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Mode: Since 1.5, 2, and 2.5 appear twice, the modes
are 1.5, 2, and 2.5 inches of rain.
Chapter Review
Find the mean, median, and mode for each set
of data. Round to the nearest tenth, if
necessary.
1. number of students in each math class: 22, 23, 24, 22,
21
SOLUTION: Mean:
The mean is 22.4 students.
Median: Order the numbers from least to greatest.
21, 22, 22, 23, 24
The median number of students in the math classes is
22.
Mode: There are two classes with 22 students, so
the mode is 22 students.
ANSWER: 22.4 students; 22 students; 22 students
3. inches of rain last week: 1.5, 2, 2.5, 2, 1.5, 2.5, 3
SOLUTION: Mean:
The mean is approximately 2.1 inches.
Median: Order the numbers from least to greatest.
1.5, 1.5, 2, 2, 2.5, 2.5, 3
The median number of inches of rain is 2.
Mode: Since 1.5, 2, and 2.5 appear twice, the modes
are 1.5, 2, and 2.5 inches of rain.
ANSWER: 2.1 in.; 2 in.; 1.5 in., 2 in., 2.5 in.
4. At the movie theater, six movies are playing and their
lengths are 138, 117, 158, 145, 135, and 120 minutes.
Which measure of center best represents the data?
Justify your selection and then find the measure of
center.
SOLUTION: Because there are no extreme values or numbers
that are identical, the mean would best represent the
data.
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Mean:
ANSWER: 2.1 in.; 2 in.; 1.5 in., 2 in., 2.5 in.
4. At the movie theater, six movies are playing and their
lengths are 138, 117, 158, 145, 135, and 120 minutes.
Which measure of center best represents the data?
Justify your selection and then find the measure of
center.
SOLUTION: Because there are no extreme values or numbers
that are identical, the mean would best represent the
data.
Mean:
The mean is 135.5 minutes.
ANSWER: Mean; there are no extreme values or numbers that
are identical; 135.5 min
Find the measures of variability and any outliers
for each set of data.
6. The number of fish in each fish tank: 6, 5, 7, 8, 5, 6, 7,
9, 8, 6
SOLUTION: Range: The greatest value is 9 and the least value is
5. So, the range is 9 – 5 or 4 fish.
Median, First Quartile, and Third Quartile: List
the data from the least to greatest.
5, 5, 6, 6, 6, 7, 7, 8, 8, 9
There is an even number of items, so find the mean
of the middle two numbers.
The median number of fish is 6.5.
The first quartile, or median of the lower half, is 6.
The third quartile, or median of the upper half, is 8.
The interquartile range is 8 – 6 or 2.
Outliers: To determine if there are any outliers,
multiply the interquartile range, 2, by 1.5.
2 • 1.5 = 3
Subtract 3 from the first quartile and add 3 to the
third quartile.
6 – 3 = 3; 8 + 3 = 11
There are no values less than 3 or greater than 11, so
there are no outliers.
ANSWER: R: 4; M: 6.5; Q1: 6; Q3: 8; IR 2; none
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7. Claire earned $5, $7, $10, $6, and $8 doing errands
for her neighbors. Find the measures of variability
and any outliers for the set of data.
The mean is 135.5 minutes.
there are no outliers.
ANSWER: Mean; there are no extreme values or numbers that
are identical; 135.5 min
ANSWER: R: 4; M: 6.5; Q1: 6; Q3: 8; IR 2; none
Chapter Review
Find the measures of variability and any outliers
for each set of data.
6. The number of fish in each fish tank: 6, 5, 7, 8, 5, 6, 7,
9, 8, 6
SOLUTION: Range: The greatest value is 9 and the least value is
5. So, the range is 9 – 5 or 4 fish.
Median, First Quartile, and Third Quartile: List
the data from the least to greatest.
5, 5, 6, 6, 6, 7, 7, 8, 8, 9
There is an even number of items, so find the mean
of the middle two numbers.
7. Claire earned $5, $7, $10, $6, and $8 doing errands
for her neighbors. Find the measures of variability
and any outliers for the set of data.
SOLUTION: Range: The greatest value is 10 and the least value
is 5. So, the range is 10 – 5 or $5.
Median, First Quartile, and Third Quartile: List
the data from the least to greatest.
5, 6, 7, 8, 10
The median amount Claire spent is $7.
The first quartile, or median of the lower half, is
or 5.5.
The third quartile, or median of the upper half, is
The median number of fish is 6.5.
The first quartile, or median of the lower half, is 6.
The third quartile, or median of the upper half, is 8.
The interquartile range is 8 – 6 or 2.
Outliers: To determine if there are any outliers,
multiply the interquartile range, 2, by 1.5.
2 • 1.5 = 3
Subtract 3 from the first quartile and add 3 to the
third quartile.
6 – 3 = 3; 8 + 3 = 11
There are no values less than 3 or greater than 11, so
there are no outliers.
ANSWER: R: 4; M: 6.5; Q1: 6; Q3: 8; IR 2; none
7. Claire earned $5, $7, $10, $6, and $8 doing errands
for her neighbors. Find the measures of variability
and any outliers for the set of data.
SOLUTION: Range: The greatest value is 10 and the least value
is 5. So, the range is 10 – 5 or $5.
Median, First Quartile, and Third Quartile: List
the data from the least to greatest.
5, 6, 7, 8, 10
The median amount Claire spent is $7.
The first quartile, or median of the lower half, is
or 5.5.
The third quartile, or median of the upper half, is
or 9.
The interquartile range is 9 – 5.5 or 3.5.
Outliers: To determine if there are any outliers,
multiply the interquartile range, 3.5, by 1.5.
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3.5 • 1.5
= -5.25
Subtract 5.25 from the first quartile and add 5.25 to
the third quartile.
or 9.
The interquartile range is 9 – 5.5 or 3.5.
Outliers: To determine if there are any outliers,
multiply the interquartile range, 3.5, by 1.5.
3.5 • 1.5 = 5.25
Subtract 5.25 from the first quartile and add 5.25 to
the third quartile.
5.5 – 5.25 = 0.25; 9 + 5.25 = 14.25
There are no values less than 0.25 or greater than
14.25, so there are no outliers.
ANSWER: R: 5; M: 7; Q1: 5.5; Q3: 9; IR 3.5; none
8. The scores Mr. Han’s students earned on their last
test are shown in the table. Use the measures of
variability to describe the data in the table.
SOLUTION: The greatest value is 99 and the least value is 58. So,
the spread of the data, or range, is 41 points. The
data ordered from the least to greatest is 58, 73, 74,
74, 74, 76, 76, 77, 80, 81, 82, 82, 82, 83, 84, 85, 86, 87,
88, 89, 92, 92, 99, and 99. There is an even number
of items, so find the mean of the middle two
numbers.
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The median test score is 82 points. The lower
quartile, or median of the lower half, is
or There are no values less than 0.25 or greater than
14.25, so there are no outliers.
ANSWER: R: 5; M: 7; Q1: 5.5; Q3: 9; IR 3.5; none
Chapter Review
8. The scores Mr. Han’s students earned on their last
test are shown in the table. Use the measures of
variability to describe the data in the table.
The spread of the data is 41 points. The median is 82
points. One fourth of the students earned 76 points or
less. One fourth of the students earned 87.5 points or
more. Half of the students earned between 76 and
87.5 points.
Find the mean absolute deviation of each data
set.
9. 25, 70, 75, 100
SOLUTION: Find the mean.
Find the absolute differences between each data
value and the mean.
SOLUTION: The greatest value is 99 and the least value is 58. So,
the spread of the data, or range, is 41 points. The
data ordered from the least to greatest is 58, 73, 74,
74, 74, 76, 76, 77, 80, 81, 82, 82, 82, 83, 84, 85, 86, 87,
88, 89, 92, 92, 99, and 99. There is an even number
of items, so find the mean of the middle two
numbers.
Find the average of these differences.
The average distance between each data value and
the mean is 21.25.
The median test score is 82 points. The lower
quartile, or median of the lower half, is
or 76. One fourth of the students earned 76 points or
less. The upper quartile, or median of the upper half,
is
or 87.5. One fourth of the students earned
87.5 points or more. The interquartile range is 87.5 –
76 or 11.5. One half of the students earned between
76 and 87.5 points.
ANSWER: The spread of the data is 41 points. The median is 82
points. One fourth of the students earned 76 points or
less. One fourth of the students earned 87.5 points or
more. Half of the students earned between 76 and
87.5 points.
Find the mean absolute deviation of each data
set.
9. 25, 70, 75, 100
SOLUTION: Find the mean.
Find the absolute differences between each data
value and the mean.
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ANSWER: 21.25
11. 35, 50, 40, 55, 45
SOLUTION: Find the mean.
Find the absolute differences between each data
value and the mean.
Find the average of these differences.
The average distance between each data value and
the mean is 6.
ANSWER: 6
14. The table shows the numbers of two colors of
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candies in several bags. Which data set has a greater
mean absolute deviation? Justify your answer.
The average distance between each data value and
the mean is 6.
ANSWER: Chapter
Review
6
14. The table shows the numbers of two colors of
candies in several bags. Which data set has a greater
mean absolute deviation? Justify your answer.
The set of brown candies has a greater mean
absolute deviation.
ANSWER: Brown candies; the mean absolute deviation is 2.32,
which is greater than 1.92, the mean absolute
deviation of the red candies.
15. The double box plot shows the numbers of floors for
buildings on two streets. Compare the centers and
variations. Write an inference you can draw about
the two populations.
SOLUTION: Find the mean of the brown candies.
Find the absolute value of the differences between
each data value and the mean.
Find the average of these differences.
The mean absolute deviation of the brown candies is
2.32.
Find the mean of the red candies.
Find the absolute value of the differences between
each data value and the mean.
Find the average of these differences.
The mean absolute deviation of the red candies is
1.92.
The set of brown candies has a greater mean
absolute deviation.
ANSWER: Brown candies; the mean absolute deviation is 2.32,
which is greater than 1.92, the mean absolute
deviation of the red candies.
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15. The double box plot shows the numbers of floors for
buildings on two streets. Compare the centers and
SOLUTION: The median number of floors on Main St. is 40 with
an interquartile range of 10. The median number of
floors on Grand Ave. is 18 with an interquartile range
of 5. Main St. has more variability and centers
around a higher number of floors than Grand Ave.
ANSWER: The median number of floors on Main St. is 40 with
an interquartile range of 10. The median number of
floors on Grand Ave. is 18 with an interquartile range
of 5. Main St. has more variability and centers
around a higher number of floors than Grand Ave.
16. To determine the weekly top ten songs, the local
radio station asks people to log onto their Web site
and vote for their favorite song. Identify the sample
as biased or unbiased and describe its type.
Explain.
SOLUTION: The sample is biased because it is a voluntary
response survey. Only those people who are
interested in participating in the survey are part of the
sample.
ANSWER: biased, voluntary response survey; only those people
who are interested in participating in the survey are
part of the sample.
17. Forty-five out of 60 people at a steakhouse said their
favorite meal was steak. Is this sampling
representative of the entire town? If so, how many of
the 13,000 residents would say steak was their
favorite meal?
SOLUTION: This is a biased, convenience sample because only
customers in one restaurant specializing in steak
were surveyed. So, this sampling method is not
representative of the entire town.
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ANSWER: No; The sample is biased. Diners at a steakhouse are
more likely to choose steak as their favorite meal.
ANSWER: biased, voluntary response survey; only those people
who are interested in participating in the survey are
part of the sample.
Chapter Review
17. Forty-five out of 60 people at a steakhouse said their
favorite meal was steak. Is this sampling
representative of the entire town? If so, how many of
the 13,000 residents would say steak was their
favorite meal?
The probability is 0 or 0%. Because the probability of
the event is 0%, the event is impossible.
ANSWER: 0 or 0%; impossible
19. P(two-digit number)
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: 16, 25, 36, 49, 64, and 81
SOLUTION: This is a biased, convenience sample because only
customers in one restaurant specializing in steak
were surveyed. So, this sampling method is not
representative of the entire town.
The probability is
ANSWER: No; The sample is biased. Diners at a steakhouse are
more likely to choose steak as their favorite meal.
Mike rolls a ten-sided solid whose identical
faces are numbered with the first ten square
numbers. Find each probability. Then describe
the likelihood of the event as impossible,
unlikely, equally likely, likely, or certain.
18. P(ones digit is 2)
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: none
The probability is 0 or 0%. Because the probability of
the event is 0%, the event is impossible.
or 60%. Because the probability
of the event is greater than 50%, the event is likely to
occur.
ANSWER: or 60%; likely
20. P(multiple of 8)
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: 16 and 64
The probability is
or 20%. Because the probability
of the event is less than 50%, the event is unlikely to
occur.
ANSWER: 0 or 0%; impossible
ANSWER: 19. P(two-digit number)
or 20%; unlikely
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: 16, 25, 36, 49, 64, and 81
21. P(ones digit not 3)
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: 1, 4, 9, 16, 25, 36, 49, 64, 81, and
100
The probability is
or 60%. Because the probability
of the event is greater than 50%, the event is likely to
occur.
ANSWER: eSolutions Manual - Powered by Cognero
or 60%; likely
The probability is 1 or 100%. Because the probability
of the event is 100%, the event is certain.
ANSWER: 1 or 100%; certain
22. P(odd)
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occur.
occur.
ANSWER: ANSWER: or 20%;
unlikely
Chapter
Review
or 50%; equally likely
21. P(ones digit not 3)
23. P(one-digit number)
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: 1, 4, 9, 16, 25, 36, 49, 64, 81, and
100
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: 1, 4, and 9
The probability is
The probability is 1 or 100%. Because the probability
of the event is 100%, the event is certain.
ANSWER: 1 or 100%; certain
or 30%. Because the
probability of the event is less than 50%, the event is
unlikely to occur.
ANSWER: or 30%; unlikely
22. P(odd)
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: 1, 9, 25, 49, and 81
In a survey of randomly selected students,
students chose their preference from among
three meal options for a school event.
25. What was the experimental probability that a student
chose burgers?
SOLUTION: The probability is
or 50%. Because the probability
of the event is 50%, the event is equally likely to
occur.
ANSWER: ANSWER: or 50%; equally likely
23. P(one-digit number)
SOLUTION: sample space: {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
favorable outcomes: 1, 4, and 9
The probability is
26. What was the experimental probability that a student
chose the vegetarian option?
SOLUTION: or 30%. Because the
probability of the event is less than 50%, the event is
unlikely to occur.
ANSWER: ANSWER: 30%;- Powered
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In a survey of randomly selected students,
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28. Out of a similar group of 450 students, predict how
many would choose the vegetarian option.
ANSWER: Chapter Review
26. What was the experimental probability that a student
chose the vegetarian option?
SOLUTION: Out of a similar group of 450 students, 99 would
choose the vegetarian option.
ANSWER: 99
A box contains 8 red markers, 5 green
markers, and 5 blue markers. Once a marker is
pulled from the box, it is replaced, and then
another marker is pulled. Find each probability.
29. P(red, then green)
SOLUTION: When drawing the first marker, there are 8 + 5 + 5
or 18 markers, and 8 of them are red.
ANSWER: The first marker is replaced. When drawing the
second marker, there are 18 markers, and 5 of them
are green.
28. Out of a similar group of 450 students, predict how
many would choose the vegetarian option.
SOLUTION: Use the percent proportion to find 22% of 450.
The probability of drawing a red marker then a green
marker is
Out of a similar group of 450 students, 99 would
choose the vegetarian option.
.
ANSWER: ANSWER: 99
A box contains 8 red markers, 5 green
markers, and 5 blue markers. Once a marker is
pulled from the box, it is replaced, and then
another marker is pulled. Find each probability.
29. P(red, then green)
SOLUTION: When drawing the first marker, there are 8 + 5 + 5
or 18 markers, and 8 of them are red.
The first marker is replaced. When drawing the
second marker, there are 18 markers, and 5 of them
are green.
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