Solutions
Chapter 1 35
30
U.K.
U.S.
25
Introduction
Percent
20
15
Answers to Check Your Understanding 10
5
page 4: 1. The cars in the student parking lot. 2. He measured 0
y
e
ty
th
y
tim
Fl
th
the car’s model (categorical), year (quantitative), color (categori-
ili
ng
pa
sib
tre
ze
le
vi
rs
ee
Te
In
pe
Fr
cal), number of cylinders (quantitative), gas mileage (quantitative),
Su
Superpower preference
weight (quantitative), and whether it has a navigation system
(categorical).
Answers to Odd-Numbered Section 1.1 Exercises
Answers to Odd-Numbered Introduction Exercises 1.9 (a) 1% (b) A bar graph is given below. (c) Yes, because the
1.1 Type of wood, type of water repellent, and paint color are cat- numbers in the table refer to parts of a single whole.
egorical. Paint thickness and weathering time are quantitative. 25
1.3 (a) AP® Statistics students who completed a questionnaire on 20
Percent
the first day of class. (b) Categorical: gender, handedness, and
15
10
favorite type of music. Quantitative: height, homework time, and 5
0
the total value of coins in a student’s pocket. (c) The individual is a
te
k
r
y
d
ue
e
d
en
er
e
ig
ac
ra
Re
ol
hi
lv
th
Bl
re
be
G
/g
Bl
W
Si
O
G
w
n/
female who is right-handed. She is 58 inches tall, spends 60 min-
llo
wo
Ye
Br
Color
utes on homework, prefers Alternative music, and has 76 cents in
her pocket.
1.11 (a) A bar graph is given below. A pie chart would also be appro-
1.5 Student answers will vary. For example, quantitative variables
priate because the numbers in the table refer to parts of a single
could be graduation rate and student-faculty ratio, and categorical
whole. (b) Perhaps induced or C-section births are scheduled for
variables could be region of the country and type of institution
weekdays so doctors don’t have to work as much on the weekend.
(2-year college, 4-year college, university).
1.7 b 14,000
12,000
10,000
Section 1.1
Births
8000
6000
Answers to Check Your Understanding 4000
page 14: 1. Fly: 99/415 = 23.9%, Freeze time: 96/415 = 23.1%, 2000
0
Invisibility: 67/415 = 16.1%, Superstrength: 43/415 = 10.4%,
ay
ay
y
y
y
y
y
sda
sda
da
da
da
Telepathy: 110/415 = 26.5%. 2. A bar graph is shown below. It
nd
nd
urs
Fri
tur
e
ne
Su
Mo
Tu
Sa
appears that telepathy, ability to fly, and ability to freeze time were Th
d
We
the most popular choices, with about 25% of students choosing Day
each one. Invisibility was the 4th most popular and superstrength
was the least popular. 1.13 About 63% are Mexican and 9% are Puerto Rican.
1.15 (a) The given percents represent fractions of different age
30 groups, rather than parts of a single whole. (b) A bar graph is given
25 below.
20
Percent
25
15
20
Percent
10
15
5
10
0 5
y
e
ty
th
hy
0
tim
Fl
li
ng
at
bi
ep
tre
ze
isi
s
rs
rs
rs
rs
rs
s
l
rs
ee
ear
an year
Te
v
ver
yea
yea
yea
yea
yea
In
pe
Fr
Su
7y
do
24
34
44
54
64
65
1
Superpower preference
12–
18–
25–
45–
35–
55–
Age group
page 18: 1. For the U.K. students: 54/200 = 27% said fly, 52/200
= 26% said freeze time, 30/200 = 15% said invisibility, 20/200 = 1.17 (a) The areas of the pictures should be proportional to the
10% said superstrength, and 44/200 = 22% said telepathy. For the numbers of students they represent. (b) A bar graph is given below.
U.S. students: 45/215 = 20.9% said fly, 44/215 = 20.5% said freeze
14
Number of students
time, 37/215 = 17.2% said invisibility, 23/215 = 10.7% said super- 12
strength, and 66/215 = 30.7% said telepathy. 2. A bar graph is 10
8
shown in the next column. 3. There is an association between 6
country of origin and superpower preference. Students in the U.K. 4
2
are more likely to choose flying and freezing time, while students in 0
the U.S. are more likely to choose invisibility or telepathy. Cycle Car Bus Walk
Superstrength is about equally unpopular in both countries. Mode of transportation
S-1
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, S-2 Solutions
1.19 (a) 133 people; 36 buyers of coffee filters made of recycled page 32: 1. Both males and females have distributions that are
paper. (b) 36.8% said “higher,” 24.1% said “the same,” and 39.1% skewed to the right, though the distribution for the males is more
said “lower.” Overall, 60.9% of the members of the sample think heavily skewed. The midpoint for the males (9 pairs) is less than the
the quality is the same or higher. midpoint for the females (26 pairs). The number of shoes owned by
1.21 For buyers, 55.6% said higher, 19.4% said the same, and 25% females varies more (from 13 to 57) than for males (from 4 to 38). The
said lower. For the nonbuyers, 29.9% said higher, 25.8% said the male distribution has three likely outliers at 22, 35, and 38. The
same, and 44.3% said lower. We see that buyers are much more females do not have any likely outliers. 2. b 3. e 4. c
likely to consider recycled filters higher in quality and much less page 38: 1. One possible histogram is shown below. 2. The dis-
likely to consider them lower in quality than nonbuyers. tribution is roughly symmetric and bell-shaped. The typical IQ
1.23 Americans are much more likely to choose white/pearl and appears to be between 110 and 120 and the IQs vary from 80 to 150.
red, while Europeans are much more likely to choose silver, black, There do not appear to be any outliers.
or gray. Preferences for blue, beige/brown, green, and yellow/gold
18
are about the same for both groups. 16
1.25 A table and a side-by-side bar graph comparing the 14
Frequency
12
distributions of snowmobile use for environmental club members 10
8
and nonmembers are shown below. There appears to be an associa- 6
4
tion between environmental club membership and snowmobile 2
use. The visitors who are members of an environmental club are 0
80 90 100 110 120 130 140 150
much more likely to have never used a snowmobile and less likely IQ
to have rented or owned a snowmobile than visitors who are not in
an environmental club. page 39: 1. This is a bar graph because field of study is a categori-
cal variable. 2. No, because the variable is categorical and the cat-
Not a member Member egories could be listed in any order on the horizontal axis.
Never used 445/1221 = 36.4% 212/305 = 69.5%
Snowmobile 497/1221 = 40.7% 77/305 = 25.2% Answers to Odd-Numbered Section 1.2 Exercises
renter 1.37 (a) The graph is shown below. (b) The distribution is roughly
Snowmobile 279/1221 = 22.9% 16/305 = 5.2% symmetric with a midpoint of 6 hours. The hours of sleep vary from
owner 3 to 11. There do not appear to be any outliers.
70
60
50 3 4 5 6 7 8 9 10 11
Percent
40 Hours of sleep
30
20 1.39 (a) This dot represents a game where the opposing team won
10
by 1 goal. (b) All but 4 of the 25 values are positive, which indicates
0
Club No Yes No Yes No Yes that the U.S. women’s soccer team had a very good season. They
Snowmobile Never Snowmobile Snowmobile won 21/25 = 84% of their games.
used renter owner
1.41 As coins get older, they are taken out of circulation and new
1.27 d coins are introduced, meaning that most coins will be from recent
1.29 d years with a few from previous years.
1.31 b 1.43 Both distributions are roughly symmetric and have about the
1.33 d same amount of variability. The center of the internal distribution
is greater than the center of the external distribution, indicating
1.35 Answers will vary. Two possible tables are given below.
that external rewards do not promote creativity. Neither distribution
appears to have outliers.
10 40 30 20 1.45 (a) Otherwise, most of the data would appear on just a few
stems, making it hard to identify the shape of the distribution.
50 0 30 20 (b) Key: 12 | 1 means that 12.1% of that state’s residents are aged
25 to 34. (c) The distribution of percent of residents aged 25−34
is roughly symmetric with a possible outlier at 16.0%. The center
is around 13%. Other than the outlier at 16.0%, the values vary
from 11.4% to 15.1%.
Section 1.2 1.47 (a) The stemplots are given in the next column. The stemplot
Answers to Check Your Understanding with split stems makes it easier to see the shape of the distribution.
page 29: 1. This distribution is skewed to the right and uni- (b) The distribution is slightly skewed to the right with a center near
modal. 2. The midpoint of the 28 values is between 1 and 2. 780 mm, and values that vary from around 600 mm to 960 mm.
3. The number of siblings varies from 0 to 6. 4. There are two There do not appear to be any outliers. (c) In El Niño years, there
potential outliers at 5 and 6 siblings. is typically less rain than in other years (18 of 23 years).
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, Solutions S-3
Without splitting stems With splitting stems 1.57 (a) The histogram is given below. The distribution of word
6 03557 6 03 lengths is skewed to the right and single-peaked. The center is
7 0124488999 6 557
around 4 letters, with words that vary from 1 to 15 letters. There do
8 113667 7 01244
not appear to be any outliers. (b) There are more short words in
9 06 7 88999
Shakespeare’s plays and more very long words in Popular Science
articles.
8 113
8 667 20
Percent of words
Key: 6 | 3 = 630 mm of rain 9 0 15
9 6 10
5
1.49 (a) Most people will round their answers to the nearest 10
0
minutes (or 30 or 60). The students who claimed 300 and 360 min- 2 4 6 8 10 12 14
utes of studying on a typical weeknight may have been exaggerat- Length of words (number of letters)
ing. (b) The stemplots suggest that women (claim to) study more
1.59 The scale on the horizontal axis is very different from one
than men. The center for women (about 175 minutes) is greater
graph to the other.
than the center for men (about 120 minutes). 1.61 A bar graph should be used because birth month is a categori-
Women Men cal variable. A possible bar graph is given below.
0 0 3 3 3 3 9
8
9 6 0 5 6 6 6 8 9 9 9 7
6
Percent
2 2 2 2 2 2 2 2 1 0 2 2 2 2 2 2 5
4
8 8 8 8 8 8 8 8 8 8 7 5 5 5 5 1 5 5 8 3
2
4 4 4 0 2 0 0 3 4 4 1
0
2
Fe ary
Ma y
rch
ril
y
e
A ly
t
Oc er
r
De mber
r
Se ugus
No tobe
be
ar
Ma
Jun
Ju
mb
Ap
Key: 2 | 3 = 230 minutes 3 0
u
bru
cem
Jan
pte
ve
6 3
Month
1.51 (a) The distribution is slightly skewed to the left and uni-
1.63 (a) The percents for women sum to 100.1% due to round-
modal. (b) The center is between 0% and 2.5%. (c) The highest
ing errors. (b) Relative frequency histograms are shown below
return was between 10% and 12.5%. Ignoring the low outliers, the
because there are considerably more men than women. (c) Both
lowest return was between −12.5% and −10%. (d) About 37% of
histograms are skewed to the right. The center of the women’s
these months (102 out of 273) had negative returns. distribution of salaries is less than the men’s. The distributions
1.53 (a) The histogram is given below. (b) The distribution of of salaries are about equally variable, and the table shows that
travel times is roughly symmetric. The center is near 23 minutes there are some outliers in each distribution who make between
and the values vary from 15.5 to 30.9 minutes. There do not appear $65,000 and $70,000.
to be any outliers.
35
Percent of men
30
14
25
12
20
Frequency
10 15
8 10
6 5
4 0
2 10 20 30 40 50 60 70
0
14 16 18 20 22 24 26 28 30 32
Salary ($ thousands)
Travel time (minutes) 35
Percent of women
30
25
1.55 The histogram is given below. The distribution of DRP 20
scores is roughly symmetric with the center around 35. The 15
10
DRP scores vary from 14 to 54. There do not appear to be any 5
outliers. 0
10 20 30 40 50 60 70
9 Salary ($ thousands)
8
7
1.65 The distribution of age is skewed to the right for both males
Frequency
6
5 and females, meaning that younger people outnumber older peo-
4
3 ple. Among the younger Vietnamese, there are more males than
2 females. After age 35, however, females seem to outnumber the
1
0
males, making the center of the female distribution a little greater
12 18 24 30 36 42 48 54 60 than the male distribution. Both distributions have about the same
DRP scores amount of variability and no outliers.
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, S-4 Solutions
1.67 (a) Amount of studying. We would expect some students to 1.81 (a) median = 85 (b) x = 79.33 and median = 84. The
study very little, but most students to study a moderate amount. Any median did not change much but the mean did, showing that the
outliers would likely be high outliers, leading to a right-skewed distri- median is more resistant to outliers than the mean.
bution. (b) Right- versus left-handed. About 90% of the population is 1.83 The mean is $60,954 and the median is $48,097. The distri-
right-handed (represented by the bar at 0). (c) Gender. We would bution of salaries is likely to be quite right skewed because of a few
expect a more similar percentage of males and females than for the people who have a very large income, making the mean larger than
right-handed and left-handed students. (d) Heights. We expect many the median.
heights near the average and a few very short or very tall people. 1.85 The team’s annual payroll is 1.2(25) = 30 or $30 million. No,
1.69 a because the median only describes the middle value in the distribu-
1.71 c tion. It doesn’t provide specific information about any of the other
1.73 d values.
1.75 (a) Major League Baseball players who were on the roster on 1.87 (a) Estimating the frequencies of the bars (from left to right)
opening day of the 2012 season. (b) 6. Two variables are categorical as 10, 40, 42, 58, 105, 60, 58, 38, 27, 18, 20, 10, 5, 5, 1, and 3, the
(team, position) and the other 4 are quantitative (age, height, 3504
mean is x = = 7.01. The median is the average of the 250th
weight, and salary). 500
1.77 (a) 71/858 = 8.3% were elite soccer players and 43/858 = 5.0% and 251st values, which is 6. (b) Because the median is less than the
of the people had arthritis. (b) 10/71 = 14.1% of elite soccer players mean, we would use the median to argue that shorter domain
had arthritis and 10/43 = 23.3% of those with arthritis were elite soc- names are more popular.
cer players. 1.89 (a) IQR = 91 − 78 = 13. The middle 50% of the data have
a range of 13 points. (b) Any outliers are below 78 − 1.5(13) = 58.5
or above 91 + 1.5(13) = 110.5. There are no outliers.
Section 1.3 1.91 (a) Outliers are anything below 3 − 1.5(40) = −57 or above
Answers to Check Your Understanding 43 + 1.5(40) = 103, so 118 is an outlier. The boxplot is shown
page 53: 1. Because the distribution is skewed to the right, we below. (b) The article claims that teens send 1742 texts a month,
would expect the mean to be larger than the median. 2. Yes. The which is about 58 texts a day. Nearly all of the members of the class
mean is 31.25 minutes, which is greater than the median of 22.5 (21 of 25) sent fewer than 58 texts per day, which seems to contra-
minutes. 3. Because the distribution is skewed, the median would dict the claim in the article.
be a better measure of the center of the distribution. 120 *
page 59: 1. The data in order are: 290, 301, 305, 307, 307, 310, 324,
Number of texts
100
345. The 5-number summary is 290, 303, 307, 317, 345. 2. The IQR 80
60
is 14 pounds. The range of the middle half of the data is 14
40
pounds. 3. Any outliers occur below 303 − 1.5(14) = 282 or above
20
317 + 1.5(14) = 338, so 345 pounds is an outlier. 4. The boxplot is 0
given below.
1.93 (a) Positive numbers indicate students who had more text
messages than calls. Because the 1st quartile is about 0, roughly
75% of the students had more texts than calls, which supports the
290 300 310 320 330 340 350
article’s conclusion. (b) No. Students in statistics classes tend to be
Weight upperclassmen and their responses might differ from those of
underclassmen.
page 63: 1. The mean is 75. 2. The table is given below. 1.95 (a) About 3% and −3.5%. (b) About 0.1%. (c) The stock fund
is much more variable. It has higher positive returns, but also higher
Observation Deviation Squared deviation negative returns.
67 67 − 75 = −8 (−8)2 = 64 2.06
1.97 (a) sx = = 0.6419 mg∙dl. (b) The phosphate level
72 72 − 75 = −3 (−3)2 = 9 Å6 − 1
76 typically varies from the mean by about 0.6419 mg/dl.
76 − 75 = 1 12 = 1
76 76 − 75 = 1 12 = 1 1.99 (a) Skewed to the right, because the mean is much larger than
84 84 − 75 = 9 92 = 81 the median and Q3 is much further from the median than Q1.
Total 0 156 (b) The amount of money spent typically varies from the mean by
$21.70. (c) Any points below 19.06 − 1.5(26.66) = −20.93 or
above 45.72 + 1.5(26.66) = 85.71 are outliers. Because the maxi-
156
3. The variance is s2x = = 39 inches squared and the stan- mum of 93.34 is greater than 85.71, there is at least one outlier.
5−1 1.101 Yes. For example, in data set 1, 2, 3, 4, 5, 6, 7, 8 the IQR is 4.
dard deviation is sx = "39 = 6.24 inches. If 8 is changed to 88, the IQR will still be 4.
4. The players’ heights typically vary by about 6.24 inches from the 1.103 (a) One possible answer is 1, 1, 1, 1. (b) 0, 0, 10, 10. (c) For
mean height of 75 inches. part (a), any set of four identical numbers will have sx = 0. For part
(b), however, there is only one possible answer. We want the values
Answers to Odd-Numbered Section 1.3 Exercises to be as far from the mean as possible, so our best choice is two
1.79 x = 85 values at each extreme.
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