35)
13
≈ 0.048
Binomial
Probabilit
Formula
n = 15
X = 13
p = 0.65
q = 0.35
Simplify.
X n –X
P(X) = nCxp q
14
P(14)
15 –
= 15C14(0.65) (0.35)
14
≈ 0.013
Binomial
Probabilit
Formula
n = 15
X = 14
p = 0.65
q = 0.35
Simplify.
X n –X
P(X) = nCxp q
eSolutions Manual - Powered by Cognero
15
P(15)
15 –
= 15C15(0.65) (0.35)
15
Binomial Probabilit
Formula
n = 15
X = 15
p = 0.65
q = 0.35
q = 0.35
Simplify.
≈ 0.013
P(15)
15 –
= 15C15(0.65) (0.35)
15
≈ 0.002
Binomial Probabilit
Formula
n = 15
X = 15
p = 0.65
q = 0.35
Simplify.
P(at least 10) = P(10) + P(11) + P(12) + P(13) + P(1
= 0.212 + 0.179 + 0.111 + 0.048 + 0.013 + 0.002 = 0.
ANSWER: a. about 11.1%
b. about 56.5%
Determine whether each experiment is a
binomial experiment or can be reduced to a
binomial experiment. If so, describe a trial,
determine the random variable, and state n, p ,
and q.
4. A poll found that 65% of high school teachers own a
pet. You ask 15 high school teachers if they own a
pet.
SOLUTION: This experiment can be reduced to a binomial
experiment. Success is yes, failure is no, a trial is
asking a teacher, and the random variable is the
number of yeses; n = 15, p = 0.65, q = 0.35.
ANSWER: This experiment can be reduced to a binomial
experiment. Success is yes, failure is no, a trial is
asking a teacher, and the random variable is the
number of yeses; n = 15, p = 0.65, q = 0.35.
5. A study finds that 20% of families in a town have a
land line telephone. You ask 55 families how many
land line telephones they have.
SOLUTION: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
ANSWER: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
6. A survey found that on a scale of 1 to 5, a sandwich
received a 3.5 rating. A restaurant manager asks 150
customers to rate the sandwich on a scale of 1 to 5.
SOLUTION: This experiment cannot be reduced to a binomialPage 2
experiment because there are more than two possible
outcomes.
ANSWER: This experiment cannot be reduced to a binomial
experiment
Practice
Test - because
Chapterthere
11 are more than two possible
outcomes.
6. A survey found that on a scale of 1 to 5, a sandwich
received a 3.5 rating. A restaurant manager asks 150
customers to rate the sandwich on a scale of 1 to 5.
SOLUTION: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
ANSWER: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
7. the number of laps that Jesse swims
SOLUTION: The random variable X is the number of laps that
Jesse swims. The laps are countable, so X is
discrete.
ANSWER: The random variable X is the number of laps that
Jesse swims. The laps are countable, so X is
discrete.
ANSWER: The random variable X is the weights of pets in a pet
store. Weight can be anywhere within a certain
range. Therefore, X is continuous.
10. MULTIPLE CHOICE The table shows the
number of gift cards previously won in a mall
contest. What is the expected value of the gift card
that is won?
A $250.00
B $223.15
C $143.25
D $100.23
SOLUTION: The correct choice is C.
8. the body temperatures of patients in a hospital
SOLUTION: The random variable X is the body temperatures of
patients in a hospital. Temperature can be anywhere
within a certain range. Therefore, X is continuous.
ANSWER: The random variable X is the body temperatures of
patients in a hospital. Temperature can be anywhere
within a certain range. Therefore, X is continuous.
9. the weights of pets in a pet store
SOLUTION: The random variable X is the weights of pets in a pet
store. Weight can be anywhere within a certain
range. Therefore, X is continuous.
ANSWER: The random variable X is the weights of pets in a pet
store. Weight can be anywhere within a certain
range. Therefore, X is continuous.
10. MULTIPLE CHOICE The table shows the
eSolutions
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Cognero won in a mall
number
of -gift
cardsbypreviously
contest. What is the expected value of the gift card
that is won?
ANSWER: C
11. WEIGHTS The weights of 1500 bodybuilders are
normally distributed with a mean of 190.6 pounds and
a standard deviation of 5.8 pounds.
a. About how many bodybuilders are between 180
and 190 pounds?
b. What is the probability that a bodybuilder selected
at random has a weight greater than 195 pounds?
SOLUTION: a.
Find the z-values for X = 180 and X = 190.
z
Formula for z-values
X = 180
µ = 190.6
σ = 5.8
= –1.828
Simplify.
z
= –0.103
Formula for z-values
X = 190
Page 3
µ = 190.6
σ = 5.8
Simplify.
= –1.828
µ = 190.6
σ = 5.8
Simplify.
Practice
Test - Chapter 11
z
= –0.103
Formula for z-values
X = 190
µ = 190.6
σ = 5.8
Simplify.
P(195 < X) corresponds to z-values greater than
0.759 , so find the area between z = 0.759 and z = 4.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
P(180 < X < 190) corresponds to z-values of –1.828
and –0.103, so find the area between z = –1.828 and
z = –0.103.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
The probability of a bodybuilder being over 195
pounds is about 22.4%.
ANSWER: a. about 638 bodybuilders
b. 22.4%
A normal distribution has a mean of 16.4 and a
standard deviation of 2.6.
12. Find the range of values that represent the middle
95% of the distribution.
The probability of a bodybuilder being between 180
and 190 pounds is about 42.5%. Multiply this by 1500
bodybuilders.
SOLUTION: The middle 95% of data in a normal distribution is the
range from µ – 2σ to µ + 2σ. The standard deviation
is 2.6, so 2σ = 2 ∙ 2.6 or 5.2.
1500 ∙ 0.425 = 637.5
16.4 – 5.2 = 11.2
16.4 + 5.2 = 21.6
About 638 bodybuilders will be between 180 and 190
pounds.
b. Find the z-value for X = 195.
z
= 0.759
Formula for z-values
X = 195
µ = 190.6
σ = 5.8
Simplify.
Therefore, the range of values in the middle 95% is
11.2 < X < 21.6.
ANSWER: 11.2 < X < 21.6
13. What percent of the data will be less than 19?
P(195 < X) corresponds to z-values greater than
0.759 , so find the area between z = 0.759 and z = 4.
SOLUTION: 19 is 2.6 more than 16.4, so µ + σ represents the data
value of 19. Half of the data, or 50%, will be less
than µ.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
68% of the data can be represented by µ – σ and µ
+ σ. We are only concerned with the data from µ to
µ + σ, or half of 68%, so 34% of the data are
between µ and µ + σ.
So, the data less than µ + σ will include 50% + 34%
or 84% of the data in the distribution.
ANSWER: 84%
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The probability of a bodybuilder being over 195
pounds is about 22.4%.
Page 4
14. MULTIPLE CHOICE A survey of 300 randomly
selected members of a fitness club showed that they
spent an average of 25.3 minutes per visit in the gym
So, the data less than µ + σ will include 50% + 34%
or 84% of the data in the distribution.
ANSWER: Practice
Test - Chapter 11
84%
14. MULTIPLE CHOICE A survey of 300 randomly
selected members of a fitness club showed that they
spent an average of 25.3 minutes per visit in the gym
with a standard deviation of 8.6 minutes. Which is
the maximum error of estimate for the time spent in
the gym using a 95% confidence interval?
F 0.82
G 0.97
H 1.28
J 2.86
≈ 0.973
ANSWER: H0: μ = 5 (claim); Ha: μ ≠ 5
16. A valet says that no more than 56 cars can be
parked in the parking lot at one time.
SOLUTION: no more than 56 cars: μ ≤ 56
more than 56 cars: μ > 56
The claim is μ ≤ 56, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ > 56, which is the complement of μ ≤ 56.
SOLUTION: E
Ha: μ ≠ 5
Maximum Error of
Estimate
z = 1.96
s = 8.6
n = 300
Simplify.
H0: μ ≤ 56 (claim)
Ha: μ > 56
ANSWER: H0: μ ≤ 56 (claim); Ha: μ > 56
The maximum error of estimate is approximately
0.97. The correct choice is G.
ANSWER: G
Identify the null and alternative hypotheses for
each statement. Then identify the statement
that represents the claim.
15. A restaurant owner says that each milk shake
contains 5 strawberries.
SOLUTION: contains 5 strawberries: μ = 5 does not contain 5 strawberries: μ ≠ 5
The claim is μ = 5, and it is the null hypothesis
because it does include equality. The alternative
hypothesis is μ ≠ 5, which is the complement of μ =
5.
H0: μ = 5 (claim)
Ha: μ ≠ 5
ANSWER: H0: μ = 5 (claim); Ha: μ ≠ 5
16. A valet says that no more than 56 cars can be
parked in the parking lot at one time.
SOLUTION: no more than 56 cars: μ ≤ 56
more than 56 cars: μ > 56
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The claim is μ ≤ 56, and it is the null hypothesis because it does include equality. The alternative
Page 5
empirical rule works for z-values of –3, –2, –1, 0, 1,
2, and 3.
Study Guide and Review - Chapter 11
Choose a term from the list above that best
completes each statement.
1. A(n) for a particular random
variable is a function that maps the sample space to
the probabilities of the outcomes of the sample
space. SOLUTION: A probability distribution for a particular random
variable is a function that maps the sample space to
the probabilities of the outcomes of the sample
space. It distributes all of the possible probabilities.
ANSWER: probability distribution
2. A(n) is an error that results in a
misrepresentation of members of a population.
SOLUTION: A bias is an error that results in a misrepresentation
of members of a population.
ANSWER: bias
3. In a statistical study, data are collected and used to
answer questions about a population characteristic or
.
SOLUTION: In a statistical study, data are collected and used to
answer questions about a population characteristic or
parameter. A sample characteristic is called a
statistic.
ANSWER: parameter
4. The can be used to determine the
area under the normal curve at specific intervals.
SOLUTION: The empirical rule can be used to determine the area
under the normal curve at specific intervals. The
empirical rule works for z-values of –3, –2, –1, 0, 1,
2, and 3.
ANSWER: Empirical Rule
5. In a(n) , members of a sample are
measured or observed without being affected by the
study.
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SOLUTION: In an observational study, members of a sample are
measured or observed without being affected by the
ANSWER: Empirical Rule
5. In a(n) , members of a sample are
measured or observed without being affected by the
study.
SOLUTION: In an observational study, members of a sample are
measured or observed without being affected by the
study. In an experiment, the members are affected
by the study.
ANSWER: observational study
6. A is an estimate of a population
parameter stated as a range with a specific degree of
certainty.
SOLUTION: A confidence interval is an estimate of a population
parameter stated as a range with a specific degree of
certainty.
ANSWER: confidence interval
Determine whether each situation describes a
survey, an experiment, or an observational study.
Then identify the sample, and suggest a
population from which it may have been
selected.
7. SHOPPING Every tenth shopper coming out of a
store is asked questions about his or her satisfaction
with the store.
SOLUTION: This is a survey because data are collected from
responses to the question.
sample: every tenth shopper; population: all potential
hardware store shoppers
ANSWER: survey; sample: every tenth shopper; population: all
potential shoppers
8. MILK SHAKE A fast food restaurant gives 25 of
their customers a sample of a new milk shake and
employees monitor their reactions as they taste it.
SOLUTION: This is an observational study because the reactions
of the employees are observed and the employees
are unaffected by the study.
ANSWER: observational study; sample: the 25 customers;
population: all potential customers
Page 1
hardware store shoppers
ANSWER: survey; sample: every tenth shopper; population: all
Study
Guide shoppers
and Review - Chapter 11
potential
8. MILK SHAKE A fast food restaurant gives 25 of
their customers a sample of a new milk shake and
employees monitor their reactions as they taste it.
SOLUTION: This is an observational study because the reactions
of the employees are observed and the employees
are unaffected by the study.
ANSWER: observational study; sample: the 25 customers;
population: all potential customers
[8, 18] scl: 1 by [0, 10] scl: 1
The majority of the data are on the left of the mean
and the data to the right of the mean are distributed
over a wider range than the data to the left, so the
distribution is positively skewed.
b.
Use the five-number summary.
Press STAT ► ENTER ENTER and scroll down
to view the five-number summary.
9. SCHOOL Every fifth person coming out of a high
school is asked what their favorite class is.
SOLUTION: This is a survey because data are collected from
responses to the question.
ANSWER: survey; sample: every fifth person; population:
student body
10. DOGSLED The Iditarod is a race across Alaska.
The table shows the winning times, in days, for
recent years.
Iditarod Winning Times
9.1, 9.4, 10.3, 9.3, 9.6, 8.7, 9.5, 9.4, 9.2, 17.3, 15.4,
15.5, 14.2, 12.0, 16.6, 13.5, 13.0, 18.1, 12.4, 11.6,
11.5, 11.3, 11.3, 13.1, 11.2, 11.6, 11.6, 9.7
a. Use a graphing calculator to create a histogram.
Then describe the shape of the distribution.
b. Describe the center and spread of the data using
either the mean and standard deviation or the fivenumber summary. Justify your choice.
SOLUTION: a. Enter the data in L1, then select [STAT PLOT]
ENTER ENTER and choose „ . Finally, adjust the
window to the dimensions shown.
[8, 18] scl: 1 by [0, 10] scl: 1
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The majority of the data are on the left of the mean
and the data to the right of the mean are distributed
over a wider range than the data to the left, so the
The range is 8.7 to 18.1 days. The median is 11.55
days, and half of the data are between 9.55 and 13.3
days.
ANSWER: a.
positively skewed
b. Sample answer: The distribution is skewed, so use
the five-number summary. The values range from
8.7 to 18.1 days. The median is 11.55 days, and half
of the data are between 9.55 and 13.3 days.
11. SWIMMING Kelly’s practice times in the 400meter individual medley are shown in the table.
Times in Seconds
301, 311, 320, 308, 312, 307, 303, 305, 309, 308,
304, 302, 311, 313, 313, 316, 314, 306, 329, 326,
319, 310, 306, 309, 320, 318, 315, 318, 314, 309,
a. Use a graphing calculator to create a box-andwhisker plot. Then describe the shape of the
distribution.
b. Describe the center and spread of the data using
either the mean and standard deviation or the fivePage 2
number summary. Justify your choice.
SOLUTION: 319, 310, 306, 309, 320, 318, 315, 318, 314, 309,
a. Use a graphing calculator to create a box-andwhisker
describe
the shape
Study
Guideplot.
andThen
Review
- Chapter
11 of the
distribution.
b. Describe the center and spread of the data using
either the mean and standard deviation or the fivenumber summary. Justify your choice.
SOLUTION: a. Enter the data as L1. Press 2ND [STAT PLOT]
ENTER ENTER and choose fl. Adjust the window
to the dimensions shown.
positively skewed
b. Sample answer: The distribution is skewed, so use
the five-number summary. Kelly’s times range from
301 to 329 seconds. The median is 311, and half of
the data are between 307 and 316 seconds.
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
12. the number of ice cream sandwiches sold at an ice
cream shop
SOLUTION: The random variable X is the number of ice cream
sandwiches sold. The number of ice cream
sandwiches countable, so X is discrete.
[300, 330] scl: 1 by [0, 5] scl: 1
The data to the right of the median are distributed
over a wider range than the data to the left. There is
a longer tail to the right. Therefore, the distribution is
positively skewed.
b. The distribution is skewed, so use the five-number
summary. Press STAT ► ENTER ENTER and
scroll down to view the five-number summary.
ANSWER: The random variable X is the number of ice cream
sandwiches sold. The number of ice cream
sandwiches countable, so X is discrete.
13. the time it takes to run a 5-kilometer race
SOLUTION: The random variable X is the time it takes to run the
race. Time can be anywhere within a certain range.
Therefore, X is continuous.
ANSWER: The random variable X is the time it takes to run the
race. Time can be anywhere within a certain range.
Therefore, X is continuous.
Kelly’s times range from 301 to 329 seconds. The
median is 311, and half of the data are between 307
and 316 seconds.
14. DANCE RECITALS The probability distribution
lists the probable number of dance recitals per year
at Rena’s Dance School. Determine the expected
number of dance recitals per year.
ANSWER: a.
SOLUTION: positively skewed
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eSolutions
b. Sample answer: The distribution is skewed, so use
the five-number summary. Kelly’s times range from
The expected value is about 1.51 dance recitals.
ANSWER: 1.51
Page 3
ANSWER: The random variable X is the time it takes to run the
race.
Timeand
canReview
be anywhere
within11a certain range.
Study
Guide
- Chapter
Therefore, X is continuous.
14. DANCE RECITALS The probability distribution
lists the probable number of dance recitals per year
at Rena’s Dance School. Determine the expected
number of dance recitals per year.
SOLUTION: The expected value is about 1.51 dance recitals.
ANSWER: 1.51
15. SNOW DAYS The distribution lists the number of
snow days per year at Washington Elementary over
the past 26 years. Determine the expected number of
snow days this year.
SOLUTION: The expected value is about 1.88 snow days.
ANSWER: 1.88 snow days
Determine whether each experiment is a
binomial experiment or can be reduced to a
binomial experiment. If so, describe a trial,
determine the random variable, and state n, p ,
and q.
16. A survey found that 30% of adults like chocolate ice
cream more than any other flavor. You ask 35 adults
if they prefer chocolate ice cream more than any
other flavor.
SOLUTION: This experiment can be reduced to a binomial
experiment. Success is yes, failure is no, a trial is
asking an adult, and the random variable is the
number of yeses; n = 35, p = 0.30, q = 0.70.
ANSWER: This experiment can be reduced to a binomial
experiment. Success is yes, failure is no, a trial is
asking an adult, and the random variable is the
number of yeses; n = 35, p = 0.30, q = 0.70.
17. Thirty random guests from Sheila’s birthday party
are asked their favorite song.
SOLUTION: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
ANSWER: This experiment cannot be reduced to a binomial
experiment because there are more than two possible
outcomes.
18. WATCHES According to an online poll, 74% of adu
surveyed 25 random adults. What is the probability th
wear a watch?
SOLUTION: X n –X
P(X) = nCxp q
The expected value is about 1.88 snow days.
20
Determine whether each experiment is a
binomial experiment or can be reduced to a
binomial experiment. If so, describe a trial,
determine the random variable, and state n, p ,
and q.
16. A survey found that 30% of adults like chocolate ice
cream more than any other flavor. You ask 35 adults
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if they prefer chocolate ice cream more than any
other flavor.
25 – 20
P(20) = 25C20(0.74) (0.26)
ANSWER: 1.88 snow days
≈ 0.153
Binomial
n = 25
X = 20
p = 0.74
q = 0.26
Simplify.
ANSWER: 15.3%
19. SEASONS Of 1108 people surveyed, 68% say that s
is the probability that at least 15 of 20 randomly selec
Page 4
SOLUTION: P(at least 15) = P(15) + P(16) + P(17) + P(18) + P(1
q = 0.32
Simplify.
≈ 0.0005
ANSWER: 15.3%
Study
Guide and Review - Chapter 11
Find the sum of the probabilities.
19. SEASONS Of 1108 people surveyed, 68% say that s
is the probability that at least 15 of 20 randomly selec
SOLUTION: P(at least 15) = P(15) + P(16) + P(17) + P(18) + P(1
X n –X
P(X) = nCxp q
15
20 – 15
Binomia
n = 20
X = 15
p = 0.6
q = 0.3
Simplify
20 – 16
Binomial
n = 20
X = 16
p = 0.68
q = 0.32
Simplify.
P(15) = 20C15(0.68) (0.32)
≈ 0.1599
X n –X
P(X) = nCxp q
16
P(16) = 20C16(0.68) (0.32)
≈ 0.1062
X n –X
P(X) = nCxp q
Binomial
n = 20
17
20 – 17 X = 17
P(17) = 20C17(0.68) (0.32)
p = 0.68
q = 0.32
≈ 0.0531
Simplify.
0.1599 + 0.1062 + 0.0531 + 0.0188 + 0.0042 + 0.005
The probability is about 34.3%.
ANSWER: 34.3%
20. RUNNING TIMES The times in the 40-meter dash
for a select group of professional football players are
normally distributed with a mean of 4.74 seconds and
a standard deviation of 0.13 second.
a. About what percent of players have times
between 4.6 and 4.8 seconds?
b. About how many of a sample of 800 players will
have times below 4.5 seconds?
SOLUTION: a. Find the z-values for X = 4.6 and X = 4.8.
z
Formula for z-values
X = 4.6
µ = 4.74
σ = 0.13
= –1.077
Simplify.
z
X n –X
P(X) = nCxp q
18
20 –
= 20C18(0.68) (0.32)
P(18)
18
≈ 0.0188
Binomial Pr
n = 20
X = 18
p = 0.68
q = 0.32
Simplify.
X n –X
P(X) = nCxp q
19
P(19)
20 –
= 20C19(0.68) (0.32)
19
≈ 0.0042
Binomial
n = 20
X = 19
p = 0.68
q = 0.32
Simplify.
= 0.462
Formula for z-values
X = 4.8
µ = 4.74
σ = 0.13
Simplify.
P(4.6 < X < 4.8) corresponds to z-values of –1.077
and 0.462, so find the area between z = –1.077 and z
= 0.462.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
X n –X
P(X) = nCxp q
Binomial
n = 20
20
20 – 20 X = 20
P(20) = 20C20(0.68) (0.32)
p = 0.68
q = 0.32
≈ 0.0005
Simplify.
Find the sum of the probabilities.
0.1599 + 0.1062 + 0.0531 + 0.0188 + 0.0042 + 0.005
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The probability is about 34.3%.
ANSWER: P(4.6 < X < 4.8) ≈ 0.537
b. Find the z-value for X = 4.5.
z
Formula for z-values
X = 4.5
µ = 4.74
Page 5
σ = 0.13
= –1.846
Simplify.
P(4.6 < X < 4.8) ≈ 0.537
Study
Review
b. Guide
Find theand
z-value
for X- Chapter
= 4.5. 11
z
Formula for z-values
X = 4.5
µ = 4.74
σ = 0.13
= –1.846
Simplify.
= 0.481
σ = 13.1
Simplify.
P(X < 75) corresponds to z-values less than 0.481, so
find the area between z = –4 and z = 0.481.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
P(X < 4.5) corresponds to z-values less than –1.846,
so find the area between z = –4 and z = –1.846.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
The probability is about 68.5%.
b. Find the z-value for X = 100.
z
The probability is about 0.0324. Multiply this by 800
players.
800 ∙ 0.0324 ≈ 26.0
About 26 players will have times below 4.5 seconds.
= 2.389
Formula for z-values
X = 100
µ = 68.7
σ = 13.1
Simplify.
P(X > 100) corresponds to z-values more than 2.389,
so find the area between z = 2.389 and z = 4.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
ANSWER: a. 53.7%
b. about 26 players
21. ATTENDANCE The number of tickets sold at high
school basketball games in a particular conference
are normally distributed with a mean of 68.7 and a
standard deviation of 13.1.
a. About what percent of the games sell fewer than
75 tickets?
b. About how many of a sample of 200 games will
sell more than 100 tickets?
SOLUTION: a. Find the z-value for X = 75.
z
Formula for z-values
X = 75
µ = 68.7
σ = 13.1
= 0.481
Simplify.
P(X < 75) corresponds to z-values less than 0.481, so
find the area between z = –4 and z = 0.481.
Select ` ú. Then, select normalcdf(lower z
value,
upper
z value).
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The probability is about 0.84%. Multiply this by 200
games.
0.0084 ∙ 200 = 1.68
About 2 games will sell more than 200 tickets.
ANSWER: a. 68.5%
b. about 2 games
22. COMMUTING The number of minutes it takes
Phil to commute to work each day are normally
distributed with a mean of 18.6 and a standard
deviation of 3.5.
a. About what percent of the time will it take Phil
more than 20 minutes to commute to work?
b. About how many of a sample of 50 days will it
take Phil less than 15 minutes to commute to work?
Page 6
SOLUTION: a. Find the z-value for X = 20.
z
Formula for z-values
deviation of 3.5.
a. About what percent of the time will it take Phil
more than 20 minutes to commute to work?
Study
and Review
b. Guide
About how
many of -a Chapter
sample of11
50 days will it
take Phil less than 15 minutes to commute to work?
SOLUTION: a. Find the z-value for X = 20.
z
Formula for z-values
X = 20
µ = 18.6
σ = 3.5
= 0.4
Simplify.
P(X > 20) corresponds to z-values more than 0.4, so
find the area between z = 0.4 and z = 4.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
The probability is about 15.17%. Multiply this by 50.
0.1517 ∙ 50 = 7.585
About 8 days
ANSWER: a. 34.5%
b. about 8 days
23. INTERNET A sample of 300 students was asked
the average amount of time they spend online during
a weeknight. The mean time was 64.3 minutes with a
standard deviation of 7.3 minutes. Determine a 95%
confidence interval.
SOLUTION: Find the maximum error of estimate.
Maximum Error of
E
Estimate
z = 1.96
s = 7.3
n = 300
≈ 0.8
Simplify.
Use the maximum error of estimate to find the
confidence interval CI.
CI =
The probability is about 34.5%.
±E
= 64.3 ± 0.8
b. Find the z-value for X = 15.
z
= –1.029
Formula for z-values
X = 15
µ = 18.6
σ = 3.5
Simplify.
P(X < 15) corresponds to z-values less than –1.029,
so find the area between z = –4 and z = –1.029.
Select ` ú. Then, select normalcdf(lower z
value, upper z value).
The 95% confidence interval is 63.5 ≤ µ ≤ 65.1.
ANSWER: Identify the null and alternative hypotheses for
each statement. Then identify the statement
that represents the claim.
24. A cupcake shop owner says that they sell 200
cupcakes everyday.
SOLUTION: sell 200: μ = 200
do not sell 200: μ ≠ 1
The claim is μ = 200, and it is the null hypothesis
because it does include equality. The alternative
hypothesis is μ ≠ 1, which is the complement of μ =
200.
The probability is about 15.17%. Multiply this by 50.
0.1517 ∙ 50 = 7.585
H0: μ = 200 (claim)
About 8 days
ANSWER: a. 34.5%
b. about 8 days
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= 64.3
E = 0.8
Confidence Interval for
Population Mean
Ha: μ ≠ 200
ANSWER: H0: μ = 200 (claim); Ha: μ ≠ 200
Page 7
25. A technician says that it takes at least 45 minutes to
The 95% confidence interval is 63.5 ≤ µ ≤ 65.1.
Study
Guide and Review - Chapter 11
ANSWER: Identify the null and alternative hypotheses for
each statement. Then identify the statement
that represents the claim.
24. A cupcake shop owner says that they sell 200
cupcakes everyday.
SOLUTION: sell 200: μ = 200
do not sell 200: μ ≠ 1
The claim is μ = 200, and it is the null hypothesis
because it does include equality. The alternative
hypothesis is μ ≠ 1, which is the complement of μ =
200.
H0: μ = 200 (claim)
Ha: μ ≠ 200
Ha: μ < 45
ANSWER: H0: μ ≥ 45 (claim); Ha: μ < 45
26. A member of a swim team says that she practices no
more than 35 minutes on Mondays.
SOLUTION: no more than 35 minutes: μ ≤ 35
more than 35 minutes: μ > 35
The claim is μ ≤ 35, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ > 35, which is the complement of μ ≤ 35.
H0: μ ≤ 35 (claim)
Ha: μ > 35
ANSWER: H0: μ ≤ 35 (claim); Ha: μ > 35
ANSWER: H0: μ = 200 (claim); Ha: μ ≠ 200
25. A technician says that it takes at least 45 minutes to
diagnose a computer problem.
SOLUTION: at least 45 minutes: μ ≥ 45
less than 45 minutes: μ < 45
The claim is μ ≥ 45, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ < 45, which is the complement of μ ≥ 45.
H0: μ ≥ 45 (claim)
Ha: μ < 45
ANSWER: H0: μ ≥ 45 (claim); Ha: μ < 45
26. A member of a swim team says that she practices no
more than 35 minutes on Mondays.
SOLUTION: no more than 35 minutes: μ ≤ 35
more than 35 minutes: μ > 35
The claim is μ ≤ 35, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ > 35, which is the complement of μ ≤ 35.
H : μ ≤ 35 (claim)
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Ha: μ > 35
Page 8