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21
CHAPTER 2 SOLUTIONS
Exercises 2.1
2.1
(a) Hair color, model of car, and brand of popcorn are qualitative
variables.
(b) Number of eggs in a nest, number of cases of flu, and number of
employees are discrete, quantitative variables.
(c) Temperature, weight, and time are quantitative continuous variables.
2.2
(a) A qualitative variable is a nonnumerically valued variable. Its
possible โvaluesโ are descriptive (e.g., color, name, gender).
(b) A discrete, quantitative variable is one whose possible values can be
listed. It is usually obtained by counting rather than by measuring.
(c) A continuous, quantitative variable is one whose possible values form
some interval of numbers. It usually results from measuring.
2.3
(a) Qualitative data result from observing and recording values of a
qualitative variable, such as, color or shape.
(b) Discrete, quantitative data are values of a discrete quantitative
variable. Values usually result from counting something.
(c) Continuous, quantitative data are values of a continuous variable.
Values are usually the result of measuring something such as
temperature that can take on any value in a given interval.
2.4
The classification of data is important because it will help you choose the
correct statistical method for analyzing the data.
2.5
Of qualitative and quantitative (discrete and continuous) types of data,
only qualitative yields nonnumerical data.
2.6
(a) The first column consists of quantitative, discrete data. This column
provides ranks of the highest recorded temperature for each continent.
(b) The second column consists of qualitative data since continent names
are nonnumerical.
(c) The fourth column consists of quantitative, continuous data.
This
column provides the highest recorded temperatures for the continents in
degrees Farenheit.
(d) The information that Death Valley is in the United States is
qualitative data since country in which a place is located is
nonnumerical.
2.7
(a) The first column consists of quantitative, continuous data.
column provides the time that the earthquake occurred.
This
(b) The second column consists of quantitative, continuous data.
column provides the magnitude of each earthquake.
This
(c) The third column consists of quantitative, continuous data.
column provides the depth of each earthquake in kilometers.
This
(d) The fourth column consists of quantitative, discrete data.
This
column provides the number of stations that reported activity on the
earthquake.
(e) The fifth column consists of qualitative data since the region of the
location of each earthquake is nonnumerical.
2.8
(a) The first column consists of quantitative, discrete data.
provides ranks of the top ten IPOs in the United States.
Copyright ยฉ 2020 Pearson Education, Inc.
This column
22
Chapter 2
(b) The second column consists of qualitative data since company names are
nonnumerical.
(c) The third column consists of quantitative, discrete data.
Since money
involves discrete units, such as dollars and cents, the data is
discrete, although, for all practical purposes, this data might be
considered quantitative continuous data.
(d) The information that Facebook is a social networking business is
qualitative data since type of business is nonnumerical.
2.9
(a) The first column consists of quantitative, discrete data. This column
provides the ranks of the deceased celebrities with the top 10
earnings.
(b) The second column consists of qualitative data since names are
nonnumerical.
(c) The third column consists of quantitative, discrete data, the earnings
of the celebrities. Since money involves discrete units, such as
dollars and cents, the data is discrete, although, for all practical
purposes, this data might be considered quantitative continuous data
2.10
(a) The first column consists of quantitative, discrete data. This column
provides the ranks of the top 10 universities for 2012-2013.
(b) The second column consists of qualitative data since names of the
institutions are nonnumerical.
(c) The third column consists of quantitative, continuous data.
This
column provides the overall score of the top 10 universities for 20122013.
2.11
(a) The first column contains types of products.
since they are nonnumerical.
They are qualitative data
(b) The second column contains number of units shipped in the millions.
These are whole numbers and are quantitative, discrete.
(c) The third column contains money values. Technically, these are
quantitative, discrete data since there are gaps between possible
values at the cent level. For all practical purposes, however, these
are quantitative, continuous data.
2.12
Player name, team, and position are nonnumerical and are therefore
qualitative data. The number of runs batted in, or RBI, are whole numbers
and are therefore quantitative, discrete. Weight is quantitative,
continuous.
2.13
The first column contains quantitative, discrete data in the form of ranks.
These are whole numbers. The second and third columns contain qualitative
data in the form of names. The last column contains the rating of the
program which is quantitative, continuous.
2.14
The first column is qualitative since it is nonnumerical. The second and
third columns are quantitative, discrete since they report the number of
grants and applications received. The last column is quantitative,
continuous since it reports the success rate of the grants.
2.15
The first column is quantitative, discrete since it is reporting a rank.
The second and third columns are qualitative since make/model and type are
nonnumerical. The last column is quantitative, continuous since it is
reporting mileage.
2.16
Of the eight items presented, only high school class rank involves ordinal
data. The rank is ordinal data.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
Exercises
23
2.2
2.17
A frequency distribution of qualitative data is a table that lists the
distinct values of data and their frequencies. It is useful to organize the
data and make it easier to understand.
2.18
(a) The frequency of a class is the number of observations in the class,
whereas, the relative frequency of a class is the ratio of the class
frequency to the total number of observations.
(b) The percentage of a class is 100 times the relative frequency of the
class. Equivalently, the relative frequency of a class is the
percentage of the class expressed as a decimal.
2.19
(a) True. Having identical frequency distributions implies that the total
number of observations and the numbers of observations in each class
are identical. Thus, the relative frequencies will also be identical.
(b) False. Having identical relative frequency distributions means that
the ratio of the count in each class to the total is the same for both
frequency distributions. However, one distribution may have twice (or
some other multiple) the total number of observations as the other.
For example, two distributions with counts of 5, 4, 1 and 10, 8, 2
would be different, but would have the same relative frequency
distribution.
(c) If the two data sets have the same number of observations, either a
frequency distribution or a relative-frequency distribution is
suitable. If, however, the two data sets have different numbers of
observations, using relative-frequency distributions is more
appropriate because the total of each set of relative frequencies is 1,
putting both distributions on the same basis for comparison.
2.20
(a)-(b)
The classes are presented in column 1. The frequency distribution of
the classes is presented in column 2. Dividing each frequency by the
total number of observations, which is 5, results in each class’s
relative frequency. The relative frequency distribution is presented
in column 3.
Relative Frequency
Class
Frequency
A
2
0.40
B
2
0.40
C
1
0.20
5
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
24
Chapter 2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each class. The result using
Minitab is
Pie Chart
Category
A
B
C
C
20.0%
A
40.0%
B
40.0%
(d) We use the bar chart to show the relative frequency with which each
class occurs. The result is
Bar Graph
40
Percent
30
20
10
0
A
B
C
Class
Percent within all data.
2.21
(a)-(b)
The classes are presented in column 1. The frequency distribution of
the classes is presented in column 2. Dividing each frequency by the
total number of observations, which is 5, results in each class’s
relative frequency. The relative frequency distribution is presented
in column 3.
Class
Frequency
Relative Frequency
A
3
0.60
B
1
0.20
C
1
0.20
5
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each class. The result using
Minitab is
Pie Chart
Category
A
B
C
C
20.0%
B
20.0%
A
60.0%
(d) We use the bar chart to show the relative frequency with which each
class occurs. The result is
Bar Graph
60
50
Percent
40
30
20
10
0
A
B
C
Class
Percent within all data.
2.22
(a)-(b)
The classes are presented in column 1. The frequency distribution of
the classes is presented in column 2. Dividing each frequency by the
total number of observations, which is 10, results in each class’s
relative frequency. The relative frequency distribution is presented
in column 3.
Class
Frequency
Relative Frequency
A
4
0.40
B
1
0.10
C
1
0.10
D
4
0.40
10
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
25
26
Chapter 2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each class. The result using
Minitab is
Pie Chart
Category
A
B
C
D
D
40.0%
A
40.0%
C
10.0%
B
10.0%
(d) We use the bar chart to show the relative frequency with which each
class occurs. The result is
Bar Chart
40
Percent
30
20
10
0
A
B
C
D
Class
Percent within all data.
2.23
(a)-(b)
The classes are presented in column 1. The frequency distribution of
the classes is presented in column 2. Dividing each frequency by the
total number of observations, which is 10, results in each class’s
relative frequency. The relative frequency distribution is presented
in column 3.
Relative Frequency
Class
Frequency
A
4
0.40
B
3
0.30
C
1
0.10
D
2
0.20
10
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each class. The result using
Minitab is
Pie Chart
Category
A
B
C
D
D
20.0%
A
40.0%
C
10.0%
B
30.0%
(d) We use the bar chart to show the relative frequency with which each
class occurs. The result is
Bar Chart
40
Percent
30
20
10
0
A
B
C
D
Class
Percent within all data.
2.24
(a)-(b)
The classes are presented in column 1. The frequency distribution of
the classes is presented in column 2. Dividing each frequency by the
total number of observations, which is 20, results in each class’s
relative frequency. The relative frequency distribution is presented
in column 3.
Class
Frequency
Relative Frequency
A
3
0.15
B
6
0.30
C
4
0.20
D
3
0.15
E
4
0.20
20
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
27
28
Chapter 2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each class. The result using
Minitab is
Pie Chart
Category
A
B
C
D
E
A
15.0%
E
20.0%
D
15.0%
B
30.0%
C
20.0%
(d) We use the bar chart to show the relative frequency with which each
class occurs. The result is
Bar Chart
30
25
Percent
20
15
10
5
0
A
B
C
D
E
Class
Percent within all data.
2.25
(a)-(b)
The classes are presented in column 1. The frequency distribution of
the classes is presented in column 2. Dividing each frequency by the
total number of observations, which is 20, results in each class’s
relative frequency. The relative frequency distribution is presented
in column 3.
Class
Frequency
Relative Frequency
A
1
0.05
B
3
0.15
C
7
0.35
D
7
0.35
E
2
0.10
20
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each class. The result using
Minitab is
Pie Chart
Category
A
B
C
D
E
A
5.0%
E
10.0%
B
15.0%
D
35.0%
C
35.0%
(d) We use the bar chart to show the relative frequency with which each
class occurs. The result is
Bar Chart
40
Percent
30
20
10
0
A
B
C
D
E
Class
Percent within all data.
2.26
(a)-(b)
The classes are the networks and are presented in column 1. The
frequency distribution of the networks is presented in column 2.
Dividing each frequency by the total number of shows, which is 20,
results in each class’s relative frequency. The relative frequency
distribution is presented in column 3.
Network
Frequency
Relative Frequency
ABC
2
0.10
CBS
11
0.55
Fox
2
0.10
NBC
5
0.25
20
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
29
30
Chapter 2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each network. The result is
Pie Chart of NETWORK
Category
ABC
CBS
FOX
NBC
ABC
10.0%
NBC
25.0%
FOX
10.0%
CBS
55.0%
(d) We use the bar chart to show the relative frequency with which each
network occurs. The result is
Bar Chart of NETWORK
60
50
Percent
40
30
20
10
0
ABC
CBS
FOX
NBC
NETWORK
Percent within all data.
2.27
(a)-(b)
The classes are the NCAA wrestling champions and are presented in
column 1. The frequency distribution of the champions is presented in
column 2. Dividing each frequency by the total number of champions,
which is 25, results in each class’s relative frequency. The relative
frequency distribution is presented in column 3.
Champion
Frequency
Relative Frequency
Iowa
12
0.48
Penn State
3
0.12
Minnesota
3
0.12
Oklahoma St.
7
0.28
25
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
31
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each team. The result is
Pie Chart of CHAMPION
Category
Iowa
Minnesota
Oklahoma St.
Penn State
Penn State
12.0%
Iowa
48.0%
Oklahoma St.
28.0%
Minnesota
12.0%
(c) We use the bar chart to show the relative frequency with which each
TEAM occurs. The result is
Bar Chart of CHAMPION
50
Percent
40
30
20
10
0
Iowa
Minnesota
Oklahoma St.
Penn State
CHAMPION
Percent within all data.
2.28
(a)-(b) The classes are the colleges and are presented in column 1. The
frequency distribution of the colleges is presented in column 2.
Dividing each frequency by the total number of students in the section
of Introduction to Computer Science, which is 25, results in each
class’s relative frequency. The relative frequency distribution is
presented in column 3.
College
Frequency
Relative Frequency
BUS
9
0.36
ENG
12
0.48
LIB
4
0.16
25
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
32
Chapter 2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each college. The result is
COLLEGE
Category
ENG
BUS
LIB
LIB
16.0%
ENG
48.0%
BUS
36.0%
(b) We use the bar chart to show the relative frequency with which each
COLLEGE occurs. The result is
COLLEGE
50
Percent
40
30
20
10
0
BUS
ENG
COLLEGE
LIB
Percent within all data.
2.29
(a)-(b)
The classes are the class levels and are presented in column 1. The
frequency distribution of the class levels is presented in column 2.
Dividing each frequency by the total number of students in the
introductory statistics class, which is 40, results in each class’s
relative frequency. The relative frequency distribution is presented
in column 3.
Class Level
Frequency
Relative Frequency
Fr
6
0.150
So
15
0.375
Jr
12
0.300
Sr
7
0.175
40
1.000
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
33
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each class level. The result is
CLASS
C ategory
So
Jr
Sr
Fr
Fr
15.0%
So
37.5%
Sr
17.5%
Jr
30.0%
(d) We use the bar chart to show the relative frequency with which each
CLASS level occurs. The result is
CLASS
40
Percent
30
20
10
0
Fr
So
Jr
Sr
CLASS
Percent within all data.
2.30
(a)-(b)
The classes are the regions and are presented in column 1. The
frequency distribution of the regions is presented in column 2.
Dividing each frequency by the total number of states, which is 50,
results in each class’s relative frequency. The relative frequency
distribution is presented in column 3.
Class Level
Frequency
Relative Frequency
NE
9
0.18
MW
12
0.24
SO
16
0.32
WE
13
0.26
50
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
34
Chapter 2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each region. The result is
REGION
C ategory
NE
MW
WE
SO
NE
18.0%
SO
32.0%
MW
24.0%
WE
26.0%
(d) We use the bar chart to show the relative frequency with which each
REGION occurs. The result is
REGION
35
30
Percent
25
20
15
10
5
0
NE
MW
WE
SO
REGION
Percent within all data.
2.31
(a)-(b)
The classes are the days and are presented in column 1. The frequency
distribution of the days is presented in column 2. Dividing each
frequency by the total number road rage incidents, which is 69, results
in each class’s relative frequency. The relative frequency
distribution is presented in column 3.
Class Level
Frequency
Relative Frequency
Su
5
0.0725
M
5
0.0725
Tu
11
0.1594
W
12
0.1739
Th
11
0.1594
F
18
0.2609
Sa
7
0.1014
69
1.0000
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
(c) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each day. The result is
DAY
C ategory
F
W
Tu
Th
Sa
Su
M
M
7.2%
Su
7.2%
F
26.1%
Sa
10.1%
Th
15.9%
W
17.4%
Tu
15.9%
(d) We use the bar chart to show the relative frequency with which each
DAY occurs. The result is
DAY
25
Percent
20
15
10
5
0
Su
M
Tu
W
DAY
Th
F
Sa
Percent within all data.
2.32
(a) We find each of the relative frequencies by dividing each of the
frequencies by the total frequency of 291,176. Due to rounding, the
sum of the relative frequency column is 0.9999.
Robbery Type
Frequency
Street/highway
127,403
Relative
Frequency
0.4375
Commercial house
37,885
0.1301
Gas or service station
7,009
0.0241
Convenience store
14,863
0.0510
Residence
49,361
0.1695
Bank
5,777
0.0198
Miscellaneous
48,878
0.1679
291,176
0.9999
Copyright ยฉ 2020 Pearson Education, Inc.
35
36
Chapter 2
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each robbery type. The result is
Pie Chart of ROBBERY TYPE
Category
Street/highway
Commercial house
Gas or service station
Convenience store
Residence
Bank
Miscellaneous
Miscellaneous
16.8%
Bank
2.0%
Street/highway
43.8%
Residence
17.0%
Convenience store
5.1%station
Gas or service
2.4%
Commercial house
13.0%
(c) We use the bar chart to show the relative frequency with which each
robbery type occurs. The result is
Bar Chart
Percent of FREQUENCY
50
40
30
20
10
0
/
et
re
St
y
wa
gh
hi
Co
m
m
c
er
lh
ia
se
ou
s
Ga
or
e
ic
rv
se
n
io
at
st
C
ce
en
ni
ve
on
o
st
re
e
nc
de
si
Re
nk
Ba
s
ou
ne
la
el
isc
M
ROBBERY TYPE
Percent within all data.
2.33
(a) We find the relative frequencies by dividing each of the frequencies by
the total sample size of 509.
Color
Frequency
Brown
152
Relative
Frequency
0.2986
Yellow
114
0.2240
Red
106
0.2083
Orange
51
0.1002
Green
43
0.0845
Blue
43
0.0845
509
1.0000
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
37
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each color of M&M. The result is
Pie Chart of RELATIVE FREQUENCY vs COLOR
BLUE
0.0844794, 8.4%
GREEN
0.0844794, 8.4%
BRO WN
0.298625, 29.9%
ORA NGE
0.100196, 10.0%
RED
0.208251, 20.8%
YELLO W
0.223969, 22.4%
(c) We use the bar chart to show the relative frequency with which each
color occurs. The result is
Chart of RELATIVE FREQUENCY vs COLOR
0.30
RELATIVE FREQUENCY
0.25
0.20
0.15
0.10
0.05
0.00
2.34
BROWN
YELLOW
RED
ORANGE
COLOR
GREEN
BLUE
(a) We find the relative frequencies by dividing each of the frequencies by
the total sample size of 500.
Political View
Frequency
Relative
Frequency
Liberal
147
0.294
Moderate
237
0.474
Conservative
116
0.232
500
1.000
Copyright ยฉ 2020 Pearson Education, Inc.
38
Chapter 2
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each political view. The result
is
Pie Chart of VIEW
Conservative
23.2%
Liberal
29.4%
Category
Liberal
Moderate
Conservative
Moderate
47.4%
(c) We use the bar chart to show the relative frequency with which each
political view occurs. The result is
Bar Chart
Percent of FREQUENCY
50
40
30
20
10
0
Liberal
Moderate
Conservative
POLITICAL VIEW
Percent within all data.
2.35
(a) We find the relative frequencies by dividing each of the frequencies by
the total sample size of 137,925.
Rank
Frequency
Professor
32,511
Relative
Frequency
0.2357
Associate professor
28,572
0.2072
Assistant professor
59,277
0.4298
Instructor
14,289
0.1036
Other
3,276
0.0238
137,925
1.0000
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
39
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each rank. The result is
Pie Chart of RANK
Instructor
10.4%
Category
Professor
Associate professor
Assistant professor
Instructor
other
other
2.4%
Professor
23.6%
Associate professor
20.7%
Assistant professor
43.0%
(c) We use the bar chart to show the relative frequency with which each
rank occurs. The result is
Bar Chart
Percent of FREQUENCY
40
30
20
10
0
Professor
Associate professor
Assistant professor
Instructor
Other
RANK
Percent within all data.
2.36
(a) We find the relative frequencies by dividing each of the frequencies by
the total sample size of 226. The sum of the relative frequency
columns is 0.9999 due to rounding.
Drug Sold
Frequency
Marijuana
73
Relative
Frequency
0.3230
Crack cocaine
62
0.2743
Powder cocaine
45
0.1991
Ecstasy
20
0.0885
Methamphetamine
17
0.0752
Heroin
5
0.0221
Other
4
0.0177
226
0.9999
Copyright ยฉ 2020 Pearson Education, Inc.
40
Chapter 2
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each drug type. The result is
Pie Chart of DRUG
Methamphetamine
7.5%
Category
Marijuana
Crack cocaine
Powder cocaine
Ecstasy
Methamphetamine
Heroin
other
other
Heroin
2.2%1.8%
Ecstasy
8.8%
Marijuana
32.3%
Powder cocaine
19.9%
Crack cocaine
27.4%
(c) We use the bar chart to show the relative frequency with which each
rank occurs. The result is
Bar Chart
Percent of FREQUENCY
35
30
25
20
15
10
5
0
ij
ar
M
na
ua
k
ac
Cr
c
co
ne
ai
w
Po
oc
rc
de
ne
ai
y
as
st
Ec
a
et
ph
m
ha
t
e
M
e
in
m
H
n
oi
er
r
he
Ot
DRUG
Percent within all data.
2.37
(a) We first find the relative frequencies by dividing each of the
frequencies by the total sample size of 200.
Color
Frequency
Relative
Frequency
0.44
Red
88
Black
102
0.51
Green
10
0.05
200
1.00
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
41
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each color. The result is
COLOR
C ategory
Black
Red
Green
Green
5.0%
Black
51.0%
Red
44.0%
(c) We use the bar chart to show the relative frequency with which each
color occurs. The result is
COLOR
Percent of FREQUENCY
50
40
30
20
10
0
Red
Black
NUMBER
Green
Percent within all data.
2.38
(a) The classes are the different levels of health status and are presented
in column 1. The frequency distribution of health status is presented
in column 2. Dividing each frequency by the total number of people,
which is 188.3 million for persons aged 19-64 and 41.5 million for
persons aged 65 and over, results in each class’s relative frequency.
The relative frequency distribution is presented in column 3.
For People aged 19-64
Health Status
Frequency
Excellent or very good
119.7
Relative
Frequency
0.636
Good
48.4
0.257
Fair or poor
20.2
0.107
188.3
1.000
Copyright ยฉ 2020 Pearson Education, Inc.
42
Chapter 2
For People aged 65 and over
Health Status
Frequency
Excellent or very good
18.1
Relative
Frequency
0.436
Good
14.2
0.342
Fair or poor
9.2
0.222
41.5
1.000
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of each pie represented by each health status. The results
are
Pie Chart of HEALTH STATUS
1 9-64
65+
Fair or poor
10.7%
Category
Excellent /very good
Good
Fair or poor
Fair or poor
22.2%
Excellent /very good
43.6%
Good
25.7%
Excellent /very good
63.6%
Good
34.2%
2.39
(c) A bigger proportion of persons aged 19-64 have a health status of
excellent or very good than persons aged 65 and over.
A smaller
proportion of persons aged 19-64 have a health status of fair or poor
than persons aged 65 and over.
(a) Using Minitab, retrieve the data from the WeissStats Resource Site.
Column 1 contains the type of the vehicle. From the tool bar, select
๏ผ
๏ผ
Stat
Tables
Tally Individual Variables, double-click on TYPE in
the first box so that TYPE appears in the Variables box, put a check
mark next to Counts and Percents under Display, and click OK. The
result is
TYPE
Bus
Car
Truck
N=
Count
21
667
62
750
Percent
2.80
88.93
8.27
(b) The relative frequencies were calculated in part(a) by putting a check
mark next to Percents. 2.8% of the vehicles were busses, 88.93% of the
vehicles were cars, and 8.27% of the vehicles were trucks.
๏ผ
(c) Using Minitab, select Graph
Pie Chart, check Chart counts of unique
values, double-click on TYPE in the first box so that TYPE appears in
the Categorical Variables box. Click Pie Options, check decreasing
volume, click OK. Click Labels, enter TYPE in for the title, click
Slice Labels, check Category Name, Percent, and Draw a line from label
to slice, Click OK twice. The result is
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
43
Pie Chart of TYPE
Truck
8.3%
Category
Car
Truck
Bus
Bus
2.8%
Car
88.9%
๏ผ
(d) Using Minitab, select Graph
Bar Chart, select Counts of unique
values, select Simple option, click OK. Double-click on TYPE in the
first box so that TYPE appears in the Categorical Variables box.
Select Chart Options, check decreasing Y, check show Y as a percent,
click OK. Select Labels, enter in TYPE as the title. Click OK twice.
The result is
TYPE
90
80
70
Percent
60
50
40
30
20
10
0
Car
Truck
Bus
TYPE
Percent within all data.
2.40
(a) Using Minitab, retrieve the data from the WeissStats Resource Site.
Column 2 contains the marital status and column 3 contains the number
๏ผ
๏ผ
Tables
Tally
of drinks per month. From the tool bar, select Stat
Individual Variables, double-click on STATUS and DRINKS in the first
box so that both STATUS and DRINKS appear in the Variables box, put a
check mark next to Counts and Percents under Display, and click OK.
The results are
STATUS Count
Single
354
Married 1173
Widowed
143
Divorced 102
N=
1772
Percent
19.98
66.20
8.07
5.76
DRINKS Count Percent
Abstain
590
33.30
1-60
957
54.01
Over 60
225
12.70
N=
1772
(b) The relative frequencies were calculated in part(a) by putting a check
mark next to Percents. For the STATUS variable; 19.98% of the US
Adults are single, 66.20% are married, 8.07% are widowed, and 5.76% are
divorced. For the DRINKS variable, 33.30% of US Adults abstain from
drinking, 54.01% have 1-60 drinks per month, and 12.70% have over 60
drinks per month.
Copyright ยฉ 2020 Pearson Education, Inc.
44
Chapter 2
๏ผ
(c) Using Minitab, select Graph
Pie Chart, check Chart counts of unique
values, double-click on STATUS and DRINKS in the first box so that
STATUS and DRINKS appear in the Categorical Variables box. Click Pie
Options, check decreasing volume, click OK. Click Multiple Graphs,
check On the Same Graphs, Click OK. Click Labels, click Slice Labels,
check Category Name, Percent, and Draw a line from label to slice,
Click OK twice. The results are
STATUS and DRINKS
STATUS
DRINKS
Div orced
Widowed 5.8%
8.1%
Category
Married
Single
Widowed
Div orced
A bstain
1-60
Ov er 60
Ov er 60
12.7%
Single
20.0%
1-60
54.0%
A bstain
33.3%
Married
66.2%
๏ผ
(d) Using Minitab, select Graph
Bar Chart, select Counts of unique
values, select Simple option, click OK. Double-click on STATUS and
DRINKS in the first box so that STATUS and DRINKS appear in the
Categorical Variables box.
Select Chart Options, check decreasing Y,
check show Y as a percent, click OK. Click OK twice.
The results are
Chart of DRINKS
Chart of STATUS
70
60
60
50
40
Percent
Percent
50
40
30
20
20
10
10
0
30
Married
Single
Widowed
Divorced
0
1-60
Abstain
DRINKS
STATUS
Percent within all data.
Percent within all data.
2.41
Over 60
(a) Using Minitab, retrieve the data from the WeissStats Resource Site.
Column 2 contains the preference for how the members want to receive
the ballots and column 3 contains the highest degree obtained by the
๏ผ
๏ผ
members. From the tool bar, select Stat
Tables
Tally Individual
Variables, double-click on PREFERENCE and DEGREE in the first box so
that both PREFERENCE and DEGREE appear in the Variables box, put a
check mark next to Counts and Percents under Display, and click OK.
The results are
PREFERENCE
Both
Email
Mail
N/A
N=
Count
112
239
86
129
566
Percent
19.79
42.23
15.19
22.79
DEGREE
MA
Other
PhD
N=
Count
167
11
388
566
Percent
29.51
1.94
68.55
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.2
45
(b) The relative frequencies were calculated in part(a) by putting a check
mark next to Percents. For the PREFERENCE variable; 19.79% of the
members prefer to receive the ballot by both e-mail and mail, 42.23%
prefer e-mail, 15.19% prefer mail, and 22.79% didnโt list a preference.
For the Degree variable; 29.51% obtained a Masterโs degree, 68.55%
obtained a PhD, and 1.94% received a different degree.
๏ผ
(c) Using Minitab, select Graph
Pie Chart, check Chart counts of unique
values, double-click on PREFERENCE and DEGREE in the first box so that
PREFERENCE and DEGREE appear in the Categorical Variables box. Click
Pie Options, check decreasing volume, click OK. Click Multiple Graphs,
check On the Same Graphs, Click OK. Click Labels, click Slice Labels,
check Category Name, Percent, and Draw a line from label to slice,
Click OK twice. The results are
Pie Chart of PREFERENCE, DEGREE
PREFERENCE
DEGREE
Category
Email
N/A
Both
Mail
MA
other
PhD
other
1.9%
Mail
15.2%
MA
29.5%
Email
42.2%
Both
19.8%
PhD
68.6%
N/A
22.8%
๏ผ
(d) Using Minitab, select Graph
Bar Chart, select Counts of unique
values, select Simple option, click OK. Double-click on PREFERENCE and
DEGREE in the first box so that PREFERENCE and DEGREE appear in the
Categorical Variables box.
Select Chart Options, check decreasing Y,
check show Y as a percent, click OK. Click OK twice.
The results are
Chart of DEGREE
Chart of PREFERENCE
70
40
60
50
Percent
Percent
30
20
40
30
20
10
10
0
Email
N/A
Both
Mail
0
PhD
PREFERENCE
Percent within all data.
Percent within all data.
Copyright ยฉ 2020 Pearson Education, Inc.
MA
DEGREE
Other
46
Chapter 2
Exercises 2.3
2.42
One important reason for grouping data is that grouping often makes a large
and complicated set of data more compact and easier to understand.
2.43
For class limits, marks, cutpoints and midpoints to make sense, data must be
numerical. They do not make sense for qualitative data classes because such
data are nonnumerical.
2.44
The most important guidelines in choosing the classes for grouping a data
set are: (1) the number of classes should be small enough to provide an
effective summary, but large enough to display the relevant characteristics
of the data; (2) each observation must belong to one, and only one, class;
and (3) whenever feasible, all classes should have the same width.
2.45
In the first method for depicting classes called cutpoint grouping, we used
the notation a โ under b to mean values that are greater than or equal to a
and up to, but not including b, such as 30 โ under 40 to mean a range of
values greater than or equal to 30, but strictly less than 40. In the
alternate method called limit grouping, we used the notation a-b to
indicate a class that extends from a to b, including both. For example, 3039 is a class that includes both 30 and 39. The alternate method is
especially appropriate when all of the data values are integers. If the
data include values like 39.7 or 39.93, the first method is more
advantageous since the cutpoints remain integers; whereas, in the alternate
method, the upper limits for each class would have to be expressed in
decimal form such as 39.9 or 39.99.
2.46
(a) For continuous data displayed to one or more decimal places, using the
cutpoint grouping is best since the description of the classes is
simpler, regardless of the number of decimal places displayed.
(b) For discrete data with relatively few distinct observations, the single
value grouping is best since either of the other two methods would
result in combining some of those distinct values into single classes,
resulting in too few classes, possibly less than 5.
2.47
For limit grouping, we find the class mark, which is the average of the
lower and upper class limit. For cutpoint grouping, we find the class
midpoint, which is the average of the two cutpoints.
2.48
A frequency histogram shows the actual frequencies on the vertical axis;
whereas, the relative frequency histogram always shows proportions (between
0 and 1) or percentages (between 0 and 100) on the vertical axis.
2.49
An advantage of the frequency histogram over a frequency distribution is
that it is possible to get an overall view of the data more easily. A
disadvantage of the frequency histogram is that it may not be possible to
determine exact frequencies for the classes when the number of observations
is large.
2.50
By showing the lower class limits (or cutpoints) on the horizontal axis, the
range of possible data values in each class is immediately known and the
class mark (or midpoint) can be quickly determined. This is particularly
helpful if it is not convenient to make all classes the same width. The use
of the class mark (or midpoint) is appropriate when each class consists of a
single value (which is, of course, also the midpoint). Use of the class
marks (or midpoints) is not appropriate in other situations since it may be
difficult to determine the location of the class limits (or cutpoints) from
the values of the class marks (or midpoints), particularly if the class
marks (or midpoints) are not evenly spaced. Class Marks (or midpoints)
cannot be used if there is an open class.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
47
2.51
If the classes consist of single values, stem-and-leaf diagrams and
frequency histograms are equally useful. If only one diagram is needed and
the classes consist of more than one value, the stem-and-leaf diagram allows
one to retrieve all of the original data values whereas the frequency
histogram does not. If two or more sets of data of different sizes are to
be compared, the relative frequency histogram is advantageous because all of
the diagrams to be compared will have the same total relative frequency of
1.00. Finally, stem-and-leaf diagrams are not very useful with very large
data sets and may present problems with data having many digits in each
number.
2.52
The histogram (especially one using relative frequencies) is generally
preferable. Data sets with a large number of observations may result in a
stem of the stem-and-leaf diagram having more leaves than will fit on the
line. In that case, the histogram would be preferable.
2.53
You can reconstruct the stem-and-leaf diagram using two lines per stem. For
example, instead of listing all of the values from 10 to 19 on a โ1โ stem,
you can make two โ1โ stems. On the first, you record the values from 10 to
14 and on the second, the values from 15 to 19. If there are still two few
stems, you can reconstruct the diagram using five lines per stem, recording
10 and 11 on the first line, 12 and 13 on the second, and so on.
2.54
For the number of bedrooms per single-family dwelling, single-value grouping
is probably the best because the data is discrete with relatively few
distinct observations.
2.55
For the ages of householders, given as a whole number, limit grouping is
probably the best because the data are given as whole numbers and there are
probably too many distinct observations to list them as single-value
grouping.
2.56
For additional sleep obtained by a sample of 100 patients by using a
particular brand of sleeping pill, cutpoint grouping is probably the best
because the data is continuous and the data was recorded to the nearest
tenth of an hour.
2.57
For the number of automobiles per family, single-value grouping is probably
the best because the data is discrete with relatively few distinct
observations.
2.58
For gas mileages, rounded to the nearest number of miles per gallon, limit
grouping is probably the best because the data are given as whole numbers
and there are probably too many distinct observations to list them as
single-value grouping.
2.59
For carapace length for a sample of giant tarantulas, cutpoint grouping is
probably the best because the data is continuous and the data was recorded
to the nearest hundredth of a millimeter.
2.60
(a) Since the data values range from 1 to 4, we construct a table with
classes based on a single value. The resulting table follows.
Single Value Class
1
2
3
4
Frequency
2
0
5
3
10
Copyright ยฉ 2020 Pearson Education, Inc.
48
Chapter 2
(b) To get the relative frequencies, divide each frequency by the sample
size of 10.
Single Value Class
1
2
3
4
(c)
Relative Frequency
0.20
0.00
0.50
0.30
1.00
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. Column
1 demonstrates that the data are grouped using classes based on a
single value. These single values in column 1 are used to label the
horizontal axis of the frequency histogram. When classes are based on
a single value, the middle of each histogram bar is placed directly
over the single numerical value represented by the class. Also, the
height of each bar in the frequency histogram matches the respective
frequency in column 2.
Figure (b)
5
50
4
40
3
30
Percent
Frequency
Figure (a)
2
20
1
10
0
1
2
3
0
4
1
2
3
4
DATA
DATA
(d) The relative-frequency histogram in Figure (b) is constructed using the
relative-frequency distribution presented in part (b) of this exercise.
It has the same horizontal axis as the frequency histogram. The middle
of each histogram bar is placed directly over the single numerical
value represented by the class. Also, the height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 2.
2.61
(a) Since the data values range from 1 to 4, we construct a table with
classes based on a single value. The resulting table follows.
Single Value Class
1
2
3
4
Frequency
2
2
5
1
10
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
49
(b) To get the relative frequencies, divide each frequency by the sample
size of 10.
Single Value Class
1
2
3
4
(c)
Relative Frequency
0.20
0.20
0.50
0.10
1.00
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. Column
1 demonstrates that the data are grouped using classes based on a
single value. These single values in column 1 are used to label the
horizontal axis of the frequency histogram. When classes are based on
a single value, the middle of each histogram bar is placed directly
over the single numerical value represented by the class. Also, the
height of each bar in the frequency histogram matches the respective
frequency in column 2.
Figure (a)
Figure (b)
50
5
40
Percent
Frequency
4
3
30
2
20
1
10
0
1
2
3
0
4
1
2
3
4
DATA
DATA
(d) The relative-frequency histogram in Figure (b) is constructed using the
relative-frequency distribution presented in part (b) of this exercise.
It has the same horizontal axis as the frequency histogram. The middle
of each histogram bar is placed directly over the single numerical
value represented by the class. Also, the height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 2.
2.62
(a) Since the data values range from 0 to 4, we construct a table with
classes based on a single value. The resulting table follows.
Single Value Class
0
1
2
3
4
Frequency
2
8
7
2
1
20
Copyright ยฉ 2020 Pearson Education, Inc.
50
Chapter 2
(b) To get the relative frequencies, divide each frequency by the sample
size of 20.
Single Value Class
0
1
2
3
4
(c)
Relative Frequency
0.10
0.40
0.35
0.10
0.05
1.00
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. Column
1 demonstrates that the data are grouped using classes based on a
single value. These single values in column 1 are used to label the
horizontal axis of the frequency histogram. When classes are based on
a single value, the middle of each histogram bar is placed directly
over the single numerical value represented by the class. Also, the
height of each bar in the frequency histogram matches the respective
frequency in column 2.
Figure (a)
Figure (b)
9
40
8
7
30
Percent
Frequency
6
5
4
20
3
10
2
1
0
0
1
2
3
0
4
0
1
DATA
2
3
4
DATA
(d) The relative-frequency histogram in Figure (b) is constructed using the
relative-frequency distribution presented in part (b) of this exercise.
It has the same horizontal axis as the frequency histogram. The middle
of each histogram bar is placed directly over the single numerical
value represented by the class. Also, the height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 2.
2.63
(a) Since the data values range from 0 to 4, we construct a table with
classes based on a single value. The resulting table follows.
Single Value Class
0
1
2
3
4
Frequency
1
1
4
8
6
20
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
51
(b) To get the relative frequencies, divide each frequency by the sample
size of 20.
Single Value Class
0
1
2
3
4
(c)
Relative Frequency
0.05
0.05
0.20
0.40
0.30
1.00
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. Column
1 demonstrates that the data are grouped using classes based on a
single value. These single values in column 1 are used to label the
horizontal axis of the frequency histogram. When classes are based on
a single value, the middle of each histogram bar is placed directly
over the single numerical value represented by the class. Also, the
height of each bar in the frequency histogram matches the respective
frequency in column 2.
Figure (a)
Figure (b)
9
40
8
7
30
5
Percent
Frequency
6
4
20
3
10
2
1
0
0
1
2
3
4
0
0
DATA
1
2
3
4
DATA
(d) The relative-frequency histogram in Figure (b) is constructed using the
relative-frequency distribution presented in part (b) of this exercise.
It has the same horizontal axis as the frequency histogram. The middle
of each histogram bar is placed directly over the single numerical
value represented by the class. Also, the height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 2.
2.64
(a) The first class to construct is 0-9. The width of all the classes is
10, so the next class would be 10-19. The classes are presented in
column 1. The last class to construct is 40-49, since the largest
single data value is 41. The tallied results are presented in column
2, which lists the frequencies.
Limit Grouping Classes
0-9
10-19
20-29
30-39
40-49
Frequency
3
3
5
8
1
20
Copyright ยฉ 2020 Pearson Education, Inc.
52
Chapter 2
(b)
Dividing each frequency by the total number of observations, which is
20, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2. The resulting
table follows.
Limit Grouping Classes
0-9
10-19
20-29
30-39
40-49
Relative Frequency
0.15
0.15
0.25
0.40
0.05
1.00
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. The
lower class limits of column 1 are used to label the horizontal axis
of the frequency histogram. The height of each bar in the frequency
histogram matches the respective frequency in column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (a)
Figure (b)
9
40
8
7
30
Percent
Frequency
6
5
4
20
3
10
2
1
0
0
10
20
30
40
50
0
0
10
DATA
2.65
20
30
40
50
DATA
(a) The first class to construct is 0-4. The width of all the classes is
5, so the next class would be 5-9. The classes are presented in column
1. The last class to construct is 25-29, since the largest single data
value is 26. The tallied results are presented in column 2, which
lists the frequencies.
Limit Grouping Classes
0-4
5-9
10-14
15-19
20-24
25-29
(b)
Frequency
5
5
1
4
2
3
20
Dividing each frequency by the total number of observations, which is
20, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2. The resulting
table follows.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
Limit Grouping Classes
0-4
5-9
10-14
15-19
20-24
25-29
Relative Frequency
0.25
0.25
0.05
0.20
0.10
0.15
1.00
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. The
lower class limits of column 1 are used to label the horizontal axis
of the frequency histogram. The height of each bar in the frequency
histogram matches the respective frequency in column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (b)
5
25
4
20
3
15
Percent
Frequency
Figure (a)
2
10
1
5
0
0
5
10
15
20
25
30
0
0
5
DATA
2.66
53
10
15
20
25
30
DATA
(a) The first class to construct is 30-36. The width of all the classes is
7, so the next class would be 37-43. The classes are presented in
column 1. The last class to construct is 72-78, since the largest
single data value is 78. The tallied results are presented in column
2, which lists the frequencies.
Limit Grouping Classes
30-36
37-43
44-50
51-57
58-64
65-71
72-78
(b)
Frequency
7
2
5
4
3
2
2
25
Dividing each frequency by the total number of observations, which is
25, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2. The resulting
table follows.
Copyright ยฉ 2020 Pearson Education, Inc.
54
Chapter 2
Limit Grouping Classes
30-36
37-43
44-50
51-57
58-64
65-71
72-78
Relative Frequency
0.28
0.08
0.20
0.16
0.12
0.08
0.08
1.00
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. The
lower class limits of column 1 are used to label the horizontal axis
of the frequency histogram. The height of each bar in the frequency
histogram matches the respective frequency in column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (a)
Figure (b)
30
7
25
6
20
Percent
Frequency
5
4
3
15
10
2
5
1
0
30
37
44
51
58
65
72
78
0
30
37
44
DATA
2.67
51
58
65
72
78
DATA
(a) The first class to construct is 50-59. The width of all the classes is
10, so the next class would be 60-69. The classes are presented in
column 1. The last class to construct is 90-99, since the largest
single data value is 98. The tallied results are presented in column
2, which lists the frequencies.
Limit Grouping Classes
50-59
60-69
70-79
80-89
90-99
(b)
Frequency
6
3
6
7
3
25
Dividing each frequency by the total number of observations, which is
25, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2. The resulting
table follows.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
Limit Grouping Classes
50-59
60-69
70-79
80-89
90-99
55
Relative Frequency
0.24
0.12
0.24
0.28
0.12
1.00
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. The
lower class limits of column 1 are used to label the horizontal axis
of the frequency histogram. The height of each bar in the frequency
histogram matches the respective frequency in column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (a)
Figure (b)
30
7
25
6
20
Percent
Frequency
5
4
3
15
10
2
5
1
0
50
60
70
80
90
1 00
0
50
60
DATA
2.68
70
80
90
1 00
DATA
(a) The first class to construct is 10 โ under 15. Since all classes are
to be of equal width 5, the second class is 15 โ under 20. All of the
classes are presented in column 1. The last class to construct is
35 โ under 40, since the largest single data value is 38.8. The
results of the tallying are presented in column 2, which lists the
frequencies.
Cutpoint Grouping Classes
10 โ under 15
15 โ under 20
20 โ under 25
25 โ under 30
30 โ under 35
35 โ under 40
Frequency
4
5
7
3
0
1
20
(b) Dividing each frequency by the total number of observations, which is
20, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
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Chapter 2
Cutpoint Grouping Classes
10 โ under 15
15 โ under 20
20 โ under 25
25 โ under 30
30 โ under 35
35 โ under 40
Relative Frequency
0.20
0.25
0.35
0.15
0.00
0.05
1.00
(c) The frequency histogram in Figure (a) is constructed using the
frequency distribution obtained in part (a) of this exercise. The
cutpoints are used to label the horizontal axis. Also, the height of
each bar in the frequency histogram matches the respective frequency in
column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution obtained in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (a)
Figure (b)
40
7
6
30
Percent
Frequency
5
4
20
3
2
10
1
0
10
15
20
25
30
35
40
0
10
15
20
2.69
25
30
35
40
DATA
DATA
(a) The first class to construct is 40 โ under 46. Since all classes are
to be of equal width 6, the second class is 46 โ under 52. All of the
classes are presented in column 1. The last class to construct is
64 โ under 70, since the largest single data value is 65.4. The
results of the tallying are presented in column 2, which lists the
frequencies.
Cutpoint Grouping Classes
40 โ under 46
46 โ under 52
52 โ under 58
58 โ under 64
64 โ under 70
Frequency
3
6
10
0
1
20
(b) Dividing each frequency by the total number of observations, which is
20, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
57
Cutpoint Grouping
40 โ under 46
46 โ under 52
52 โ under 58
58 โ under 64
64 โ under 70
Relative Frequency
0.15
0.30
0.50
0.00
0.05
1.00
(c) The frequency histogram in Figure (a) is constructed using the
frequency distribution obtained in part (a) of this exercise. The
cutpoints are used to label the horizontal axis. Also, the height of
each bar in the frequency histogram matches the respective frequency in
column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution obtained in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (a)
Figure (b)
10
50
40
6
Percent
Frequency
8
30
4
20
2
10
0
40
46
52
58
64
70
0
40
46
DATA
2.70
52
58
64
70
DATA
(a) The first classโ midpoint is 0.5 with a width of 1. Therefore, the
first class to construct is 0 โ under 1. Since all classes are to be
of equal width 1, the second class is 1 โ under 2. All of the classes
are presented in column 1. The last class to construct is 7 โ under 8,
since the largest single data value is 7.69. The results of the
tallying are presented in column 2, which lists the frequencies.
Cutpoint Grouping Classes
0 โ under 1
1 โ under 2
2 โ under 3
3 โ under 4
4 โ under 5
5 โ under 6
6 โ under 7
7 โ under 8
Frequency
3
11
4
1
2
1
2
1
25
(b) Dividing each frequency by the total number of observations, which is
25, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
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Chapter 2
Cutpoint Grouping Classes
0 โ under 1
1 โ under 2
2 โ under 3
3 โ under 4
4 โ under 5
5 โ under 6
6 โ under 7
7 โ under 8
Relative Frequency
0.12
0.44
0.16
0.04
0.08
0.04
0.08
0.04
1.00
(c) The frequency histogram in Figure (a) is constructed using the
frequency distribution obtained in part (a) of this exercise. The
cutpoints are used to label the horizontal axis. Also, the height of
each bar in the frequency histogram matches the respective frequency in
column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution obtained in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (a)
Figure (b)
12
50
10
40
Percent
Frequency
8
6
30
20
4
10
2
0
0
2
4
6
8
0
0
2
DATA
2.71
4
6
8
DATA
(a) The first classโ cutpoint is 25 with a width of 3. Therefore, the
first class to construct is 25 โ under 28. Since all classes are to be
of equal width 3, the second class is 28 โ under 31. All of the
classes are presented in column 1. The last class to construct is 43 โ
under 46, since the largest single data value is 43.01. The results of
the tallying are presented in column 2, which lists the frequencies.
Cutpoint Grouping Classes
25 โ under 28
28 โ under 31
31 โ under 34
34 โ under 37
37 โ under 40
40 โ under 43
43 โ under 46
Frequency
1
2
5
7
5
4
1
25
(b) Dividing each frequency by the total number of observations, which is
25, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
59
Cutpoint Grouping Classes
25 โ under 28
28 โ under 31
31 โ under 34
34 โ under 37
37 โ under 40
40 โ under 43
43 โ under 46
Relative Frequency
0.04
0.08
0.20
0.28
0.20
0.16
0.04
1.00
(c) The frequency histogram in Figure (a) is constructed using the
frequency distribution obtained in part (a) of this exercise. The
cutpoints are used to label the horizontal axis. Also, the height of
each bar in the frequency histogram matches the respective frequency in
column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution obtained in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (a)
Figure (b)
30
7
6
25
20
4
Percent
Frequency
5
3
15
10
2
5
1
0
25
28
31
34
37
40
43
0
46
25
28
31
DATA
2.72
34
37
40
43
46
DATA
The horizontal axis of this dotplot displays a range of possible values. To
complete the dotplot, we go through the data set and record data value by
placing a dot over the appropriate value on the horizontal axis.
Dotplot of DATA
1
2
3
4
DATA
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60
Chapter 2
2.73
The horizontal axis of this dotplot displays a range of possible values. To
complete the dotplot, we go through the data set and record data value by
placing a dot over the appropriate value on the horizontal axis.
Dotplot of DATA
0
1
2
3
DATA
2.74
The horizontal axis of this dotplot displays a range of possible values. To
complete the dotplot, we go through the data set and record data value by
placing a dot over the appropriate value on the horizontal axis.
Dotplot of DATA
10
11
12
13
14
15
DATA
2.75
The horizontal axis of this dotplot displays a range of possible values. To
complete the dotplot, we go through the data set and record data value by
placing a dot over the appropriate value on the horizontal axis.
Dotplot of DATA
30
31
32
33
34
35
36
37
38
DATA
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
2.76
61
Each data value consists of 2 or 3 digit numbers ranging from 60 to 125. The
last digit becomes the leaf and the remaining digits are the stems, so we
have stems of 6 to 12. The resulting stem-and-leaf diagram is
6| 0
7|
8|
9| 79
10| 068
11| 278
12| 5
2.77
2.78
2.79
2.80
Each data value consists of 2 digit numbers ranging from 20 to 62. The last
digit becomes the leaf and the remaining digits are the stems, so we have
stems of 2 to 6. The resulting stem-and-leaf diagram is
2| 01
3| 2278
4| 13
5| 5
6| 2
Each data value consists of 1 or 2 digit numbers ranging from 5 to 23. The
last digit becomes the leaf and the remaining digits are the stems, so we
have stems of 0 to 2. Splitting the stems into five lines per stem, the
resulting stem-and-leaf diagram is
0| 5
0|
0| 89
1| 001
1| 2
1| 44455
1| 667
1|
2| 0
2| 2233
Each data value consists of 2 digit numbers ranging from 22 to 46. The last
digit becomes the leaf and the remaining digits are the stems, so we have
stems of 2 to 4. Splitting the stems into two lines per stem, the resulting
stem-and-leaf diagram is
2| 224
2| 5577789
3| 12234
3| 67
4| 0
4| 56
(a) Since the data values range from 0 to 4, we construct a table with
classes based on a single value. The resulting table follows.
Number of Siblings
0
1
2
3
4
Frequency
8
17
11
3
1
40
Copyright ยฉ 2020 Pearson Education, Inc.
Chapter 2
(b) To get the relative frequencies, divide each frequency by the sample
size of 40.
Relative Frequency
0.200
0.425
0.275
0.075
0.025
1.000
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. Column 1
demonstrates that the data are grouped using classes based on a single
value. These single values in column 1 are used to label the horizontal
axis of the frequency histogram. Suitable candidates for vertical axis
units in the frequency histogram are the integers within the range 0
through 17, since these are representative of the magnitude and spread of
the frequencies presented in column 2. When classes are based on a
single value, the middle of each histogram bar is placed directly over
the single numerical value represented by the class. Also, the height of
each bar in the frequency histogram matches the respective frequency in
column 2.
Number of Siblings
0
1
2
3
4
Figure (a)
Histogram of SIBLINGS
Histogram of SIBLINGS
18
40
16
14
30
Percent
12
Frequency
62
10
8
20
6
10
4
2
0
0
1
2
SIBLINGS
3
0
4
0
1
2
SIBLINGS
3
4
Figure
(b)
(d) The relative-frequency histogram in Figure (b) is constructed using the
relative-frequency distribution presented in part (b) of this exercise.
It has the same horizontal axis as the frequency histogram. We notice
that the relative frequencies presented in column 2 range in size from
0.025 to 0.425. Thus, suitable candidates for vertical-axis units in
the relative-frequency histogram are increments of 0.05 (or 5%), from
zero to 0.45 (or 45%). The middle of each histogram bar is placed
directly over the single numerical value represented by the class.
Also, the height of each bar in the relative-frequency histogram
matches the respective relative frequency in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
2.81
63
(a) Since the data values range from 1 to 7, we construct a table with
classes based on a single value. The resulting table follows.
Number of Persons
Frequency
1
7
2
13
3
9
4
5
5
4
6
1
7
1
40
(b) To get the relative frequencies, divide each frequency by the sample
size of 40.
Number of Persons
Relative Frequency
1
0.175
2
0.325
3
0.225
4
0.125
5
0.100
6
0.025
7
0.025
1.000
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. Column
1 demonstrates that the data are grouped using classes based on a
single value. These single values in column 1 are used to label the
horizontal axis of the frequency histogram. Suitable candidates for
vertical axis units in the frequency histogram are the integers within
the range 0 through 13, since these are representative of the
magnitude and spread of the frequencies presented in column 2. When
classes are based on a single value, the middle of each histogram bar
is placed directly over the single numerical value represented by the
class. Also, the height of each bar in the frequency histogram
matches the respective frequency in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
64
Chapter 2
Figure (a)
(d)
2.82
Figure (b)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 2 range in
size from 0.025 to 0.325. Thus, suitable candidates for vertical-axis
units in the relative-frequency histogram are increments of 0.05 (or
5%), from zero to 0.35 (or 35%). The middle of each histogram bar is
placed directly over the single numerical value represented by the
class. Also, the height of each bar in the relative-frequency
histogram matches the respective relative frequency in column 2.
(a) Since the data values range from 1 to 8, we construct a table with
classes based on a single value. The resulting table follows.
Litter Size
Frequency
1
1
2
0
3
1
4
3
5
7
6
6
7
4
8
2
24
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
65
(b) To get the relative frequencies, divide each frequency by the sample
size of 24.
Relative Frequency
Litter Size
1
0.042
2
0.000
3
0.042
4
0.125
5
0.292
6
0.250
7
0.167
8
0.083
1.000
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. Column
1 demonstrates that the data are grouped using classes based on a
single value. These single values in column 1 are used to label the
horizontal axis of the frequency histogram. When classes are based on
a single value, the middle of each histogram bar is placed directly
over the single numerical value represented by the class. Also, the
height of each bar in the frequency histogram matches the respective
frequency in column 2.
Figure (a)
Figure (b)
Litter Size
Litter Size
30
7
25
6
20
Percent
Frequency
5
4
3
10
2
5
1
0
15
1
2
3
4
5
6
7
8
0
1
2
3
SIZE
(d)
4
5
6
7
8
SIZE
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 2 range in
size from 0.000 to 0.292. Thus, suitable candidates for vertical-axis
units in the relative-frequency histogram are increments of 0.05 (or
5%), from zero to 0.30 (or 30%). The middle of each histogram bar is
placed directly over the single numerical value represented by the
class. Also, the height of each bar in the relative-frequency
histogram matches the respective relative frequency in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
66
Chapter 2
2.83
(a) Since the data values range from 0 to 5, we construct a table with
classes based on a single value. The resulting table follows.
Number of Computers
0
1
2
3
4
5
Frequency
5
22
11
2
4
1
45
(b) To get the relative frequencies, divide each frequency by the sample
size of 45. The sum of the relative frequency column is 0.999 due to
rounding
Number of Computers
0
1
2
3
4
5
(c)
Relative Frequency
0.111
0.489
0.244
0.044
0.089
0.022
0.999
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. Column
1 demonstrates that the data are grouped using classes based on a
single value. These single values in column 1 are used to label the
horizontal axis of the frequency histogram. When classes are based on
a single value, the middle of each histogram bar is placed directly
over the single numerical value represented by the class. Also, the
height of each bar in the frequency histogram matches the respective
frequency in column 2.
Figure (a)
Figure (b)
25
50
40
15
Percent
Frequency
20
10
20
5
0
30
10
0
1
2
3
4
5
0
0
1
COMPUTERS
2
3
4
5
COMPUTERS
(d) The relative-frequency histogram in Figure (b) is constructed using the
relative-frequency distribution presented in part (b) of this exercise.
It has the same horizontal axis as the frequency histogram. The middle
of each histogram bar is placed directly over the single numerical
value represented by the class. Also, the height of each bar in the
relative-frequency histogram matches the respective relative
frequencies in column 2.
2.84
(a) The first class to construct is 40-49. Since all classes are to be of
equal width, and the second class begins with 50, we know that the
width of all classes is 50 – 40 = 10. All of the classes are presented
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
67
in column 1. The last class to construct is 150-159, since the largest
single data value is 155. Having established the classes, we tally the
energy consumption figures into their respective classes. These
results are presented in column 2, which lists the frequencies.
Consumption (mil. BTU)
40-49
50-59
60-69
70-79
80-89
90-99
100-109
110-119
120-129
130-139
140-149
150-159
(b)
Frequency
1
7
7
3
6
10
5
4
2
3
0
2
50
Dividing each frequency by the total number of observations, which is
50, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2. The resulting
table follows.
Consumption (mil. BTU)
40-49
50-59
60-69
70-79
80-89
90-99
100-109
110-119
120-129
130-139
140-149
150-159
Relative Frequency
0.02
0.14
0.14
0.06
0.12
0.20
0.10
0.08
0.04
0.06
0.00
0.04
1.00
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. The
lower class limits of column 1 are used to label the horizontal axis
of the frequency histogram. Suitable candidates for vertical-axis
units in the frequency histogram are the even integers 0 through 10,
since these are representative of the magnitude and spread of the
frequency presented in column 2. The height of each bar in the
frequency histogram matches the respective frequency in column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 2 vary in
size from 0.00 to 0.20. Thus, suitable candidates for vertical axis
Copyright ยฉ 2020 Pearson Education, Inc.
68
Chapter 2
units in the relative-frequency histogram are increments of 0.05 (or
5%), from zero to 0.20 (or 20%). The height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 2.
Figure (a)
Figure (b)
Histogram of ENERGY
Histogram of ENERGY
20
10
15
6
Percent
Frequency
8
5
2
0
2.85
10
4
40
50
60
70
80
90
100 110
ENERGY
120
130
140
150
160
0
40
50
60
70
80
90
100 110
ENERGY
120
130
140
150
160
(a) The first class to construct is 40-44. Since all classes are to be of
equal width, and the second class begins with 45, we know that the
width of all classes is 45 – 40 = 5. All of the classes are presented
in column 1. The last class to construct is 60-64, since the largest
single data value is 61. Having established the classes, we tally the
age figures into their respective classes. These results are presented
in column 2, which lists the frequencies.
Age
40-44
45-49
50-54
55-59
60-64
(b)
Dividing each frequency by the total number of observations, which is
21, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2. The resulting
table follows.
Age
40-44
45-49
50-54
55-59
60-64
(c)
Frequency
4
3
4
8
2
21
Relative Frequency
0.190
0.143
0.190
0.381
0.095
1.000
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. The
lower class limits of column 1 are used to label the horizontal axis
of the frequency histogram. Suitable candidates for vertical-axis
units in the frequency histogram are the even integers 2 through 8,
since these are representative of the magnitude and spread of the
frequency presented in column 2. The height of each bar in the
frequency histogram matches the respective frequency in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
(d)
69
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 2 vary in
size from 0.095 to 0.381. Thus, suitable candidates for vertical axis
units in the relative-frequency histogram are increments of 0.10 (or
10%), from zero to 0.40 (or 40%). The height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 2.
Figure (a)
Figure (b)
Early-Onset Dementia
Early-Onset Dementia
9
40
8
7
30
5
Percent
Frequency
6
4
20
3
10
2
1
0
40
45
50
55
60
65
0
40
45
AGE
2.86
50
55
60
65
AGE
(a) The first class to construct is 20-22. Since all classes are to be of
equal width, and the second class begins with 23, we know that the
width of all classes is 23 – 20 = 3. All of the classes are presented
in column 1. The last class to construct is 44-46, since the largest
single data value is 46. Having established the classes, we tally the
cheese consumption figures into their respective classes. These
results are presented in column 2, which lists the frequencies.
Cheese Consumption
20-22
23-25
26-28
29-31
32-34
35-37
38-40
41-43
44-46
(b)
Frequency
1
3
5
3
5
9
2
4
3
35
Dividing each frequency by the total number of observations, which is
35, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2. The resulting
table follows. The sum of the relative frequency column is 1.001 due
to rounding.
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70
Chapter 2
Cheese Consumption
20-22
23-25
26-28
29-31
32-34
35-37
38-40
41-43
44-46
Relative Frequency
0.029
0.086
0.143
0.086
0.143
0.257
0.057
0.114
0.086
1.001
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. The
lower class limits of column 1 are used to label the horizontal axis
of the frequency histogram. Suitable candidates for vertical-axis
units in the frequency histogram are the even integers 2 through 7,
since these are representative of the magnitude and spread of the
frequency presented in column 2. The height of each bar in the
frequency histogram matches the respective frequency in column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 2 vary in
size from 0.057 to 0.200. Thus, suitable candidates for vertical axis
units in the relative-frequency histogram are increments of 0.05 (or
5%), from zero to 0.20 (or 20%). The height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 2.
Figure (a)
Figure (b)
9
25
8
20
6
Percent
Frequency
7
5
4
15
10
3
2
5
1
0
20
26
32
38
44
0
20
26
Cheese Consumption
2.87
32
38
44
Cheese Consumption
(a) The first class to construct is 12-17. Since all classes are to be of
equal width, and the second class begins with 18, we know that the
width of all classes is 18 – 12 = 6. All of the classes are presented
in column 1. The last class to construct is 60-65, since the largest
single data value is 61. Having established the classes, we tally the
anxiety questionnaire score figures into their respective classes.
These results are presented in column 2, which lists the frequencies.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
Anxiety
12-17
18-23
24-29
30-35
36-41
42-47
48-53
54-59
60-65
(b)
71
Frequency
2
3
6
5
10
4
0
0
1
31
Dividing each frequency by the total number of observations, which is
31, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2. The resulting
table follows.
Anxiety
12-17
18-23
24-29
30-35
36-41
42-47
48-53
54-59
60-65
Relative Frequency
0.065
0.097
0.194
0.161
0.323
0.129
0.000
0.000
0.032
1.000
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution presented in part (a) of this exercise. The
lower class limits of column 1 are used to label the horizontal axis
of the frequency histogram. Suitable candidates for vertical-axis
units in the frequency histogram are the even integers 0 through 10,
since these are representative of the magnitude and spread of the
frequency presented in column 2. The height of each bar in the
frequency histogram matches the respective frequency in column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 2 vary in
size from 0.000 to 0.323. Thus, suitable candidates for vertical axis
units in the relative-frequency histogram are increments of 0.05 (or
5%), from zero to 0.35 (or 35%). The height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
72
Chapter 2
Figure (a)
Figure (b)
Chronic Hemodialysis and Axiety
Chronic Hemodialysis and Axiety
35
10
30
8
6
Percent
Frequency
25
4
20
15
10
2
0
2.88
5
12
18
24
30
36
42
ANXIETY
48
54
60
0
66
12
18
24
30
36
42
ANXIETY
48
54
60
66
(a) The first class to construct is 4 โ under 5. Since all classes are to
be of equal width 1, the second class is 5 โ under 6. All of the
classes are presented in column 1. The last class to construct is
10 โ under 11, since the largest single data value is 10.360. Having
established the classes, we tally the audience sizes into their
respective classes. These results are presented in column 2, which
lists the frequencies.
Audience (Millions)
4 โ under 5
5 โ under 6
6 โ under 7
7 โ under 8
8 โ under 9
9 โ under 10
10 โ under 11
Frequency
1
6
7
1
3
1
1
20
(b) Dividing each frequency by the total number of observations, which is
20, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2.
Audience (Millions)
4 โ under 5
5 โ under 6
6 โ under 7
7 โ under 8
8 โ under 9
9 โ under 10
10 โ under 11
Relative
0.05
0.30
0.35
0.05
0.15
0.05
0.05
1.00
(c) The frequency histogram in Figure (a) is constructed using the
frequency distribution obtained in part (a) of this exercise Column 1
demonstrates that the data are grouped using classes with class widths
of 1. Also, the height of each bar in the frequency histogram matches
the respective frequency in column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
(d)
73
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution obtained in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
The height of each bar in the relative-frequency histogram matches the
respective relative frequency in column 2.
Figure (a)
Figure (b)
40
7
6
30
Percent
Frequency
5
4
3
2
20
10
1
0
4
5
6
7
8
9
10
11
0
4
5
6
AUDIENCE
2.89
7
8
9
10
11
AUDIENCE
(a) The first class to construct is 52 โ under 54. Since all classes are
to be of equal width 2, the second class is 54 โ under 56. All of the
classes are presented in column 1. The last class to construct is
74 โ under 76, since the largest single data value is 75.3. Having
established the classes, we tally the cheetah speeds into their
respective classes. These results are presented in column 2, which
lists the frequencies.
Speed
52 โ under 54
54 โ under 56
56 โ under 58
58 โ under 60
60 โ under 62
62 โ under 64
64 โ under 66
66 โ under 68
68 โ under 70
70 โ under 72
72 โ under 74
74 โ under 76
Frequency
2
5
6
8
7
3
2
1
0
0
0
1
35
(b) Dividing each frequency by the total number of observations, which is
35, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2
Copyright ยฉ 2020 Pearson Education, Inc.
74
Chapter 2
Speed
52 โ under 54
54 โ under 56
56 โ under 58
58 โ under 60
60 โ under 62
62 โ under 64
64 โ under 66
66 โ under 68
68 โ under 70
70 โ under 72
72 โ under 74
74 โ under 76
Relative Frequency
0.057
0.143
0.171
0.229
0.200
0.086
0.057
0.029
0.000
0.000
0.000
0.029
1.000
(c)
The frequency histogram in Figure (a) is constructed using the
frequency distribution obtained in part (a) of this exercise Column 1
demonstrates that the data are grouped using classes with class widths
of 2. Suitable candidates for vertical axis units in the frequency
histogram are the integers within the range 0 through 8, since these
are representative of the magnitude and spread of the frequencies
presented in column 2. Also, the height of each bar in the frequency
histogram matches the respective frequency in column 2.
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution obtained in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 3 range in
size from 0.000 to 0.229. Thus, suitable candidates for vertical axis
units in the relative-frequency histogram are increments of 0.05 (5%),
from zero to 0.25 (25%). The height of each bar in the relativefrequency histogram matches the respective relative frequency in
column 2.
Figure (a)
Figure (b)
Speeds of Cheetahs
Speeds of Cheetahs
25
9
8
20
7
Percent
Frequency
6
5
4
15
10
3
2
5
1
0
2.90
52
54
56
58
60
62
64
66
SPEED
68
70
72
74
76
0
52
54
56
58
60
62
64
66
SPEED
68
70
72
74
76
(a) The first class to construct is 12 โ under 14. Since all classes are
to be of equal width 2, the second class is 14 โ under 66. All of the
classes are presented in column 1. The last class to construct is
26 โ under 28, since the largest single data value is 26.4. Having
established the classes, we tally the fuel tank capacities into their
respective classes. These results are presented in column 2, which
lists the frequencies.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
75
Fuel Tank Capacity
12 โ under 14
14 โ under 16
16 โ under 18
18 โ under 20
20 โ under 22
22 โ under 24
24 โ under 26
26 โ under 28
Frequency
2
6
7
6
6
3
3
2
35
(b) Dividing each frequency by the total number of observations, which is
35, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2
Fuel Tank Capacity
12 โ under 14
14 โ under 16
16 โ under 18
18 โ under 20
20 โ under 22
22 โ under 24
24 โ under 26
26 โ under 28
Relative Frequency
0.057
0.171
0.200
0.171
0.171
0.086
0.086
0.057
1.000
The frequency histogram in Figure (a) is constructed using the
frequency distribution obtained in part (a) of this exercise Column 1
demonstrates that the data are grouped using classes with class widths
of 2. Suitable candidates for vertical axis units in the frequency
histogram are the integers within the range 2 through 7, since these
are representative of the magnitude and spread of the frequencies
presented in column 2. Also, the height of each bar in the frequency
histogram matches the respective frequency in column 2.
(c)
(d)
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution obtained in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 3 range in
size from 0.057 to 0.200. Thus, suitable candidates for vertical axis
units in the relative-frequency histogram are increments of 0.05 (5%),
from zero to 0.20 (20%). The height of each bar in the relativefrequency histogram matches the respective relative frequency in
column 2.
Figure (a)
Figure (b)
Fuel Tank Capacity
Fuel Tank Capacity
20
7
6
15
Percent
Frequency
5
4
3
2
10
5
1
0
12
14
16
18
20
CAPACITY
22
24
26
28
0
12
14
16
18
Copyright ยฉ 2020 Pearson Education, Inc.
20
CAPACITY
22
24
26
28
76
Chapter 2
2.91
(a) The first class to construct is 0 โ under 1. Since all classes are to
be of equal width 1, the second class is 1 โ under 2. All of the
classes are presented in column 1. The last class to construct is
7 – under 8, since the largest single data value is 7.6. Having
established the classes, we tally the fuel tank capacities into their
respective classes. These results are presented in column 2, which
lists the frequencies.
Oxygen Distribution
0 โ under 1
1 โ under 2
2 โ under 3
3 โ under 4
4 โ under 5
5 โ under 6
6 โ under 7
7 โ under 8
Frequency
1
10
5
4
0
0
1
1
22
(b) Dividing each frequency by the total number of observations, which is
22, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 2
Oxygen Distribution
0 โ under 1
1 โ under 2
2 โ under 3
3 โ under 4
4 โ under 5
5 โ under 6
6 โ under 7
7 โ under 8
(c)
(d)
Relative Frequency
0.045
0.455
0.227
0.182
0.000
0.000
0.045
0.045
1.000
The frequency histogram in Figure (a) is constructed using the
frequency distribution obtained in part (a) of this exercise Column 1
demonstrates that the data are grouped using classes with class widths
of 2. Suitable candidates for vertical axis units in the frequency
histogram are the integers within the range 0 through 10, since these
are representative of the magnitude and spread of the frequencies
presented in column 2. Also, the height of each bar in the frequency
histogram matches the respective frequency in column 2.
The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution obtained in part (b) of this
exercise. It has the same horizontal axis as the frequency histogram.
We notice that the relative frequencies presented in column 3 range in
size from 0.000 to 0.455. Thus, suitable candidates for vertical axis
units in the relative-frequency histogram are increments of 0.05 (5%),
from zero to 0.50 (50%). The height of each bar in the relativefrequency histogram matches the respective relative frequency in
column 2.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
Figure (a)
Figure (b)
Oxygen Distribution
Oxygen Distribution
50
10
40
6
2.92
Percent
Frequency
8
30
4
20
2
10
0
77
0
1
2
3
4
OXYGEN
5
6
7
0
8
0
1
2
3
4
OXYGEN
5
6
7
8
The horizontal axis of this dotplot displays a range of possible exam
scores. To complete the dotplot, we go through the data set and record each
exam score by placing a dot over the appropriate value on the horizontal
axis.
Dotplot of SCORE
36
45
54
63
72
81
90
99
SCORE
2.93
The horizontal axis of this dotplot displays of range of possible ages of
the passenger cars. To complete the dotplot, we go through the data set and
record each age by placing a dot over the appropriate value on the
horizontal axis.
Dotplot of AGE
2
3
4
5
6
7
8
9
10 11
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 20 21
22 23 24
AGE of Passenger Cars
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78
Chapter 2
2.94
(a)
The data values range from 52 to 84, so the scale must accommodate
those values. We stack dots above each value on two different lines
using the same scale for each line. The result is
Dotplot of INTERVENTION, CONTROL
INTERVENTION
CONTROL
55
60
65
70
75
80
85
Data
2.95
(b)
The two sets of pulse rates are both centered near 68, but the
Intervention data are more concentrated around the center than are the
Control data.
(a)
The data values range from 7 to 18, so the scale must accommodate
those values. We stack dots above each value on two different lines
using the same scale for each line. The result is
Dotplot of DYNAMIC, STATIC
DYNAMIC
STATIC
(b)
6
8
10
12
Data
14
16
18
The Dynamic system does seem to reduce acute postoperative days in the
hospital on the average. The Dynamic data are centered at about 7
days, whereas the Static data are centered at about 11 days and are
much more spread out than the Dynamic data.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
2.96
The data values consist of 2 digit numbers followed by a one digit decimal
place ranging from 45.5 to 57.0.
The last digit following the decimal will
be the leaves and the stems will be the remaining digits.
Therefore, the
stem-and-leaf diagram will have stems of 45 through 57.
45|
46|
47|
48|
49|
50|
51|
52|
53|
54|
55|
56|
57|
2.97
5
4
2
0
1
3578
35589
011
355689
13
033
0
0
Since each data value consists of 2 digits, each beginning with 1, 2, 3, or
4, we will construct the stem-and-leaf diagram with these four values as the
stems. The result is
1|
2|
3|
4|
2.98
79
(a)
Since each data value consists of a 2 digit number with a one digit
decimal, we will make the leaf the decimal digit and the stems the
remaining two digit numbers of 28, 29, 30, and 31. The result is
28|
29|
30|
31|
(b)
8
368
1247
02
Splitting into two lines per stem, leafs of 0-4 belong in the first
stem and leafs of 5-9 belong in the second stem. The result is
28|
29|
29|
30|
30|
31|
(c)
238
1678899
34459
04
8
3
68
124
7
02
The stem-and-leaf diagram in part (b) is more useful because by
splitting the stems into two lines per stem, you have created more
lines. Part (a) had too few lines.
Copyright ยฉ 2020 Pearson Education, Inc.
80
Chapter 2
2.99
(a) Since each data value lies between 2 and 93, we will construct the
stem-and-leaf diagram with one line per stem. The result is
0|
1|
2|
3|
4|
5|
6|
7|
8|
9|
(b)
2234799
11145566689
023479
004555
19
5
9
9
3
Using two lines per stem, the same data result in the following diagram:
0| 2234
0| 799
1| 1114
1| 5566689
2| 0234
2| 79
3| 004
3| 555
4| 1
4| 9
5|
5| 5
6|
6| 9
7|
7| 9
8|
8|
9| 3
(c) The stem with one line per stem is more useful. One gets the same
impression regarding the shape of the distribution, but the two lines
per stem version has numerous lines with no data, making it take up
more space than necessary to interpret the data and giving it too many
lines.
2.100 (a) Since we have two digit numbers, the last digit becomes the leaf and
the first digit becomes the stem. For this data, we have a stem of 7.
Splitting the data into five lines per stem, we put the leaves 0-1 in
the first stem, 2-3 in the second stem, 4-5 in the third stem, 6-7 in
the fourth stem, and 8-9 in the fifth stem. The result is
7|
7|
7|
7|
001111111
2222223333333
444444445555
66666666677
(b) Using one or two lines per stem would have given us too few lines.
2.101 (a) Since we have two digit numbers, the last digit becomes the leaf and
the first digit becomes the stem. For this data, we have stems of 6,
7, and 8. Splitting the data into five lines per stem, we put the
leaves 0-1 in the first stem, 2-3 in the second stem, 4-5 in the third
stem, 6-7 in the fourth stem, and 8-9 in the fifth stem. The result is
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
6|
7|
7|
7|
7|
7|
8|
(b)
81
8
11111111
222222222233333
4444555555
66666777777
8
0
Using one or two lines per stem would have given us too few lines.
2.102 The heights of the bars of the relative-frequency histogram indicate that:
(a) About 27.5% of the returns had an adjusted gross income between $10,000
and $19,999, inclusive.
(b) About 37.5% were between $0 and $9,999; 27.5% were between $10,000 and
$19,999; and 19% were between $20,000 and $29,999. Thus, about 84%
(i.e., 37.5% + 27.5% + 19%) of the returns had an adjusted gross income
less than $30,000.
(c) About 11% were between $30,000 and $39,999; and 5% were between $40,000
and $49,999. Thus, about 16% (i.e., 11% + 5%) of the returns had an
adjusted gross income between $30,000 and $49,999. With 89,928,000
returns having an adjusted gross income less than $50,000, the number
of returns having an adjusted gross income between $30,000 and $49,999
was 14,388,480 (i.e., 0.16 x 89,928,000).
2.103 The graph indicates that:
(a) 20% of the patients have cholesterol levels between 205 and 209,
inclusive.
(b) 20% are between 215 and 219; and 5% are between 220 and 224. Thus, 25%
(i.e., 20% + 5%) have cholesterol levels of 215 or higher.
(c) 35% of the patients have cholesterol levels between 210 and 214,
inclusive. With 20 patients in total, the number having cholesterol
levels between 210 and 214 is 7 (i.e., 0.35 x 20).
2.104 The graph indicates that:
(a) 15 states had at least three but less than four hospital beds per 1000
people available.
(b) 3 states had at least four and a half hospital beds per 1000 people
available.
2.105 The graph indicates that:
(a) 1 of these 14 patients was under 45 years old at the onset of symptoms.
(b) 3 of these 14 patients were at least 65 years old at the onset of
symptoms.
(c) 8 of these 14 patients were between 50 and 64 years old, inclusive, at
the onset of symptoms.
2.106 (a) Using Minitab, retrieve the data from the WeissStats Resource Site.
Column 1 contains the numbers of pups borne in a lifetime for each of
80 female Great White Sharks.
From the tool bar, select Stat
๏ผ Tables
๏ผ Tally Individual Variables, double-click on PUPS in the first box so
that PUPS appears in the Variables box, put a check mark next to Counts
and Percents under Display, and click OK. The result is
Copyright ยฉ 2020 Pearson Education, Inc.
82
Chapter 2
PUPS
Count
Percent
3
2
2.50
4
5
6.25
5
10
12.50
6
11
13.75
7
17
21.25
8
17
21.25
9
11
13.75
10
4
5.00
11
2
2.50
12
1
1.25
(b) After retrieving the data from the WeissStats Resource Site, select
๏ผ
Histogram, choose Simple and click OK. Double click on PUPS in
Graph
the first box to enter PUPS in the Graphs variables box, and click OK.
The frequency histogram is
Histogram of PUPS
18
16
14
Frequency
12
10
8
6
4
2
0
4
6
8
10
12
PUPS
To change to a relative-frequency histogram, before clicking OK the
second time, click on the Scale button and the Y-Scale type tab, and
choose Percent and click OK. The graph will look like the frequency
histogram, but will have relative frequencies on the vertical scale
instead of counts.
The numbers of pups range from 1 to 12 per female with 7 and 8 pups
occurring more frequently than any other values.
2.107 (a) Using Minitab, there is not a direct way to get a grouped frequency
distribution. However, you can use an option in creating your
histogram that will report the frequencies in each of the classes,
essentially creating a grouped frequency distribution. Retrieve the
data from the WeissStats Resource Site. Column 2 contains the length
๏ผ
of the Beatles songs, in seconds. From the tool bar, select Graph
Histogram, choose Simple and click OK. Double click on LENGTH to enter
LENGTH in the Graph variables box. Click Labels, click the Data Labels
tab, then check Use y-value labels. Click OK twice. The result is
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
83
70
70
60
54
Frequency
50
42
40
30
20
17
13
9
10
3
1
0
8
3
3
75
1 50
1
225
1
0
1
0
300
375
1
0
1
1
450
LENGTH
Above each bar is a label for each of the frequencies. Also, the
labeling on the horizontal axis is the midpoint for class, where each
class has a width of 25. To change the width of the bars, right Click
on the bars and choose Edit Bars, click Binning, click
Midpoint/Cutpoint positions:, type 0 50 100 150 200 250 300 350 400 450
500. The result is
1 40
124
1 20
Frequency
1 00
80
60
42
40
32
18
20
1
0
5
0
2
1 00
200
300
1
1
400
2
1
500
LENGTH
To get the relative-frequency distribution, follow the same steps as
above, but also Click the Scale button, click the Y-scale Type, check
Percent, then click OK twice. The result is
60
54.1485
50
Percent
40
30
18.3406
20
13.9738
7.86026
10
0
0.436681
0
2.18341
0.873362 0.436681 0.436681 0.873362 0.436681
1 00
200
300
400
500
LENGTH
Above each bar is the percentage for that class, essentially creating a
relative-frequency distribution. You could also transfer these results
into a table.
(b) A frequency histogram and a relative frequency distribution were
created in part (a).
Copyright ยฉ 2020 Pearson Education, Inc.
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Chapter 2
๏ผ
(c) To obtain the dotplot, select Graph
Dotplot, select Simple in the One
Y row, and click OK. Double click on LENGTH to enter LEGNTH in the
Graph variables box and click OK. The result is
Dotplot of LENGTH
70
1 40
21 0
280
350
420
490
LENGTH
(d) The graphs are similar, but not identical. This is because the dotplot
preserves the raw data by plotting individual dots and the histogram
looses the raw data because it groups observations into grouped
classes. The overall impression, however, remains the same. They both
are generally the same shape with outliers to the right.
2.108 (a) After entering the data from the WeissStats Resource Site, in Minitab,
๏ผ
select Graph
Stem-and-Leaf, double click on PERCENT to enter PERCENT
in the Graph variables box and enter a 10 in the Increment box, and
click OK. The result is
Stem-and-leaf of PERCENT
Leaf Unit = 1.0
(35)
16
8
9
N
= 51
00122223334444455566777777888889999
0000000001111122
(b) Repeat part (a), but this time enter a 5 in the Increment box.
result is
Stem-and-leaf of PERCENT
Leaf Unit = 1.0
15
(20)
16
8
8
9
N
The
= 51
001222233344444
55566777777888889999
0000000001111122
(c) Repeat part (a) again, but this time enter a 2 in the Increment box.
The result is
Stem-and-leaf of PERCENT
Leaf Unit = 1.0
3
10
18
(8)
25
16
2
8
8
8
8
8
9
9
N
= 51
001
2222333
44444555
66777777
888889999
00000000011111
22
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
85
(d) The last graph is the most useful since it gives a better idea of the
shape of the distribution. Typically, we like to have five to fifteen
classes and this is the only one of the three graphs that satisfies
that condition.
2.109 (a) After entering the data from the WeissStats Resource Site, in Minitab,
๏ผ
select Graph
Stem-and-Leaf, double click on PERCENT to enter PERCE T
in the Graph variables box and enter a 10 in the Increment box, and
click OK. The result is
Stem-and-leaf of PERCENT
Leaf Unit = 1.0
2
(32)
17
1
1
2
3
4
N
= 51
79
01122333444455555566666777778999
0001112223456668
9
(b) Repeat part (a), but this time enter a 5 in the Increment box.
result is
Stem-and-leaf of PERCENT
Leaf Unit = 1.0
2
14
(20)
17
6
1
1
1
2
2
3
3
4
4
N
The
= 51
79
011223334444
55555566666777778999
00011122234
56668
9
(c) Repeat part (a) again, but this time enter a 2 in the Increment box.
The result is
Stem-and-leaf of PERCENT
Leaf Unit = 1.0
1
2
5
10
20
(10)
21
17
11
7
5
2
1
1
1
1
1
1
1
2
2
2
2
2
3
3
3
3
3
4
4
4
4
4
N
= 51
7
9
011
22333
4444555555
6666677777
8999
000111
2223
45
666
8
9
(d) The second graph is the most useful. The third one has more classes
than necessary to comprehend the shape of the distribution and has a
number of empty stems. Typically, we like to have five to fifteen
classes and the first and second diagrams satisfy that condition, but
the second one provides a better idea of the shape of the distribution.
Copyright ยฉ 2020 Pearson Education, Inc.
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Chapter 2
2.110 (a) After entering the data from the WeissStats Resource Site, in Minitab,
๏ผ
Histogram, select Simple and click OK. double click on
select Graph
TEMP to enter TEMP in the Graph variables box and click OK. The result
is
Histogram of TEMP
20
Frequency
15
10
5
0
97.0
97.5
98.0
TEMP
98.5
99.0
99.5
๏ผ
Dotplot, select Simple in the One Y row, and click
(b) Now select Graph
OK. Double click on TEMP to enter TEMP in the Graph variables box and
click OK. The result is
Dotplot of TEMP
96.8
97.2
97.6
98.0
TEMP
98.4
98.8
99.2
๏ผ
Stem-and-Leaf, double click on TEMP to enter TEMP in
(c) Now select Graph
the Graph variables box and click OK. Leave the Increment box blank to
allow Minitab to choose the number of lines per stem. The result is
Stem-and-leaf of TEMP N
Leaf Unit = 0.10
1
96 7
3
96 89
8
97 00001
13
97 22233
19
97 444444
26
97 6666777
31
97 88889
45
98 00000000000111
(10) 98 2222222233
38
98 4444445555
28
98 66666666677
17
98 8888888
10
99 00001
5
99 2233
1
99 4
= 93
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
87
(d) The dotplot shows all of the individual values. The stem-and-leaf
diagram used five lines per stem and therefore each line contains
leaves with possibly two values. The histogram chose classes of width
0.25. This resulted in, for example, the class with midpoint 97.0
including all of the values 96.9, 97.0, and 97.1, while the class with
midpoint 97.25 includes only the two values 97.2 and 97.3. Thus the
โsmoothingโ effect is not as good in the histogram as it is in the
stem-and-leaf diagram. Overall, the dotplot gives the truest picture
of the data and allows recovery of all of the data values.
2.111 (a)
The classes are presented in column 1. With the classes established,
we then tally the exam scores into their respective classes. These
results are presented in column 2, which lists the frequencies.
Dividing each frequency by the total number of exam scores, which is
20, results in each class’s relative frequency. The relative
frequencies for all classes are presented in column 3. The class mark
of each class is the average of the lower and upper limits. The class
marks for all classes are presented in column 4.
Score
Frequency
Relative Frequency
30-39
40-49
50-59
60-69
70-79
80-89
90-100
2
0
0
3
3
8
4
0.10
0.00
0.00
0.15
0.15
0.40
0.20
20
1.00
Class Mark
34.5
44.5
54.5
64.5
74.5
84.5
95.0
(b)
The first six classes have width 10; the seventh class had width 11.
(c)
Answers will vary, but one choice is to keep the first six classes the
same and make the next two classes 90-99 and 100-109. Another
possibility is 31-40, 41-50, โฆ, 91-100.
2.112 Answers will vary, but by following the steps we first decide on the
approximate number of classes. Since there are 40 observations, we should
have 7-14 classes. This exercise states we should have approximately seven
classes. Step 2 says that we calculate an approximate class width as
(99 โ 36)/7 = 9. A convenient class width close to 9 would be a class width
of 10. Step 3 says that we choose a number for the lower class limit which
is less than or equal to our minimum observation of 36. Letโs choose 35.
Beginning with a lower class limit of 35 and width of 10, we have a first
class of 35-44, a second class of 45-54, a third class of 55-64, a fourth
class of 65-74, a fifth class of 75-84, a sixth class of 85-94, and a
seventh class of 95-104. This would be our last class since the largest
observation is 99.
2.113 Answers will vary, but by following the steps we first decide on the
approximate number of classes. Since there are 37 observations, we should
have 7-14 classes. This exercise states we should have approximately eight
classes. Step 2 says that we calculate an approximate class width as
(278.8 โ 129.2)/8 = 18.7. A convenient class width close to 18.7 would be a
class width of 20. Step 3 says that we choose a number for the lower
cutpoint which is less than or equal to our minimum observation of 129.2.
Letโs choose 120. Beginning with a lower cutpoint of 120 and width of 20,
we have a first class of 120 โ under 140, a second class of 140 โ under 160,
a third class of 160 โ under 180, a fourth class of 180 – under 200, a fifth
Copyright ยฉ 2020 Pearson Education, Inc.
88
Chapter 2
class of 200 โ under 220, a sixth class of 220 โ under 240, a seventh class
of 240 โ under 260, and an eighth class of 260 โ under 280. This would be
our last class since the largest observation is 278.8.
2.114 (a)
Tally marks for all 50 students, where each student is categorized by
age and gender, are presented in the contingency table given in part
(b).
(b)
Tally marks in each box appearing in the following chart are counted.
These counts, or frequencies, replace the tally marks in the
contingency table. For each row and each column, the frequencies are
added, and their sums are recorded in the proper “Total” box.
Age (yrs)
Gender
Under 21
21 – 25
Over 25
Male
|||||
|||
|||||
||
|||||
||
Female
|||||
||
|||||
|||||
|||
|||||
|||
Total
Total
Age (yrs)
Gender
Under 21
21-25
Over 25
Total
Male
8
12
2
22
Female
12
13
3
28
Total
20
25
5
50
(c)
The row and column totals represent the total number of students in
each of the corresponding categories. For example, the row total of
22 indicates that 22 of the students in the class are male.
(d)
The sum of the row totals is 50, and the sum of the column totals is
50. The sums are equal because they both represent the total number
of students in the class.
(e)
Dividing each frequency reported in part (b) by the grand total of 50
students results in a contingency table that gives relative
frequencies.
Age (yrs)
Under 21
21-25
Over 25
Total
Male
0.16
0.24
0.04
0.44
Female
0.24
0.26
0.06
0.56
Total
0.40
0.50
0.10
1.00
Gender
(f)
The 0.16 in the upper left-hand cell indicates that 16% of the
students in the class are males and under 21. The 0.40 in the lower
left-hand cell indicates that 40% of the students in the class are
under age 21. A similar interpretation holds for the remaining
entries.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
89
2.115 Consider columns 1 and 2 of the energy-consumption data given in Exercise
2.56 part (b). Compute the class mark for each class presented in column 1.
Pair each class mark with its corresponding relative frequency found in
column 2. Construct a horizontal axis, where the units are in terms of
class marks and a vertical axis where the units are in terms of relative
frequencies. For each class mark on the horizontal axis, plot a point whose
height is equal to the relative frequency of the class. Then join the
points with connecting lines. The result is a relative-frequency polygon.
Residential Energy Consumption
Relative Frequency
0.25
0.20
0.15
0.10
0.05
0.00
35 45
55 65
75 85
95 105 115 125 135 145 155 165
BTU (Millions)
2.116 Consider columns 1 and 2 of the Cheetah speed data given in Exercise 2.61
part (b). Compute the midpoint for each class presented in column 1. Pair
each midpoint with its corresponding relative frequency found in column 2.
Construct a horizontal axis, where the units are in terms of midpoints and a
vertical axis where the units are in terms of relative frequencies. For
each midpoint on the horizontal axis, plot a point whose height is equal to
the relative frequency of the class. Then join the points with connecting
lines. The result is a relative-frequency polygon.
Cheetah Speeds
0.25
Relative Frequency
0.20
0.15
0.10
0.05
0.00
53
55
57
59
61
63
65
Speed (mph)
67
69
71
73
75
2.117 In single value grouping the horizontal axis would be labeled with the value
of each class.
2.118 (a)
Consider parts (a) and (b) of the energy-consumption data given in
Exercise 2.56. The classes are now reworked to present just the lower
class limit of each class. The frequencies are reworked to sum the
frequencies of all classes representing values less than the specified
lower class limit. These successive sums are the cumulative
frequencies. The relative frequencies are reworked to sum the
relative frequencies of all classes representing values less than the
specified class limits. These successive sums are the cumulative
relative frequencies. (Note: The cumulative relative frequencies can
also be found by dividing the each cumulative frequency by the total
number of data values.)
Copyright ยฉ 2020 Pearson Education, Inc.
90
Chapter 2
Less than
(b)
Cumulative
Frequency
Cumulative
Relative Frequency
40
0
0.00
50
1
0.02
60
8
0.16
70
15
0.30
80
18
0.36
90
24
0.48
100
34
0.68
110
39
0.78
120
43
0.86
130
45
0.90
140
48
0.96
150
48
0.96
160
50
1.00
Pair each class limit with its corresponding cumulative relative
frequency found in column 3. Construct a horizontal axis, where the
units are in terms of the class limits and a vertical axis where the
units are in terms of cumulative relative frequencies. For each class
limit on the horizontal axis, plot a point whose height is equal to
the cumulative relative frequency. Then join the points with
connecting lines. The result, presented in the following figure, is
an ogive using cumulative relative frequencies. (Note: A similar
procedure could be followed using cumulative frequencies.)
Cumulative Relative
Frequency
Residential Energy Consum ption
1.2
1
0.8
0.6
0.4
0.2
0
20 30 40 50 60 70 80 90 10 11 12 13 14 15 16
0 0 0 0 0 0 0
BTU (Millions)
2.119 (a)
Consider parts (a) and (b) of the Cheetah speed data given in Exercise
2.61. The classes are now reworked to present just the lower cutpoint
of each class. The frequencies are reworked to sum the frequencies of
all classes representing values less than the specified lower
cutpoint. These successive sums are the cumulative frequencies. The
relative frequencies are reworked to sum the relative frequencies of
all classes representing values less than the specified cutpoints.
These successive sums are the cumulative relative frequencies. (Note:
The cumulative relative frequencies can also be found by dividing the
each cumulative frequency by the total number of data values.)
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.3
(b)
Less than
Cumulative
Frequency
52
54
56
58
60
62
64
66
68
70
72
74
76
0
2
7
13
21
28
31
33
34
34
34
34
35
91
Cumulative
Relative Frequency
0.000
0.057
0.200
0.371
0.600
0.800
0.886
0.943
0.971
0.971
0.971
0.971
1.000
Pair each cutpoint with its corresponding cumulative relative
frequency found in column 3. Construct a horizontal axis, where the
units are in terms of the cutpoints and a vertical axis where the
units are in terms of cumulative relative frequencies. For each
cutpoint on the horizontal axis, plot a point whose height is equal to
the cumulative relative frequency. Then join the points with
connecting lines. The result, presented in the following figure, is
an ogive using cumulative relative frequencies. (Note: A similar
procedure could be followed using cumulative frequencies.)
Clocking the Cheetah
Cumulative Relative Frequency
1.0
0.8
0.6
0.4
0.2
0.0
52
54
56
58
60
62
64
66
Speed (mph)
68
70
72
74
76
2.120 (a) After rounding each observation to the nearest year, the stem-and-leaf
diagram for the rounded ages is
5|
6|
7|
8|
9|
334469
6
256689
234678
8
(b) After truncating each weight by dropping the decimal part, the stemand-leaf diagram for the rounded weights is
5|
6|
7|
8|
9|
223468
5
245688
123678
7
Copyright ยฉ 2020 Pearson Education, Inc.
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Chapter 2
(c) Although there are minor differences between the two diagrams, the
overall impression of the distribution of weights is the same for both
diagrams.
2.121 Minitab used truncation. Note that there was a data point of 5.8 in the
sample. It would have been plotted with a stem of 0 and a leaf of 6 if it
had been rounded. Instead Minitab plotted the observation with a stem of 0
and a leaf of 5.
Section 2.4
2.122 The distribution of a data set is a table, graph, or formula that provides
the values of the observations and how often they occur.
2.123 Sample data are the values of a variable for a sample of the population.
2.124 Population data are the values of a variable for the entire population.
2.125 A sample distribution is the distribution of sample data.
2.126 A population distribution is the distribution of population data.
2.127 A distribution of a variable is the same as a population distribution.
2.128 A smooth curve makes it a little easier to see the shape of a distribution
and to concentrate on the overall pattern without being distracted by minor
differences in shape.
2.129 A large simple random sample from a bell-shaped distribution would be
expected to have roughly a bell-shaped distribution since more sample values
should be obtained, on average, from the middle of the distribution.
2.130 (a) Yes. We would expect both simple random samples to have roughly a
reverse J-shaped distribution.
(b) Yes. We would expect some variation in shape between the two sample
distributions since it is unlikely that the two samples would produce
exactly the same frequency table. It should be noted, however, that as
the sample size is increased, the difference in shape for the two
samples should become less noticeable.
2.131 Three distribution shapes that are symmetric are bell-shaped, triangular,
and Uniform (or rectangular), shown in that order below. It should be noted
that there are others as well.
Bell-shaped
Triangular
2.132 (a) The shape of the distribution is unimodal.
(b) The shape of the distribution is symmetric.
2.133 (a) The shape of the distribution is unimodal.
(b) The shape of the distribution is symmetric.
2.134 (a) The shape of the distribution is unimodal.
(b) The shape of the distribution is not symmetric.
(c) The shape of the distribution is right-skewed.
Copyright ยฉ 2020 Pearson Education, Inc.
Uniform (or rectangular)
Section 2.4
93
2.135 (a) The shape of the distribution is unimodal.
(b) The shape of the distribution is not symmetric.
(c) The shape of the distribution is left-skewed.
2.136 (a) The shape of the distribution is unimodal.
(b) The shape of the distribution is not symmetric.
(c) The shape of the distribution is right-skewed.
2.137 (a) The shape of the distribution is unimodal.
(b) The shape of the distribution is not symmetric.
(c) The shape of the distribution is left-skewed.
2.138 (a) The shape of the distribution is bimodal.
(b) The shape of the distribution is symmetric.
2.139 (a) The shape of the distribution is multimodal.
(b) The shape of the distribution is not symmetric.
2.140 The overall shape of the distribution of the number of children of U.S.
presidents is right skewed.
2.141 Except for the one data value between 74 and 76, this distribution is close
to bell-shaped. That one value makes the distribution slightly right
skewed.
2.142 The distribution of weights of the male Ethiopian born school children is
roughly symmetric.
2.143 The distribution of depths of the burrows is left skewed.
2.144 The distribution of heights of the Baltimore Ravens is roughly symmetric.
2.145 The distribution of PCB concentration is symmetric.
2.146 The distribution of adjusted gross incomes is right skewed.
2.147 The distribution of cholesterol levels appears to be slightly left skewed.
2.148 The distribution of hemoglobin levels for patients with sickle cell disease
is roughly symmetric.
2.149 The distribution of length of stay is right skewed.
2.150 (a) The frequency distribution for this data is shown in the following
table.
Time Between Eruptions (minutes) Frequency
60 โ under 70
3
70 โ under 80
1
80 โ under 90
6
90 โ under 100
12
100 โ under 110
9
110 โ under 120
1
Copyright ยฉ 2020 Pearson Education, Inc.
94
Chapter 2
The histogram for the distribution is shown below.
12
10
Frequency
8
6
4
2
0
60
70
80
90
1 00
110
1 20
TIME
(b) This distribution is unimodal.
(c) This distribution is not symmetric.
(d) This distribution is left skewed.
2.151 (a)
(b)
(c)
(d)
2.152
The distributions for Year 1 and Year 2 are both unimodal.
Both distributions are not symmetric.
Both distributions are right skewed.
The distribution for Year 1 has a longer right tail indicating more
variation than the distribution for Year 2. They are also not centered
in the same place.
(a) After entering the data from the WeissStats Resource Site, in
๏ผ
Minitab, select Graph
Histogram, select Simple and click OK. Double
click on PUPS to enter PUPS in the Graph variables box and click OK.
Our result is as follows. Results may vary depending on the type of
technology used and graph obtained.
The overall shape of the distribution is unimodal and symmetric.
Histogram of PUPS
18
16
14
Frequency
12
10
8
6
4
2
0
4
6
8
10
12
PUPS
2.153
(a) After entering the data from the WeissStats Resource Site, in
๏ผ
Minitab, select Graph
Histogram, select Simple and click OK. Double
click on LENGTH to enter LENGTH in the Graph variables box and click
OK. Our result is as follows. Results may vary depending on the type
of technology used and graph obtained.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.4
95
70
60
Frequency
50
40
30
20
10
0
75
1 50
225
300
375
450
LENGTH
The distribution of LENGTH is unimodal and not symmetric.
(b) The distribution is right skewed.
2.154
(a) In Exercise 2.108, we used Minitab to obtain a stem-and-leaf
diagram using 5 lines per stem. That diagram is shown below
Stem-and-leaf of PERCENT
Leaf Unit = 1.0
3
8 001
10 8 2222333
18 8 44444555
(8) 8 66777777
25 8 888889999
16 9 00000000011111
2
9 22
N
= 51
The overall shape of this distribution is unimodal and not symmetric.
(b) The distribution is left skewed.
2.155
(a) In Exercise 2.109, we used Minitab to obtain a stem-and-leaf
diagram using 2 lines per stem.
That diagram is shown below.
Stem-and-leaf of PERCENT
Leaf Unit = 1.0
2
14
(20)
17
6
1
1
1
2
2
3
3
4
4
N
= 51
79
011223334444
55555566666777778999
00011122234
56668
9
The overall shape of this distribution is unimodal and roughly
symmetric.
2.156
๏ผ
(a) After entering the data in Minitab, select Graph
Dotplot, select
Simple in the One Y row, and click OK. Double click on TEMP to enter
TEMP in the Graph variables box and click OK. Our result is as
follows. Results may vary depending on the type of technology used and
graph obtained.
Copyright ยฉ 2020 Pearson Education, Inc.
96
Chapter 2
Dotplot of TEMP
96.8
97.2
97.6
98.0
TEMP
98.4
98.8
99.2
The overall distribution of temperatures is roughly unimodal and
roughly symmetric.
(a) After entering the data from the WeissStats Resource Site, in
2.157
๏ผ
Minitab, select Graph
Histogram, select Simple and click OK. Double
click on LENGTH to enter LENGTH in the Graph variables box and click
OK. Our result is as follows. Results may vary depending on the type
of technology used and graph obtained.
Histogram of LENGTH
30
25
Frequency
20
15
10
5
0
16
17
18
19
LENGTH
20
21
The distribution of LENGTH is approximately unimodal and symmetric.
2.158 Class Project.
class.
The precise answers to this exercise will vary from class to
2.159 The precise answers to this exercise will vary from class to class or
individual to individual. Thus your results will likely differ from our
results shown below.
(a)
We obtained 50 random digits from a table of random numbers.
digits were
The
4 5 4 6 8 9 9 7 7 2 2 2 9 3 0 3 4 0 0 8 8 4 4 5 3
9 2 4 8 9 6 3 0 1 1 0 9 2 8 1 3 9 2 5 8 1 8 9 2 2
(b)
Since each digit is equally likely in the random number table, we
expect that the distribution would look roughly uniform.
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.4
(c)
97
Using single value classes, the frequency distribution is given by the
following table. The relative frequency histogram is shown below.
Value
Frequency
Relative-Frequency
0
5
.10
1
4
.08
2
8
.16
3
5
.10
4
6
.12
5
3
.06
6
2
.04
7
2
.04
8
7
.14
9
8
.16
0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.00
0
1
2
3
4
5
6
7
8
9
We did not expect to see this much variation.
(d)
We would have expected a histogram that was a little more โevenโ, more
like a uniform distribution, but the relatively small sample size can
result in considerable variation from what is expected.
(e)
We should be able to get a more uniformly distributed set of data if
we choose a larger set of data.
(f)
Class project.
2.160 (a-c) Your results will differ from the ones below which were obtained using
Excel. Enter a name for the data in cell A1, say RANDNO. Click on
cell A2 and enter =RANDBETWEEN(0,9). Then copy this cell into cells
A3 to A51. There are two ways to produce a histogram of the resulting
data in Excel. The easier way is to highlight A1-A51 with the mouse,
click on the toolbar, select Graphs and Plots, then choose Histogram
in the Function type box. Now click on RANDNO in the Names and
Columns box and drag the name into the Quantitative Variables box.
Then click OK. A graph and a summary table will be produced. To get
five more samples, simply go back to the spreadsheet and press the F9
key. This will generate an entire new sample in Column A and you can
repeat the procedure. The only disadvantage of this method is that
the graphs produced use white lines on a black background.
The second method is a bit more cumbersome and does not provide a
summary chart, but yields graphs that are better for reproduction and
Copyright ยฉ 2020 Pearson Education, Inc.
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Chapter 2
that can be edited. Generate the data in the same way as was done
above. In cells B1 to B10 enter the integers 0 to 9. These cells are
called the BIN. Now click on Tools, Data Analysis, Histogram.
(If Data Analysis is not in the Tools menu, you will have to add it
from the original CD.) Click on the Input box and highlight cells A2A51 with the mouse, then click on the Bin box and highlight cells B1B10. Finally click on the Output box and enter C1. This will give
you a frequency table in columns C and D. Now enter the integers 0 to
9 as text in cells E2 to E11 by entering each digit preceded by a
single quote mark, i.e., โ0, โ1, etc. In cell F2, enter =D2, and copy
this cell into F3 through F11. Now highlight the data in columns E
and F with the mouse and click the chart icon, pick the Column graph
type, pick the first sub-type, click on the Next button twice, enter
any titles desired, remove the legend, and then click on the Next
button and then the Finish button. The graph will appear on the
spreadsheet as a bar chart with spaces between the bars. Use the
mouse to point to any one of the bars and click with the right mouse
button. Choose Format Data Series. Click on the Options tab and
change the Gap Width to zero, and click OK. Repeat this sequence to
produce additional histograms, but use different cells.
[If you would like to avoid repeating most of the above steps, click
near the border of the graph and copy the graph to the Clipboard, then
go to Microsoft Word or other word processor, and click on Edit on the
Toolbar and Paste Special. Highlight Microsoft Excel Chart Object,
and click OK. The graphs can be resized in the word processor if
necessary. Now go back to Excel and hit the F9 key. This will
produce a completely new set of random numbers. Click on Tools, Data
Analysis, Histogram, leave all the boxes as they are and click OK.
Then click OK to overwrite existing data. A new table will be created
and the existing histogram will be updated automatically. We used
this process for the following histograms.]
(d) These shapes are about what we expected.
(e) The relative frequency histograms for six samples of digits of size
๏ผ
๏ผ
Random Data
1000 were obtained using Minitab. Choose Calc
Integer…, type 1000 in the Generate rows of data test box, click in
the Store in column(s) text box and type C1 C2 C3 C4 C5 C6, click in
the Minimum value text box and type 0, type in the Maximum value text
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.4
๏ผ
99
box and type 9 and click OK. Then choose Graph
Histogram, select the
Simple version and click OK, enter C1 C2 C3 C4 C5 C6 in the Graph
variables text box, C2 in the Graph 2 text box un x, and so on for C3
through C6. Click on the Multiple Graphs button. Click on the On
separate graphs button, and check the boxes for Same Y and Same X,
including same bins. Click OK and click OK. The following graphs
resulted.
100
100
80
80
60
60
40
40
20
20
0
0
2
4
Histogram of C3
6
0
8
C1
120
100
100
80
80
60
40
20
20
0
0
2
4
2
4 of C4
Histogram
6
8
4
6
8
6
8
C2
60
40
0
0
120
Frequency
Frequency
Histogram of C2
120
Frequency
Frequency
Histogram of C1
120
6
8
0
2
C4
C3
Histogram of C6
120
100
100
80
80
Frequency
Frequency
Histogram of C5
120
60
60
40
40
20
20
0
0
2
4
6
8
0
0
2
C5
4
C6
The histograms for samples of size 1000 are much closer to the rectangular
distribution we expected than are the ones for samples of size 50.
Copyright ยฉ 2020 Pearson Education, Inc.
Review Problems
2.161 (a)
100
Your result will differ from, but be similar to, the one below which
๏ผ
๏ผ
Random Data
Normal…,
was obtained using Minitab. Choose Calc
type 3000 in the Generate rows of data text box, click in the Store in
column(s) text box and type C1, and click OK.
(b)
๏ผ
Then choose Graph
Histogram, choose the Simple version, click OK,
enter C1 in the Graph variables text box, click on the Scale button
and then on the Y-Scale Type tab. Check the Percent box and click OK
twice.
Histogram of C1
8
7
Percent
6
5
4
3
2
1
0
(c)
-3
-2
-1
0
C1
1
2
3
The histogram in part (b) has the bell, unimodal and symmetric
distribution. The sample of 3000 is representative of the population
from which the sample was taken.
Section 2.5
2.162 Graphs are sometimes constructed in ways that cause them to be misleading.
2.163 (a) A truncated graph is one for which the vertical axis starts at a value
other than its natural starting point, usually zero.
(b) A legitimate motivation for truncating the axis of a graph is to place
the emphasis on the ups and downs of the distribution rather than on
the actual height of the graph.
(c) To truncate a graph and avoid the possibility of misinterpretation, one
should start the axis at zero and put slashes in the axis to indicate
that part of the axis is missing.
2.164 Answers will vary.
2.165 (a)
A large lower portion of the graph is eliminated. When this is done,
differences between district and national averages appear greater than
in the original figure.
(b)
Even more of the graph is eliminated. Differences between district
and national averages appear even greater than in part (a).
(c)
The truncated graphs give the misleading impression that, in 2013, the
district average is much greater relative to the national average than
it actually is.
2.166 (a)
A break is shown in the first bar on the left to warn the reader that
part of the first bar itself has been removed.
(b)
It was necessary to construct the graph with a broken bar to let the
reader know that the first bar is actually much taller than it
appears. If the true height of the first bar were presented, but
without the break, the height would span most of an entire page. This
Copyright ยฉ 2020 Pearson Education, Inc.
Section 2.5
101
would have used up, perhaps, more room than that desired by the person
reporting the graph.
(c)
This bar chart is potentially misleading if the reader does not pay
attention to the true magnitude of the first bar relative to the other
three bars. This is precisely the reason for the break in the first
bar, however. It is meant to alert the reader that special treatment
is to be applied to the first bar. It is actually much taller than it
appears. Supplying the numbers for each bar of the graph makes it
clear that there was no intention to mislead the reader. This was
necessary also because there is no scale on the vertical axis.
2.167 (a)
The problem with the bar chart is that it is truncated. That is, the
vertical axis, which should start at $0 (in trillions), starts with
$7.6 (in trillions) instead. The part of the graph from $0 (in
trillions) to $7.6 (in trillions) has been cut off. This truncation
causes the bars to be out of correct proportion and hence creates the
misleading impression that the money supply is changing more than it
actually is.
(b)
A version of the bar chart with an untruncated and unmodified vertical
axis is presented in Figure (a). Notice that the vertical axis starts
at $0.00 (in trillions). Increments are in trillion dollars. In
contrast to the original bar chart, this one illustrates that the
changes in money supply from week to week are not that different.
However, the “ups” and “downs” are not as easy to spot as in the
original, truncated bar chart.
(c)
A version of the bar chart in which the vertical axis is modified in
an acceptable manner is presented in Figure (b). Notice that the
special symbol “//” is used near the base of the vertical axis to
signify that the vertical axis has been modified. Thus, with this
version of the bar chart, not only are the “ups” and “downs” easy to
spot but the reader is also aptly warned by the slashes that part of
the vertical axis between $0.00 (in trillions) and $7.6 (in trillions)
has been removed.
Figure (a)
Figure (b)
Chart of Money Supply
8
7
Money Supply
6
5
4
3
2
1
0
8/4
8/11
8/18
8/25
9/1
9/8
9/15
9/22
9/29
10/6
10/13 10/20 10/27
Date
2.168 (a)
The problem with the bar chart is that it is truncated. That is, the
vertical axis, which should start at 0 for both the Total Fatalities
and for the Licensed Drivers starts with 10,000 and 145 respectively.
This truncation causes the time-plot to be out of proportion and hence
creates the misleading impression that the number of fatalities is
changing more than it actually is.
Copyright ยฉ 2020 Pearson Education, Inc.
102
Chapter 2
(b)
The truncation was most likely used to clearly display the yearly
changes in total drunk driving fatalities. Without the truncation, if
the scale continued to zero, it might be hard to notice the changes
from year to year.
(c)
An acceptable manner to modify the graph would be to use the special
symbol “//” near the base of the vertical axis to signify that the
vertical axis has been modified. Thus, with this modified version of
the bar chart, not only are the changes in total drunk driving
fatalities easy to spot but the reader is also aptly warned by the
slashes that part of the vertical axis between 0 and 10,000 Total
Fatalities has been removed.
2.169 (b) Without the vertical scale, it would appear that the happiness score
dropped about 25% between the ages of 15 and 20.
(c) The actual drop was from a happiness score of about 5.5 to a happiness
score of 5.25 between the ages of 15 and 20. This is a drop of 0.25/5.5
= 0.045 or about 4.5%.
(d) Without the vertical scale, it would look like the happiness scale has
dropped about 50% from 5.5 to 5.0 where actually it really is closer to
a 20% drop.
The peak at 74 looks like a bigger significant increase
without the vertical scale as well. Because the scale on the horizontal
axis is not evenly spaced, the time between events is misleading.
(e) The graph could be made less potentially misleading by either making
the vertical scale range from zero to 6.0 and by making the scale on
the vertical axis the same width.
2.170 A correct way in which the developer can illustrate the fact that twice as
many homes will be built in the area this year as last year is as follows:
Last Year
2.171 (a)
This Year
The brochure shows a “new” ball with twice the radius of the “old”
ball. The intent is to give the impression that the “new” ball lasts
roughly twice as long as the “old” ball. However, if the “new” ball
has twice the radius of the “old” ball, the “new” ball will have eight
times the volume of the “old” ball (since the volume of a sphere is
proportional to the cube of its radius, or the radius 23 = 8). Thus,
the scaling is improper because it gives the impression that the “new”
ball lasts eight times as long as the “old” ball rather than merely
two times as long.
Old Ball
New Ball
Copyright ยฉ 2020 Pearson Education, Inc.
Review Problems
103
(b) One possible way for the manufacturer to illustrate the fact that the
“new” ball lasts twice as long as the “old” ball is to present pictures
of two balls, side by side, each of the same magnitude as the picture
of the “old” ball and to label this set of two balls “new ball”. This
will illustrate the point that a purchaser will be getting twice as
much for the money.
Old Ball
New Ball
Review Problems For Chapter 2
1.
2.
(a)
A variable is a characteristic that varies from one person or thing to
another.
(b)
Variables are quantitative or qualitative.
(c)
Quantitative variables can be discrete or continuous.
(d)
Data are values of a variable.
(e)
The data type is determined by the type of variable being observed.
3.
A frequency distribution of qualitative data is a table that lists the
distinct values of data and their frequencies. It is useful to organize the
data and make it easier to understand. A relative-frequency distribution of
qualitative data is a table that lists the distinct values of data and a
ratio of the class frequency to the total number of observations, which is
called the relative frequency.
4.
For both quantitative and qualitative data, the frequency and relativefrequency distributions list the values of the distinct classes and their
frequencies and relative frequencies. For single value grouping of
quantitative data, it is the same as the distinct classes for qualitative
data. For class limit and cutpoint grouping in quantitative data, we create
groups that form distinct classes similar to qualitative data.
5.
The two main types of graphical displays for qualitative data are the bar
chart and the pie chart.
6.
The bars do not abut in a bar chart because there is not any continuity
between the categories. Also, this differentiates them from histograms.
7.
Answers will vary. One advantage of pie charts is that it shows the
proportion of each class to the total. One advantage of bar charts is that
it emphasizes each individual class in relation to each other. One
disadvantage of pie charts is that if there are too many classes, the chart
becomes confusing. Also, if a class is really small relative to the total,
it is hard to see the class in a pie chart.
8.
Single value grouping is appropriate when the data are discrete with
relatively few distinct observations.
Copyright ยฉ 2020 Pearson Education, Inc.
104
Chapter 2
9.
(a)
The second class would have lower and upper limits of 9 and 14. The
class mark of this class would be the average of these limits and
equal 11.5.
(b)
The third class would have lower and upper limits of 15 and 20.
(c)
The fourth class would have lower and upper limits of 21 and 26.
class would contain an observation of 23.
(a)
If the width of the class is 5, then the class limits will be four
whole numbers apart. Also, the average of the two class limits is the
class mark of 8. Therefore, the upper and lower class limits must be
two whole numbers above and below the number 8. The lower and upper
class limits of the first class are 6 and 10.
(b)
The second class will have lower and upper class limits of 11 and 15.
Therefore, the class mark will be the average of these two limits and
equal 13.
(c)
The third class will have lower and upper class limits of 16 and 20.
(d)
The fourth class has limits of 21 and 25, the fifth class has limits
of 26 and 30. Therefore, the fifth class would contain an observation
of 28.
(a)
The common class width is the distance between consecutive cutpoints,
which is 15 – 5 = 10.
(b)
The midpoint of the second class is halfway between the cutpoints 15
and 25, and is therefore 20.
(c)
The sequence of the lower cutpoints is 5, 15, 25, 35, 45, …
Therefore, the lower and upper cutpoints of the third class are 25 and
35.
(d)
Since the third class has lower and upper cutpoints of 25 and 35, an
observation of 32.4 would belong to this class.
(a)
The midpoint is halfway between the cutpoints.
is 8, 10 is halfway between 6 and 14.
(b)
The class width is also the distance between consecutive midpoints.
Therefore, the second midpoint is at 10 + 8 = 18.
(c)
The sequence of cutpoints is 6, 14, 22, 30, 38, … Therefore the
lower and upper cutpoints of the third class are 22 and 30.
(d)
An observation of 22 would go into the third class since that class
contains data greater than or equal to 22 and strictly less than 30.
(a)
If lower class limits are used to label the horizontal axis, the bars
are placed between the lower class limit of one class and the lower
class limit of the next class.
(b)
If lower cutpoints are used to label the horizontal axis, the bars are
placed between the lower cutpoint of one class and the lower cutpoint
of the next class.
(c)
If class marks are used to label the horizontal axis, the bars are
placed directly above and centered over the class mark for that class.
(d)
If midpoints are used to label the horizontal axis, the bars are
placed directly above and centered over the midpoint for that class.
10.
11.
12.
13.
Copyright ยฉ 2020 Pearson Education, Inc.
This
Since the class width
Review Problems
14.(a)Bell-shaped
(b)Triangular
(c) Reverse J shape
105
(d)Uniform
15.
Answers will vary but here is one possibility.
16.
(a)
The distribution of the large simple random sample will reflect the
distribution of the population, so it would be left-skewed as well.
(b)
No. The randomness in the samples will almost certainly produce
different sets of observations resulting in shapes that are not
identical.
(c)
Yes. We would expect both of the simple random samples to reflect the
shape of the population and be left-skewed.
(a)
The first column ranks the hydroelectric plants.
quantitative, discrete data.
(b)
The fourth column provides measurements of capacity.
consists of quantitative, continuous data.
Thus, it
(c)
The third column provides nonnumerical information.
of qualitative data.
Thus, it consists
(a)
A single value frequency histogram for the prices of DVD players would
be identical to the dotplot in example 2.16 because the classes would
be 197 through 224 and the height for each bar in the frequency
histogram would reflect the number of dots above each observation in
the dotplot.
(b)
No. The frequency histogram with cutpoint or class limit grouping
would combine some of the single values together which would change
the frequencies and the heights of the bars corresponding to those
classes.
(a)
The first class to construct is 40-44. Since all classes are to be of
equal width, and the second class begins with 45, we know that the
width of all classes is 45 – 40 = 5. All of the classes are presented
in column 1 of the grouped-data table in the figure below. The last
class to construct does not go beyond 65-69, since the largest single
data value is 69. The sum of the relative frequency column is 0.999
due to rounding.
17.
18.
19.
Copyright ยฉ 2020 Pearson Education, Inc.
Thus, it consists of
106
Chapter 2
Age at
Inauguration
Frequency
Relative
Frequency
Class
Mark
40-44
2
0.045
42
45-49
7
0.159
47
50-54
13
0.295
52
55-59
12
0.273
57
60-64
7
0.159
62
65-69
3
0.068
67
44
0.999
(b)
By averaging the lower and upper limits for each class, we arrive at
the class mark for each class. The class marks for all classes are
presented in column 4.
(c)
Having established the classes, we tally the ages into their
respective classes. These results are presented in column 2, which
lists the frequencies. Dividing each frequency by the total number of
observations, which is 44, results in each class’s relative frequency.
The relative frequencies for all classes are presented in column 3.
(d)
The frequency histogram presented below is constructed using the
frequency distribution presented above; i.e., columns 1 and 2. Notice
that the lower cutpoints of column 1 are used to label the horizontal
axis of the frequency histogram. Suitable candidates for verticalaxis units in the frequency histogram are the even integers within the
range 2 through 14, since these are representative of the magnitude
and spread of the frequencies presented in column 2. The height of
each bar in the frequency histogram matches the respective frequency
in column 2.
Histogram of AGE
14
12
Frequency
10
8
6
4
2
0
40
45
50
55
AGE
60
65
70
(e) The shape of the inauguration ages is unimodal and symmetric.
20.
The horizontal axis of this dotplot displays a range of possible ages for
the 44 Presidents of the United States. To complete the dotplot, we go
through the data set and record each age by placing a dot over the
appropriate value on the horizontal axis.
Copyright ยฉ 2020 Pearson Education, Inc.
Review Problems
107
Age at Inauguration for first 44 Presidents
44
21.
(a)
52
56
AGE
60
64
68
Using one line per stem in constructing the ordered stem-and-leaf
diagram means vertically listing the numbers comprising the stems
once. The leaves are then placed with their respective stems in
order. The ordered stem-and-leaf diagram using one line per stem is
presented in the following figure.
4|
5|
6|
(b)
48
236677899
0011112244444555566677778
0111244589
Using two lines per stem in constructing the ordered stem-and-leaf
diagram means vertically listing the numbers comprising the stems
twice. In turn, if the leaf is one of the digits 0 through 4, it is
ordered and placed with the first of the two stem lines. If the leaf
is one of the digits 5 through 9, it is ordered and placed with the
second of the two stem lines. The ordered stem-and-leaf diagram using
two lines per stem is presented in the following figure.
4|
4|
5|
5|
6|
6|
23
6677899
0011112244444
555566677778
0111244
589
(c) The stem-and-leaf diagram in part (b) corresponds to the frequency
distribution of Problem 19.
22.
(a)
Using one line per stem in constructing the ordered stem-and-leaf
diagram means vertically listing the numbers comprising the stems
once. The leaves are then placed with their respective stems in
order. The ordered stem-and-leaf diagram using one line per stem is
presented in the following figure.
3|
4|
5|
6|
(b)
4778
0467
446677778
0
Using two lines per stem in constructing the ordered stem-and-leaf
diagram means vertically listing the numbers comprising the stems
twice. In turn, if the leaf is one of the digits 0 through 4, it is
ordered and placed with the first of the two stem lines. If the leaf
is one of the digits 5 through 9, it is ordered and placed with the
second of the two stem lines. The ordered stem-and-leaf diagram using
two lines per stem is presented in the following figure.
Copyright ยฉ 2020 Pearson Education, Inc.
108
Chapter 2
3|
3|
4|
4|
5|
5|
6|
23.
4
778
04
67
44
6677778
0
(c)
The second graph is the most useful. Typically, we like to have five
to fifteen classes and the second one provides a better idea of the
shape of the distribution.
(a)
The frequency and relative-frequency distribution presented below is
constructed using classes based on a single value. Since each data
value is one of the integers 0 through 6, inclusive, the classes will
be 0 through 6, inclusive. These are presented in column 1. Having
established the classes, we tally the number of busy tellers into
their respective classes. These results are presented in column 2,
which lists the frequencies. Dividing each frequency by the total
number of observations, which is 25, results in each class’s relative
frequency. The relative frequencies for all classes are presented in
column 3.
Number
Busy
0
1
2
3
4
5
6
(b)
Frequency
1
2
2
4
5
7
4
25
Relative
Frequency
0.04
0.08
0.08
0.16
0.20
0.28
0.16
1.00
The following relative-frequency histogram is constructed using the
relative-frequency distribution presented in part (a); i.e., columns 1
and 3. Column 1 demonstrates that the data are grouped using classes
based on a single value. These single values in column 1 are used to
label the horizontal axis of the relative-frequency histogram. We
notice that the relative frequencies presented in column 3 range in
size from 0.04 to 0.28 (4% to 28%). Thus, suitable candidates for
vertical axis units in the relative-frequency histogram are increments
of 0.05, starting with zero and ending at 0.30. The middle of each
histogram bar is placed directly over the single numerical value
represented by the class. Also, the height of each bar in the
relative-frequency histogram matches the respective relative frequency
in column 3.
Copyright ยฉ 2020 Pearson Education, Inc.
Review Problems
109
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
Tellers
4
5
5
6
6
(c) The distribution is unimodal.
(d) The distribution is left skewed.
(e)
Dotplot of TELLERS
0
1
2
3
TELLERS
4
(f) Since both the histogram and the dotplot are based on single value
grouping, they both convey exactly the same information.
24.
(a) The classes will begin with the class 60 โ under 65. The second class
will be 65 โ under 70. The classes will continue like this until the
last class, which will be 90 โ under 95, since the largest observation
is 93.1. The classes can be found in column 1 of the frequency
distribution in part (c).
(b) The midpoints of the classes are the averages of the lower and upper
cutpoint for each class. For example, the first midpoint will be 62.5,
the second midpoint will be 67.5. The midpoints will continue like
this until the last midpoint, which will be 92.5. The midpoints can be
found in column 4 of the frequency distribution in part (c).
(c) The first and second columns of the following table provide the
frequency distribution. The first and third columns of the following
table provide the relative-frequency distribution for percentages of
on-time arrivals for the airlines.
Copyright ยฉ 2020 Pearson Education, Inc.
110
Chapter 2
Percent On-Time
Frequency
Relative
Frequency
Midpoint
60 โ under 65
65 โ under 70
70 โ under 75
75 โ under 80
80 โ under 85
85 โ under 90
90 โ under 95
1
0.0625
62.5
5
0.3125
67.5
5
0.3125
72.5
3
0.1875
77.5
0
0.0000
82.5
1
0.0625
87.5
1
0.0625
92.5
16
1.0000
(d) The frequency histogram is constructed using columns 1 and 2 from the
frequency distribution from part (c). The cutpoints are used to label
the horizontal axis. The vertical axis is labeled with the frequencies
that range from 0 to 5.
5
Frequency
4
3
2
1
0
60
65
70
75
80
85
90
95
PERCENTAGE
(e) After rounding each observation to the nearest whole number, the stemand-leaf diagram with two lines per stem for the rounded percentages is
6|
6|
7|
7|
8|
8|
9|
2
669
0011334
678
8
3
(f) After obtaining the greatest integer in each observation, the stem-andleaf diagram with two lines per stem is
6| 1
6| 56999
7| 01233
7| 677
8|
8| 7
9| 3
(g) The stem-and-leaf diagram in part (f) corresponds to the frequency
distribution in part (d) because the observations werenโt rounded in
constructing the frequency distribution.
25.
(a) The dotplot of the ages of the oldest player on each major league
baseball team is
Copyright ยฉ 2020 Pearson Education, Inc.
Review Problems
111
Dotplot of AGES
34
36
38
40
AGES
42
44
46
(b) The overall shape of the distribution of ages is bimodal and roughly
symmetric.
26.
(a) The classes are the types of evidence and are presented in column 1.
The frequency distribution of the champions is presented in column 2.
Dividing each frequency by the total number of submissions, which is
3436, results in each class’s relative frequency. The relative
frequency distribution is presented in column 3.
Evidence
Frequency
Relative Frequency
Type
Firearm
289
0.084
Magazine
161
0.047
Live Cartridge
2727
0.794
Spent Cartridge
259
0.075
3436
1.000
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each evidence type. The result
is
Pie Chart of EVIDENCE TYPE
Spent cartridge casing
7.5%
Firearm
8.4%
Magazine
4.7%
Category
Firearm
Magazine
Live cartridge
Spent cartridge casing
Live cartridge
79.4%
(c) We use the bar chart to show the relative frequency with which each
EVIDENCE TYPE occurs. The result is
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Chapter 2
Evidence Type
80
Percent of FREQUENCY
70
60
50
40
30
20
10
0
Firearm
Magazine
Live cartridge
Spent cartridge casing
EVIDENCE
Percent within all data.
(d) The most common type of evidence type submitted to crime labs is a live
cartridge which was the type submitted 79.4% of the time.
27.
(a) The population consists of the states in the United States.
variable is the division.
The
(b) The frequency and relative frequency distribution for Region is shown
below.
Region
Frequency
Relative
Frequency
East North Central
5
0.10
East South Central
4
0.08
Middle Atlantic
3
0.06
Mountain
8
0.16
New England
6
0.12
Pacific
5
0.10
South Atlantic
8
0.16
West North Central
7
0.14
West South Central
4
0.08
50
1.000
(c) The pie chart for Region is shown below.
Pie Chart of DIVISION
W SC
8.0%
MA C
6.0%
C ategory
MTN
SA C
W NC
NED
ENC
PA C
ESC
W SC
MA C
MTN
16.0%
ESC
8.0%
SA C
16.0%
PA C
10.0%
ENC
10.0%
WNC
14.0%
NED
12.0%
Copyright ยฉ 2020 Pearson Education, Inc.
Review Problems
(d)
113
The bar chart for Region is shown below.
Chart of DIVISION
18
16
14
Percent
12
10
8
6
4
2
0
MTN
SAC
WNC
NED
ENC
PAC
DIVISION
ESC
WSC
MAC
Percent within all data.
(e)
28.
29.
30.
(a)
The distribution of Region seems to be almost uniformly distributed
between the categories. The smallest region is the Middle Atlantic at
a frequency of 3, the largest region is the Mountain and South
Atlantic at a frequency of 8.
The break in the third bar is to emphasize that the bar as shown is
not as tall as it should be.
(b)
The bar for the space available in coal mines is at a height of about
30 billion tonnes. To accurately represent the space available in
saline aquifers (10,000 billion tonnes), the third bar would have to
be over 300 times as high as the first bar and more than 10 times as
high as the second bar. If the first two bars were kept at the sizes
shown, there wouldnโt be enough room on the page for the third bar to
be shown at its correct height. If a reasonable height were chosen
for the third bar, the first bar wouldnโt be visible. The only
apparent solution is to present the third bar as a broken bar.
(a)
Covering up the numbers on the vertical axis totally obscures the
percentages.
(b)
Having followed the directions in part (a), we might conclude that the
percentage of women in the labor force for 2000 is about three and
one-half times that for 1960.
(c)
Not covering up the vertical axis, we find that the percentage of
women in the labor force for 2000 is about 1.8 times that for 1960.
(d)
The graph is potentially misleading because it is truncated.
that vertical axis units begin at 30 rather than at zero.
(e)
To make the graph less potentially misleading, we can start it at zero
instead of 30.
Notice
(a) Using Minitab, retrieve the data from the WeissStats Resource Site.
Column 2 contains the eye color and column 3 contains the hair color
๏ผ
๏ผ
for the students. From the tool bar, select Stat
Tables
Tally
Individual Variables, double-click on EYES and HAIR in the first box so
that both EYES and HAIR appear in the Variables box, put a check mark
next to Counts and Percents under Display, and click OK. The results
are
EYES Count Percent
Blue
215
36.32
Brown
220
37.16
Green
64
10.81
Hazel
93
15.71
N=
592
HAIR Count Percent
Black
108
18.24
Blonde
127
21.45
Brown
286
48.31
Red
71
11.99
N=
592
Copyright ยฉ 2020 Pearson Education, Inc.
114
Chapter 2
๏ผ
Using Minitab, select Graph
Pie Chart, check Chart counts of unique
values, double-click on EYES and HAIR in the first box so that EYES and
HAIR appear in the Categorical Variables box. Click Pie Options, check
decreasing volume, click OK. Click Multiple Graphs, check On the Same
Graphs, Click OK. Click Labels, click Slice Labels, check Category
Name, Percent, and Draw a line from label to slice, Click OK twice.
The results are
(b)
Pie Chart of EYES, HAIR
EYES
HAIR
Green
10.8%
Category
Brown
Blue
Hazel
Green
Black
Blonde
Red
Red
12.0%
Brown
37.2%
Hazel
15.7%
Black
18.2%
Brown
48.3%
Blonde
21.5%
Blue
36.3%
๏ผ
(c) Using Minitab, select Graph
Bar Chart, select Counts of unique
values, select Simple option, click OK. Double-click on EYES and HAIR
in the first box so that EYES and HAIR appear in the Categorical
Variables box.
Select Chart Options, check decreasing Y, check show Y
as a percent, click OK. Click OK twice.
The results are
Chart of HAIR
Chart of EYES
50
40
40
Percent
Percent
30
20
10
0
30
20
10
Brown
Blue
Hazel
Green
0
Brown
Black
Red
Percent within all data.
Percent within all data.
31.
Blonde
HAIR
EYES
(a) The population consists of the states of the U.S. and the variable
under consideration is the value of the exports of each state.
(b) Using Minitab, we enter the data from the WeissStats Resource Site,
๏ผ
choose Graph
Histogram, click on Simple and click OK. Then double
click on VALUE to enter it in the Graph variables box and click OK.
The result is
Copyright ยฉ 2020 Pearson Education, Inc.
Review Problems
115
Histogram of VALUE
25
Frequency
20
15
10
5
0
0
2000
4000
VALUE
6000
8000
๏ผ
Dotplot, click on Simple from the
(c) For the dotplot, we choose Graph
One Y row and click OK. Then double click on VALUE to enter it in the
Graph variables box and click OK. The result is
Dotplot of VALUE
0
1200
2400
3600
4800
VALUE
6000
7200
8400
๏ผ
Stem-and-Leaf, double
(d) For the stem-and-leaf plot, we choose Graph
click on VALUE to enter it in the Graph variables box and click OK.
The result is
Stem-and-leaf of VALUE
Leaf Unit = 100
23
(10)
17
11
7
7
5
2
1
1
1
1
1
1
1
1
1
0
0
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
N
= 50
00000000001112222344444
5677888899
012224
5679
69
034
6
2
(e) The overall shape of the distribution is unimodal and not symmetric.
(f) The distribution is right skewed.
Copyright ยฉ 2020 Pearson Education, Inc.
116
Chapter 2
32.
(a) The population consists of countries of the world, and the variable
under consideration is the expected life in years for people in those
countries.
(b) Using Minitab, we enter the data from the WeissStats Resource Site,
๏ผ
choose Graph
Histogram, click on Simple and click OK. Then double
click on YEARS to enter it in the Graph variables box and click OK.
The result is
Histogram of YEARS
30
25
Frequency
20
15
10
5
0
54
60
66
72
78
84
90
YEARS
๏ผ
Dotplot, click on Simple from the
(c) For the dotplot, we choose Graph
One Y row and click OK. Then double click on YEARS to enter it in the
Graph variables box and click OK. The result is
Dotplot of YEARS
48
54
60
66
72
78
84
90
YEARS
๏ผ
(d) For the stem-and-leaf plot, we choose Graph
Stem-and-Leaf, double
click on YEARS to enter it in the Graph variables box and click OK.
The result is
Stem-and-leaf of YEARS
Leaf Unit = 1.0
3
21
30
50
73
(51)
99
33
1
4
5
5
6
6
7
7
8
8
N
= 223
999
000112222223344444
556677899
00001112233333333444
55556666677788888999999
000111111111112222222223333333333344444444444444444
555555555555566666666666666667777777777788888888888889999999999999
00000000000011111111111122223444
9
(e) The overall shape of the distribution is unimodal and not symmetric.
(f) This distribution is classified as left skewed.
Copyright ยฉ 2020 Pearson Education, Inc.
Review Problems
(a) The population consists of cities in the U.S., and the variables under
consideration are their annual average maximum and minimum
temperatures.
(b) Using Minitab, we enter the data from the WeissStats Resource Site,
๏ผ
choose Graph
Histogram, click on Simple and click OK. Double click
on HIGH to enter it in the Graph variables box, and double click on LOW
to enter it in the Graph variables box. Now click on the Multiple
graphs button and click to Show Graph Variables on separate graphs and
also check both boxes under Same Scales for Graphs, and click OK twice.
The result is
Histogram of HIGH
Histogram of LOW
25
25
20
20
15
15
Frequency
Frequency
33.
117
10
5
0
10
5
36
48
60
HIGH
72
0
84
36
48
60
72
LOW
๏ผ
(c) For the dotplot, we choose Graph
Dotplot, click on Simple from the
Multiple Yโs row and click OK. Then double click on HIGH and then LOW
to enter them in the Graph variables box and click OK. The result is
Dotplot of HIGH, LOW
HIGH
LOW
32
40
48
56
Data
64
72
80
๏ผ
(d) For the stem-and-leaf diagram, we choose Graph
Stem-and-Leaf, double
click on HIGH and then on LOW to enter then in the Graph variables box
and click OK. The result is
Stem-and-leaf of HIGH N = 71
Leaf Unit = 1.0
3
4 789
6
5 444
21
5 555677777888999
(17) 6 00001122222333444
33
6 55556677779
22
7 00001122234
11
7 5577889
4
8 444
1
8 5
Copyright ยฉ 2020 Pearson Education, Inc.
84
118
Chapter 2
Stem-and-leaf of LOW
Leaf Unit = 1.0
1
7
19
(20)
32
19
11
4
3
2
2
3
3
4
4
5
5
6
6
7
N
= 71
9
001234
555556779999
00011111223333344444
5567777888899
11122223
5667889
1
9
04
(e) Both variables have distributions that are unimodal.
symmetric and LOW is not symmetric.
(f) LOW is slightly right skewed.
Using the FOCUS Database:
HIGH is close to
Chapter 2
We use the Menu commands in Minitab to complete parts (a)-(e). The data sets
in the Focus database and their names have already been stored in the file
FOCUS.MTW on the WeissStats Resource Site supplied with the text.
(a)
UWEC is a school that attracts good students. HSP reflects pre-college
experience and will tend to be left-skewed since fewer students with lower
high school percentile scores will have been admitted, but exceptions are
made for older students whose high school experience is no longer relevant.
GPA will probably show left skewness tendencies since many, but not all,
students with lower cumulative GPAs (below 2.0 on a 4-point scale) will
likely have been suspended and will not appear in the database, but there
are also upper limits on these scores, so the scores will tend to bunch up
nearer to the high end than to the low end. AGE will be right skewed
because there are few students below the typical 17-22 ages, but many above
that range. ENGLISH, MATH, and COMP will be closer to bell-shaped. The ACT
typically is taken only by high school students intending to go to college,
and the scores are designed to roughly follow a bell-shaped curve.
Individual colleges may, however, have a different profile that reflects
their admission policies.
(b)
Using Minitab with FocusSample, choose Graph
Histogram…, select the
Simple version, and Click OK. Then specify HSP GPA AGE ENGLISH MATH COMP in
the Graph variables text box and click on the button for Multiple Graphs.
Click on the button for On separate graphs and click OK twice. The results
are
๏ผ
Copyright ยฉ 2020 Pearson Education, Inc.
Using the Focus Database
Histogram of AGE
Histogram of ENGLISH
40
50
30
Frequency
Frequency
40
30
20
20
10
10
0
119
18
21
24
27
AGE
30
0
33
12
16
Histogram of GPA
20
24
ENGLISH
28
32
36
Histogram of HSP
30
25
25
20
Frequency
Frequency
20
15
15
10
10
5
5
0
1.80
2.25
2.70
GPA
3.15
3.60
0
4.05
30
45
Histogram of MATH
60
HSP
75
90
Histogram of COMP
25
30
25
20
Frequency
Frequency
20
15
10
15
10
5
0
5
15
18
21
24
MATH
27
30
33
0
18
20
22
24
COMP
26
28
30
32
The graphs compare quite well with the educated guesses for all six
variables.
๏ผ
Histogram…, select the
(d) Using Minitab with Focus, choose Graph
Simple version, and Click OK. Then specify HSP GPA AGE ENGLISH MATH COMP
in the Graph variables text box and click on the button for Multiple
Graphs. Click on the button for On separate graphs and click OK twice.
The results are
Copyright ยฉ 2020 Pearson Education, Inc.
120
Chapter 2
Histogram of HSP
Histogram of GPA
350
300
300
250
200
Frequency
Frequency
250
200
150
100
100
50
50
0
150
14
28
42
56
70
84
0
98
0.90
1.35
1.80
2.25
GPA
HSP
Histogram of AGE
2.70
3.15
3.60
4.05
28
32
36
Histogram of ENGLISH
1800
700
1600
600
1400
500
Frequency
Frequency
1200
1000
800
600
400
300
200
400
100
200
0
20
24
28
AGE
32
36
0
40
8
12
16
Histogram of MATH
20
24
ENGLISH
Histogram of COMP
700
900
800
600
700
600
Frequency
Frequency
500
400
300
500
400
300
200
200
100
0
100
15
18
21
24
MATH
27
30
33
36
0
15
18
21
24
COMP
27
30
33
We were correct on the first five variables: HSP and GPA are left skewed,
AGE is right skewed, ENGLISH and MATH are fairly symmetric. COMP is close
to symmetric, but is slightly right skewed. Comparing the graphs for the
sample with those for the entire population, we see similarities between
each pair of graphs, but the outline of the histogram for the entire
population is much smoother than that of the histogram for the sample.
(d)
๏ผ
Piechart, click on
Using Minitab and the FocusSample file, choose Graph
the Chart raw data button, specify SEX CLASS RESIDENCY TYPE in the Graph
variables text box and click on the Labels button. Now click on the tab for
Copyright ยฉ 2020 Pearson Education, Inc.
Using the Focus Database
121
Slice Labels, check all four boxes and click OK, click on the button for
Multiple Graphs and ensure that the button for On the same graph is checked,
and click OK twice.
Once the graphs are displayed, we right clicked on the legend that was shown
and selected Delete since we already had provided for each slice of the
graphs to be labeled.
Pie Chart of SEX, CLASS, RESIDENCY, TYPE
SEX
CLASS Freshman
25, 12.5%
Sophomore
59, 29.5%
M
82, 41.0%
Junior
53, 26.5%
F
118, 59.0%
Senior
63, 31.5%
RESIDENCY
Nonresident
46, 23.0%
Transfer TYPE
22, 11.0%
Readmit
2, 1.0%
Resident
154, 77.0%
New
176, 88.0%
From the graph of SEX, we see that about 59% of the students are females.
From the graph of CLASS, we see that the student sample is about 12.5%
Freshmen, 29.5% Sophomores, 26.5% Juniors, and 31.5% Seniors. From the
graph of RESIDENCY, we see that about 77% of the students are Wisconsin
residents and 23$ are nonresidents. From the graph of TYPE, we see that
88.0% of the students were admitted initially as new students, 11.0% were
admitted initially as transfer students, and 1.0% are readmits, that is,
students who were initially new or transfer students, left the university,
and were later readmitted.
(e)
Now repeat part (d) using the entire Focus file.
The results are
Pie Chart of SEX, CLASS, RESIDENCY, TYPE
SEX
CLASS Freshman
12.8%
Sophomore
29.8%
M
39.3%
Junior
24.1%
F
60.7%
Senior
33.3%
Other
0.0%
Resident
76.1%
RESIDENCY
Nonresident
23.9%
Transfer TYPE
11.1%
Readmit
0.4%
New
88.5%
From the graph of SEX, we see that about 61% of the students are females.
From the graph of CLASS, we see that the student population is about 13%
Freshmen, 30% Sophomores, 24% Juniors, and 33% Seniors. From the graph of
RESIDENCY, we see that about 76% of the students are Wisconsin residents and
24$ are nonresidents. From the graph of TYPE, we see that 85.5% of the
Copyright ยฉ 2020 Pearson Education, Inc.
122
Chapter 2
students were admitted initially as new students, 11.1% were admitted
initially as transfer students, and 0.4% are readmits, that is, students who
were initially new or transfer students, left the university, and were later
readmitted. We would expect that the two sets of graphs would be
approximately the same, but not identical since the sample contains only 200
students out of a population of 6738. This is, in fact, the case. The
percentages in each sample graph are very close to the percentages in the
corresponding population graph.
Case Study: Worldโs Richest People
(a)
The first column variable is Rank and is quantitative discrete. The
second column variable is Name and is qualitative. The third column
variable is age and it is quantitative continuous. The fourth column
variable is Citizenship and is qualitative. The fifth column variable
is Wealth and is quantitative discrete since money involves discrete
units, such as dollars and cents. Although, for all practical
purposes, Wealth might be considered quantitative continuous data.
(b)
The classes are the countries of citizenship and are presented in
column 1. The frequency distribution of the champions is presented in
column 2. Dividing each frequency by the total number of observations,
which is 25, results in each class’s relative frequency. The relative
frequency distribution is presented in column 3.
Country
(c)
Frequency
Relative Frequency
Canada
1
0.04
France
2
0.08
Germany
1
0.04
Hong Kong
2
0.08
India
1
0.04
Italy
1
0.04
Mexico
1
0.04
Spain
1
0.04
Sweden
1
0.04
United States
14
0.56
25
1.00
We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each team. The result is
Pie Chart of CITIZENSHIP
Canada
4.0%
France
8.0%
Germany
4.0%
Hong Kong
8.0%
Category
Canada
France
Germany
Hong Kong
India
Italy
Mexico
Spain
Sweden
United States
India
4.0%
United States
56.0%
Italy
4.0%
Mexico
4.0%
Spain
4.0%
Sweden
4.0%
Copyright ยฉ 2020 Pearson Education, Inc.
Case Study
123
The United States is the most frequent country of citizenship amongst
the worldโs richest. Otherwise, citizenship country seems to be
randomly distributed.
(d) We use the bar chart to show the relative frequency with which each
COUNTRY occurs. The result is
Chart of CITIZENSHIP
14
12
Count
10
8
6
4
2
0
d
na
Ca
a
Fr
ce
an
y
an
m
er
G
H
g
on
Ko
ng
a
di
In
ly
Ita
o
ic
ex
M
a
Sp
in
ed
Sw
en
U
d
te
ni
es
at
St
CITIZENSHIP
The United States is the most frequent country of citizenship amongst
the worldโs richest. Otherwise, citizenship country seems to be
randomly distributed.
(e) The first class to construct is “30 โ 39โ. All of these classes are
presented in column 1. The last class to construct is “90-99โ, since
the largest data value is 93. Having established the classes, we tally
the ages into their respective classes. These results are presented in
column 2, which lists the frequencies. Dividing each frequency by the
total number of observations, which is 25, results in the relative
frequencies for each class which are presented in column 3.
(f)
Age
Frequency
Relative
Frequency
30 โ 39
1
0.04
40 โ 49
2
0.08
50 โ 59
4
0.16
60 โ 69
6
0.24
70 โ 79
6
0.24
80 โ 89
4
0.16
90 โ 99
2
0.08
25
1.00
The frequency and relative-frequency histograms for age are
constructed using the frequency and relative-frequency distribution
presented in part (e); i.e. The lower class limits of column 1 are
used to label the horizontal axis of the histograms. The heights of
each bar in the frequency histogram in Figure (a) matches the
respective frequency in column 2. The heights of each bar in the
relative-frequency histogram in Figure (b) matches the respective
relative-frequencies in column 3.
Copyright ยฉ 2020 Pearson Education, Inc.
124
Chapter 2
Figure (a)
Figure (b)
Histogram of AGE
Histogram of AGE
25
6
20
4
Percent
Frequency
5
3
15
10
2
5
1
0
30
40
50
60
70
80
90
0
1 00
30
40
50
60
70
80
90
1 00
AGE
AGE
(g)
The shape of the distribution of age in part (e) is unimodal and is roughly
symmetric.
(h)
The stem-and-leaf diagram of age using one line per stem is
3| 9
4| 09
5| 5678
6| 345589
7| 133779
8| 2458
9| 03
The stem-and-leaf diagram of age using two lines per stem is
3|
4|
4|
5|
5|
6|
6|
7|
7|
8|
8|
9|
9
0
9
5678
34
5589
133
779
24
58
03
The stem-and-leaf diagram of age using one line per stem corresponds to the
histogram in part (f).
(i)
The dotplot for age is
Dotplot of AGE
40
48
56
64
72
80
88
AGE
Copyright ยฉ 2020 Pearson Education, Inc.
Case Study
125
(j) The first class to construct is “20 โ under 25โ. All of these classes
are presented in column 1. The last class to construct is “70 โ under
75โ, since the largest data value is 73. Having established the
classes, we tally the into their respective classes. These results
are presented in column 2, which lists the frequencies. Dividing each
frequency by the total number of observations, which is 25, results in
the relative frequencies for each class which are presented in column
3.
Frequency
Relative
Frequency
20 โ under 25
6
0.24
25 โ under 30
10
0.40
30 โ under 35
4
0.16
35 โ under 40
0
0.00
40 โ under 45
1
0.04
45 โ under 50
0
0.00
50 โ under 55
1
0.04
55 โ under 60
1
0.04
60 โ under 65
0
0.00
65 โ under 70
1
0.04
70 โ under 75
1
0.04
25
1.00
Wealth
($ billions)
The frequency and relative-frequency histograms for wealth are
constructed using the frequency and relative-frequency distribution
presented in part (j); i.e. The lower cutpoints of column 1 are used
to label the horizontal axis of the histograms. The heights of each
bar in the frequency histogram in Figure (a) matches the respective
frequency in column 2. The heights of each bar in the relativefrequency histogram in Figure (b) matches the respective relativefrequencies in column 3.
(k)
Figure (a)
Figure (b)
Histogram of WEALTH
Histogram of WEALTH
40
10
30
6
Percent
Frequency
8
20
4
10
2
0
20
30
40
50
60
70
0
20
30
WEALTH
(l)
40
50
60
70
WEALTH
The shape of the distribution of wealth in part (e) is unimodal and is
not symmetric. The distribution of wealth is right skewed.
Copyright ยฉ 2020 Pearson Education, Inc.
126
Chapter 2
(m)
Truncating wealth to a whole number, the stem-and-leaf diagram using two
lines per stem is
(n)
2| 000123
2| 5666667889
3| 0144
3|
4| 3
4|
5| 3
5| 7
6|
6| 7
7| 3
Rounding wealth to a whole number, the dotplot is
Dotplot of Wealth
21
28
35
42
49
56
63
70
Wealth ($ billions)
Copyright ยฉ 2020 Pearson Education, Inc.
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