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Chapter 3
Data Description
Section 3-2
Measures of Central
Tendency
Chapter 3
Data Description
Section 3-2
Exercise #3
61, 11, 1, 3, 2, 30, 18, 3, 7
The data above are the number of
burglaries reported for a specific year for
nine western Pennsylvania universities.
Find (a) the mean.
Find (b) the median.
Find (c) the mode.
Find (d) the midrange.
61, 11, 1, 3, 2, 30, 18, 3, 7
The data above are the number of
burglaries reported for a specific year for
nine western Pennsylvania universities.
Find (a) the mean.
x

X=
n
=
136
9
= 15.1
61, 11, 1, 3, 2, 30, 18, 3, 7
The data above are the number of
burglaries reported for a specific year for
nine western Pennsylvania universities.
Find (b) the median.
MD: 1, 2, 3, 3, 7, 11, 18, 30, 61
MD: 7
61, 11, 1, 3, 2, 30, 18, 3, 7
The data above are the number of
burglaries reported for a specific year for
nine western Pennsylvania universities.
Find (c) the mode.
Mode = 3
61, 11, 1, 3, 2, 30, 18, 3, 7
The data above are the number of
burglaries reported for a specific year for
nine western Pennsylvania universities.
Find (d) the midrange.
MR =
1 + 61
2
= 31
61, 11, 1, 3, 2, 30, 18, 3, 7
Which measure of average might be the
best in this case?
Explain your answer.
x

X=
n
=
136
= 15.1
9
MD: 1, 2, 3, 3, 7, 11, 18, 30, 61
MD = 7
Mode = 3
MR = 31
61, 11, 1, 3, 2, 30, 18, 3, 7
Which measure of average might be the
best in this case?
Explain your answer.
The median is probably the best
measure of average because 61
is an extremely large data value
and makes the mean
artificially high.
Chapter 3
Data Description
Section 3-2
Exercise #5
A researcher claims that each year, there is an average of
300 victims of identity theft in major cities. Twelve were
randomly selected, and the number of victims of identity
theft in each city is shown.
574
117
229
239
663
465
372
136
1,202 88
189 75
Find (a) the mean, (b) the median,
(c) the mode, and (d) the midrange.
A researcher claims that each year, there is an average of
300 victims of identity theft in major cities. Twelve were
randomly selected, and the number of victims of identity
theft in each city is shown.
574
117
(a)
229
239
663
465
372
136
1,202 88
189 75
X

X = 4349, n = 12
X=

n
X=
4349
12
- 362
A researcher claims that each year, there is an average of
300 victims of identity theft in major cities. Twelve were
randomly selected, and the number of victims of identity
theft in each city is shown.
574
117
229
239
663
465
372
136
1,202 88
189 75
(b) Arrange data in increasing order:
75 88 117 136 189 229 239 372 465 574 663 1202
median
A researcher claims that each year, there is an average of
300 victims of identity theft in major cities. Twelve were
randomly selected, and the number of victims of identity
theft in each city is shown.
574
117
229
239
663
465
372
136
1,202 88
189 75
(b) Arrange data in increasing order:
MD =
229 + 239
2
= 234
A researcher claims that each year, there is an average of
300 victims of identity theft in major cities. Twelve were
randomly selected, and the number of victims of identity
theft in each city is shown.
574
117
229
239
663
465
372
136
(c) No data are repeated.
No mode.
1,202 88
189 75
A researcher claims that each year, there is an average of
300 victims of identity theft in major cities. Twelve were
randomly selected, and the number of victims of identity
theft in each city is shown.
574
117
229
239
663
465
372
136
1,202 88
189 75
(d) Arrange data in increasing order.
75 88 117 136 189 229 239 372 465 574 663 1202
A researcher claims that each year, there is an average of
300 victims of identity theft in major cities. Twelve were
randomly selected, and the number of victims of identity
theft in each city is shown.
574
117
229
239
663
465
372
136
1,202 88
189 75
(d) Arrange data in increasing order.
MR =
75 + 1202
2
= 638.5
A researcher claims that each year, there is an average of
300 victims of identity theft in major cities. Twelve were
randomly selected, and the number of victims of identity
theft in each city is shown.
574
117
229
239
663
465
372
136
1,202 88
189 75
Can you conclude that the researcher
was correct?
X = 362, MD = 234, no mode, MR = 638.5
It seems that the average number
of identity thefts is higher than 300.
Chapter 3
Data Description
Section 3-2
Exercise #17
Eighty randomly selected light bulbs were tested to
determine their lifetimes (in hours). This frequency
distribution was obtained.
Find the (a) mean and (b) modal class.
Class Boundaries
Frequency
52.5-63.5
6
63.5-74.5
12
74.5-85.5
25
85.5-96.5
18
96.5-107.5
14
107.5-118.5
5
Boundaries
Xm
f
52.5 – 63.5
58
6
348
63.5 – 74.5
69
12
828
74.5 – 85.5
80
25
2000
85.5 – 96.5
91
18
1638
96.5 – 107.5 102
14
1428
107.5 – 118.5 113
5
565
80
6807
f • Xm
Eighty randomly selected light bulbs were tested to
determine their lifetimes (in hours). This frequency
distribution was obtained.
Find the (a) mean and (b) modal class.
 f • X m 6807
= 80 = 85.1
X =
n
modal class: 74.5 – 85.5
Chapter 3
Data Description
Section 3-2
Exercise #19
The cost per load (in cents) of 35 laundry detergents
tested by a consumer organization is shown.
Find the (a) mean and (b) modal class.
Class Limits
13-19
20-26
27-33
34-40
41-47
48-54
55-61
62-68
Frequency
2
7
12
5
6
1
0
2
Class
limits
13 – 19
Boundarie
s
12.5 – 19.5
Xm
f
16
2
32
20 – 26
19.5 – 26.5
23
7
161
27 – 33
26.5 – 33.5
30
12
360
34 – 40
33.5 – 40.5
37
5
185
41 – 47
40.5 – 47.5
44
6
264
48 – 54
47.5 – 54.5
51
1
51
55 – 61
54.5 – 61.5
58
0
0
62 – 68
61.5 – 68.5
65
2
130
35
1183
f • Xm
The cost per load (in cents) of 35 laundry detergents
tested by a consumer organization is shown.
Find the (a) mean and (b) modal class.
 f •Xm
1183
= 35 = 33.8
X =
n
modal class: 26.5 – 33.5
Chapter 3
Data Description
Section 3-3
Exercise #7
The number of incidents where policies were needed for a
sample of ten schools in Allegheny County is
7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Assume the data
represent samples.
Find the range.
range = 48 – 0 = 48
The number of incidents where policies were needed for a
sample of ten schools in Allegheny County is
7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Assume the data
represent samples.
Use the shortcut formula for the unbiased
estimator to compute the variance and
standard deviation.
s2 =


X2 –
 X = 133

  X
n–1

2

/ n

2 = 4061 n = 10
X

The number of incidents where policies were needed for a
sample of ten schools in Allegheny County is
7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Assume the data
represent samples.
Use the shortcut formula for the unbiased
estimator to compute the variance and
standard deviation.
2
4061 –  133  / 10 


2
s =
- 254.7
10 – 1
s-
254.7 - 16
The number of incidents where policies were needed for a
sample of ten schools in Allegheny County is
7, 37, 3, 8, 48, 11, 6, 0, 10, 3.
Is the data consistent or does it vary?
Explain your answer.
range = 48 s2 = 254.7 s = 16
By any of these measures, it can
be said that the data can vary.
Chapter 3
Data Description
Section 3-3
Exercise #21
The data shows the number of murders in 25
selected cities.
Find the variance and
Number
f
standard deviation.
27-90
13
91-154
155-218
219-282
283-346
347-410
411-474
475-539
539-602
2
0
5
0
2
0
1
2
Class
Xm
f • Xm f • X 2
m
13 760.5 44,489.25
f
27-90
58.5
91-154
122.5
2
245
30,012.5
155-218
186.5
0
0
0
219-282
250.5
5 1252.5 313,751.25
283-346
314.5
0
0
0
347-410
378.5
2
757
286,524.5
411-474
442.5
0
0
0
475-538
506.5
1
506.5 256,542.25
539-602
570.5
2
1141
650,940.5
The data shows the number of murders in 25
selected cities.
Find the variance and standard deviation.
 f • X m = 4662.5
s2 =
2–
f
•
X

2 = 1,582,260.25
f
•
X
 m
f • X 
n–1
n
2
The data shows the number of murders in 25
selected cities.
Find the variance and standard deviation.
 f • X m = 4662.5
s2 =
1, 582, 260.25 –
2 = 1,582,260.25
f
•
X
 m
 4662.5 
24
= 29,696
s = 29,696 - 172.3
25
2
Chapter 3
Data Description
Section 3-3
Exercise #33
The mean of a distribution is 20 and the standard
deviation is 2. Answer each. Use Chebyshev’s theorem.
a. At least what percentage of the values will
fall between 10 and 30?
b. At least what percentage of the
values will fall between 12 and 28?
a. Subtract the mean from the larger value: 30 – 20 = 10
10
Divide by the standard deviation to get k:
=5
2
1
1–
= 0.96 or 96%
52
b. Subtract the mean from the larger
value: 28 – 20 = 8. Divide by the standard
8
deviation to get k: = 4
2
1–
1
42
= 0.9375 or 93.75%
Chapter 3
Data Description
Section 3-3
Exercise #41
The average U.S. yearly per capita consumption of citrus
fruits is 26.8 pounds. Suppose that the distribution of
fruit amounts consumed is bell-shaped with a standard
deviation equal to 4.2 pounds.
What percentage of Americans would you expect to
consume more than 31 pounds of citrus
fruit per year?
By the Empirical Rule, 68% of consumption is within 1
standard deviation of the mean. Then 1/2 of 32%, or 16%,
of consumption would be more than 31 pounds of citrus
fruit per year.
 = 4.2
 = 26.8
34% 34%
16%
22.6
26.8
16%
31
Chapter 3
Data Description
Section 3-4
Exercise #13
Which of the following exam scores has a better
relative position?
a. A score of 42 on an exam with X = 39 and s = 4
z=
42 – 39
4
3
=
4
b. A score of 76 on an exam with
X = 71 and s = 3
z=
76 – 71
3
=
5
3
Chapter 3
Data Description
Section 3-4
Exercise #22
Find the percentile ranks of each weight in the data set.
The weights are in pounds.
Data: 78, 82, 86, 88, 92, 97
Percentile =
number of values below + 0.5
total number of values
Data: 78, 82, 86, 88, 92, 97
0 + 0.5
 100% = 8th percentile
For 78,
6
th
For 82, 1 + 0.5 
= 25 percentile
100%
6
nd
For 86, 2 + 0.5 
=
100% 42 percentile
6

100%
Percentile =
number of values below + 0.5
total number of values
Data: 78, 82, 86, 88, 92, 97
For 88, 3 + 0.5  100% = 58th percentile
6
th
For 92, 4 + 0.5 
=
75
percentile
100%
6
nd
For 97, 5 + 0.5 
92
percentile
=
100%
6

100%
Chapter 3
Data Description
Section 3-4
Exercise #23
What value corresponds to the 30th percentile?
Find the percentile ranks of each weight in the data set.
The weights are in pounds.
78, 82, 86, 88, 92, 97
6(30)
c=
100
= 1.8 or 2
Therefore, the answer is the
2nd in the series, or 82.
Chapter 3
Data Description
Section 3-5
Exercise #1
Identify the five number summary and find
the interquartile range.
8, 12, 32, 6, 27, 19, 54
Data arranged in order:
6, 8, 12, 19, 27, 32, 54
Minimum: 6
Median: 19
Maximum: 54
Q1: 8
Q3: 32
Interquartile Range: 32 – 8 = 24
Chapter 3
Data Description
Section 3-5
Exercise #9
Use the boxplot to identify the maximum value, minimum
value, median, first quartile, third quartile, and
interquartile range.
50 55 60 65 70 75 80 85 90 95 100
Minimum: 55
Maximum: 95
Median: 70
Interquartile Range:
Q1: 65
Q3: 90
90 – 65 = 25
Chapter 3
Data Description
Section 3-5
Exercise #15
9.8
15.9
8.0
3.2
13.9
11.7
4.4
24.8
3.9
34.1
21.7
17.6
These data are the number of inches of
snow reported in randomly selected cities
for September 1 through January 10.
Construct a boxplot and comment on the
skewness of the data.
0
5
10
15
20
25
30 35
Data arranged in order :
9.8
15.9
8.0
3.2
13.9
11.7
Minimum: 3.2
4.4
24.8
3.9
34.1
Maximum: 34.1
MD: 11.7 + 13.9 = 12.8
2
Q1: 4.4 + 8.0 = 6.2
2
+ 21.7
17.6
Q3:
= 19.65
2
21.7
17.6
Comment on the skewness of the data.
3.2 6.2 12.8
0
5
10
34.1
19.65
15
The distribution is
positively skewed.
20
25
30
35
Chapter 3
Data Description
Section 3-5
Exercise #16
These data represent the volumes in cubic yards of the
largest dams in the United States and in South America.
Construct a boxplot of the data for each
region and compare the distributions.
United States
125,628
92,000
78,008
77,700
66,500
62,850
52,435
50,000
South America
311,539
274,026
105,944
102,014
56,242
46,563
For USA:
Min = 50,000 Max = 125,628
MD = 72,100 Q1 = 57,642.5 Q3 = 85,004
72,100
125,628
50 100
150
United States
125,628
92,000
78,008
77,700
66,500
62,850
52,435
50,000
200 250 300 350
For South Min = 46,563 Max = 311,539
America: MD = 103,979 Q = 56,242
1
Q3 = 274,026
South America
311,539
274,026
105,944
102,014
56,242
46,563
103,979
50
100
311,539
150
200
250
300
Compare the distributions:
72,100
125,628
USA
103,979
50
100
311,539
150
200
South America
250
300
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