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Ch. 3.2 P.116 # 3, 5, 7
Find (a) the mean, (b) the median, (c) the mode, and (d) the midrange.
3) Burglaries at Pennsylvania Universities the following data are the numbers of burglaries reported for a specific year
for nine western Pennsylvania universities. Which measure of average might be the best in this case? Explain your
answer.
61 11 1 3 2 30 18 3 7
(a) Mean
Mean =
 ( x)  136  15.1
n
9
(b) Median
1 2 3 3 7 11 18 30 61
Median=7
(c) Mode=3
(d) Midrange
Midrange=
1  61 62

 31
2
2
Since the data is not clustered around the mean, the median is a better choice in representing the average of the
measurements.
5). Identity Theft A researcher claims that each year, there are average or 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. Can you conclude that
the researcher was correct?
574
229
663
372
102
88
117
239
465
136
189
75
a)
b)
Mean =
75
88
 ( x)  3249  270.75
n
102
Median=
12
117
136
189
189  229 418

 209
2
2
229
239
372
465
c) Mode= none
574
663
d) midrange=
75  663 738

 369
2
2
Since both the mean and the median are below 300, it seems the number of identity thefts on average is not higher than
300.
7) Earthquake Strengths twelve major earthquakes had Richter magnitudes shown here.
5.4
5.4
a) Mean =
6.2
6.2
6.4
6.4
6.5
7.0
7.2
7.2
7.7
8.0
 ( x)  79.6  6.63
b) Median =
n
12
6.4  6.5
 6.45
2
c) Mode = none
d) Midrange =
5.4  8.0
 6.7
2
Since almost all data is around the mean value (6.63), the mean is a better choice for the average of measurement.
Section 3-3 #’s 7,9,11 13,30, 31, 33, 34, 35,37, 41, 42
For exercises 7-13, find the range, variance, and standard deviation. Assume the data represent samples, and use the
shortcut formula for the unbiased estimator to compute the variance and standard deviation.
(X ) ;
mean  X 
variance = s
n
2
 ( X  X ) or s   X

2
n 1
2
 [( X ) 2 / n]
n 1
standard deviation = s  variance
7) Police Calls in Schools the number of incidents where police were needed for a sample of 10 schools in Allegheny
County is 7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Are the data consistent or do they vary? Explain your answer.
Sum
x
x2
0
0
3
9
3
9
6
36
7
49
8
64
10
100
11
121
37
1369
48
2304
133
4061
Mean =
(X )
n

133
 13.3
10
Range = 48 – 0 = 48 n=10
2
Variance = s 
X
2
 [( X ) 2 / n]
n 1

4061  (133) / 10
9

4061  1768.9 2292.1

 254.7
9
9
2
Standard deviation =
The data vary widely.
254.7  16
9) Precipitation and High Temperatures the normal daily high temperatures (in degrees Fahrenheit) in January for 10
selected cities are as follows.
50, 37, 29, 54, 30, 61, 47, 38, 34, 61
The normal monthly precipitation (in inches) for theses same 10 cities is listed here.
4.8, 2.6, 1.5, 1.8, 1.8, 3.3, 5.1, 1.1, 1.8, 2.5
Which set is more variable?
Sum
x
x2
29
30
34
37
38
47
50
54
61
61
441
841
900
1156
1369
1444
2209
2500
2916
3721
3721
20,777
For the Temperature:
Mean =
(X )
n
Variance = s
2

441
 44.1
10
X

2
 [( X ) 2 / n]
Range = 61 – 29 = 32

n=10
20,777  [(441) 2 /10]
9
n 1
20,777  19, 448.1 1328.9


 147.6
9
9
Standard deviation = 147.6  12.15
For the Precipitation:
Sum
x
x2
4.8
2.6
1.5
1.8
1.8
3.3
5.1
1.1
1.8
2.5
26.3
23.04
6.76
2.25
3.24
3.24
10.89
26.01
1.21
3.24
6.25
86.13
Mean =
(X )
n

26.3
 2.63
10
Range = 5.1 – 1.1 = 4.0
Variance =
s2 

X
2
 [( X ) 2 / n]
n 1
n=10

86.13  [(26.3) 2 /10]
9
86.13  69.17 16.96

 1.88
9
9
Standard deviation = 1.88  1.37
The standard deviation for the temperature is larger than the standard deviation for the precipitation.
11) Heights of Tall Buildings Shown here are the numbers of stories in the 11 tallest buildings in St. Paul, Minnesota.
32, 36, 46, 20, 32, 18, 16, 34, 26, 27, 26
Shown here are the numbers of stories in the 11 tallest buildings in Chicago, Illinois.
100, 100, 83, 60, 64, 65, 66, 74, 60, 67, 57
Which data set is more variable?
For St. Paul, Minnesota:
Sum
x
x2
16
18
20
26
26
27
32
32
34
36
46
313
256
324
400
676
676
729
1024
1024
1156
1296
2116
9677
Mean =
(X )
n

313
 28.45
11
Range = 46 - 16 = 30
Variance =
s
2
X

n=11
 [( X ) 2 / n]
9677  [(313) 2 / 11]

n 1
10
9677  8906.27 770.73


 77.07
10
10
2
Standard deviation =
77.07  8.78
For Chicago, Illinois:
Sum
x
x2
57
60
60
64
65
66
67
74
83
100
100
796
3249
3600
3600
4096
4225
4356
4489
5476
6889
10000
10000
59980
Mean =
(X )
n

796
 72.36
11
Range = 100 - 57 = 43
Variance =
s2 
X
n=11
 [( X ) 2 / n]
59980  (796) 2 / 11
n 1
10
59980  57601.45 2378.55


 237.86
10
10
2
Standard deviation =

237.86  15.42
The standard deviation for Chicago is larger than the standard deviation for St. Paul, therefore the data for
Chicago is more variable.
13) Hardcover Bestsellers the number of weeks on The New York Times Best Sellers list for hardcover fiction is
1
4
2
2
3
18
5
5
10
4
3
6
2
2
22
Use the range rule of thumb to estimate the standard deviation. Compare the estimate to the actual standard
deviation.
s
x
range 22  1

 5.25 ,
4
4
(89)2
1061
[
]

X  [( X ) / n]

2
15

 38.1
s 
n 1
15  1
2
2
s  38.1  6.2
Sum
1
4
2
2
3
18
5
5
10
4
3
6
2
2
22
89
x2
1
16
4
4
9
324
25
25
100
16
9
36
4
4
484
1061
By the rule of thumb the estimation of the standard deviation is s=5.25, and by actually calculating it the standard
deviation is s = 6.2. Thus, we can say that the estimate is close to the actual standard deviation.
30) Exam Scores the average score on an English final examination was 85, with a standard deviation of 5; the average
score on a history final exam was 110, with a standard deviation of 8. Which class was more variable?
The coefficients of variation are:
English Final:
CVar =
s
5
 100%  5.9%
X 85
History Final:
CVar =
s
8

100%  7.3%
X 110
The history class was more variable.
31) Ages of accountants the average age of the accountants at Three Rivers Corp. is 26 years, with a standard deviation
of 6 years; the average salary of the accountants is $31,000, with a standard deviation of $4000. Compare the
variations of age and income.
Age:
CVar =
s
6
 100%  23.1%
X 26
Salary:
CVar =
s
4000

100%  12.9%
X 31000
Age is more variable.
33) The mean of a distribution is 20 and the standard deviation is 2. Use Chebyshev’s theorem.
a) At least what percentage of the values will fall between 10 and 30?
k 5
1
1
1
 2 
 0.04
2
5
25
k
1
1  2  1  0.04  0.96  96%
k
b) At least what percentage of the values will fall between 12 and 28?
k 4
1
1
1
 2 
 0.0625
2
4 16
k
1
1  2  1  0.0625  0.9375  93.75%
k
34) In a distribution of 200 values, the mean is 50 and the standard deviation is 5. Use Chebyshev’s theorem.
a) At least how many values will fall between 30 and 70?
k 4
1
1
1
 2 
 0.0625
2
4 16
k
1
1  2  1  0.0625  0.9375  93.75%
k
0.9375200  187.5
At least 187.5 values will fall between 30 and 70.
b) At most how many values will be less than 40 or more than 60?
k 2
1
1 1
 2   0.25
2
k
2
4
0.25200  50
At most 50 values will be less than 40 or more than 60.
35) Hourly Wages of Restaurant Employees A sample of the hourly wages of employees who work in restaurants in a
large city has a mean of $5.02 and a standard deviation of $0.09. Using Chebyshev’s theorem, find the range in
which at least 75% of the data values will fall.
1
1
1
1
1
 0.75  1  0.75  2  0.25  2  k 2 
 k2  4  k  4  k  2
2
k
k
k
0.25
At least 75% of the data values fall between $4.84 and $5.20.
37) Potassium Content of Cereal A survey of a number of the leading brands of cereal shows that the mean content of
potassium per serving is 95 milligrams, and the standard deviation is 2 milligrams. Find the range in which at
least 88.89% of the data will fall. Use Chebyshev’s theorem.
1
1
1
1
1
 0.8889  1  0.8889  2  0.1111  2  k 2 
 k 2  9.009  k  9.009  k  3
2
k
k
k
0.1111
At least 88.89% of the data will fall between 89 and 101.
41) Citrus Fruit Consumption the average U.S. yearly per capita consumption of citrus fruit 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
of 32%, or 16% of
2
consumption would be more than 31 pounds of citrus fruit per year.
42) Work Hours for College Faculty the average full-time faculty member in a post-secondary degree-granting
institution works an average of 53 hours per week.
a) If we assume the standard deviation is 2.8 hours, what percentage of faculty members work more than 58.6
hours a week?
k 2
1
1 1
 2   0.25
2
k
2
4
0.25
 0.125  12.5%
2
12.5% of faculty members work more than 58.6 hours a week.
b) If we assume a bell-shaped distribution, what percentage of faculty members work more than 58.6 hours a
week?
By the Empirical Rule, 95% of members working is within 2 standard deviations of the mean. Then
1
of 5% or
2
2.5% of faculty member work more than 58.6 hours a week.
Ch 3-4 P. 151 # 13, 15, 22, 23, 24, 25, 27
11) Exam Scores A final examination for a psychology course has a mean of 84 and a standard deviation of 4. Find the
corresponding Z score for each raw score.
a) z 
x   87  84

 0.75

4
b) z 
x   79  84

 1.25

4
c) z 
x   93  84

 2.25

4
d) z 
x   76  84

 2

4
e)
z
x   82  84

 0.5

4
13) Exam Scores 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 .
x  x 42  39

 0.75
s
4
b. A score of 76 on an exam with X = 71 and s  3
x  x 76  71

 1.67
s
3
The score for part (b) has a better relative position.
14) Test Scores A student scores 60 on a mathematics test that has a mean of 54 and a standard deviation of 3, and she
scores 80 on a history test with a mean of 75 and a standard deviation of 2. On which test did she perform better?
z
x   60  54

 2.0

3
z
x   80  75

 2.5

2
The student performed better in the history test.
22) Weights find the percentile ranks of each weight in the data set. The weights are in pounds.
78 82 86 88 92 97
Percentile =
(number of values below x)  0.5
 100%
total number of values
78: percentile =
0  0.5
 100 %  8.3 % 8 th
6
82: percentile =
1  0.5
 100 %  0.25 %  25 th
6
86: percentile =
2  0.5
 100 %  41.6 % 42 nd
6
88: percentile =
3  0.5
 100 %  58.3 % 58 th
6
92: percentile =
4  0.5
 100 %  75%  475 th
6
97: percentile =
5  0.5
 100 %  91.6% 92 nd
6
23) In exercise 22, what value corresponds to the 30th percentile?
What is 30% of n? = 0.3 x 6 = 1.8, round up to 2.
Therefore, the number located in the 2nd place which is 82 is the 30th percentile.
24) Test scores find the percentile rank for each test score in the data set.
12 28 35 42 47 49 50
Percentile =
(number of values below x)  0.5
 100%
total number of values
12: percentile =
0  0.5
 100 % 7 th
7
28: percentile =
1  0.5
 100 21 st
7
35: percentile =
2  0.5
 100 % 36 th
7
42: percentile =
3  0.5
 100 % 50 th
7
47: percentile =
4  0.5
 100 % 64 th
7
49: percentile =
5  0.5
 100 % 79 th
7
50: percentile =
6  0.5
 100 % 93 rd
7
25) In exercise 24, what value corresponds to the 60th percentile?
What is 60% of n?
0.6 x 7 = 4.2, round it up to 5. Therefore the number in the 5th location which is 47 corresponds to the 60th percentile.
27) What value in Exercise 26 corresponds to the 40th percentile?
Values in Exercise 26: 1.1 1.7 1.9 2.1 2.2 2.5 3.3 6.2 6.8 20.3
What is 40% of n?
0.40 x 10 = 4. Thus, the value halfway between the 4th and 5th place is 2.15 which corresponds to the 40% percentile.
3.5 P. 163 # 1, 3, 7, 9, 11, 13, 15
Identify the five-number summary and find the interquartile range.
1) 8, 12, 32, 6, 27, 19, 54
Reorder: 6, 8, 12, 19, 27, 32, 54
Q1  8
Low = 6
Median = Q2  19
Q3  32
High = 54
Q3  437
High = 589
IQR = Q3  Q1  32  8  24
3) 188
192
316
362
Q1  192
Low = 188
437
589
Median = 339
IQR = Q3  Q1  437  192  245
Use each box plot to identify the maximum value, minimum value, median, first quartile, third quartile, and
interquartile range.
High = 11, median = 8, Q1  5 , Q3  9 , and
7) Low = 3,
9) Low = 200,
IQR = Q3  Q1  4
High = 325, median = 275, Q1  225 , Q3  300 , and
IQR = Q3  Q1  75
11) Police Force Sizes Shown next are the sizes of the police forces in the 10 largest cities in the United States in 1993
(the numbers represent hundreds). Construct a box plot for the data and comment on the shape of the distribution.
29.3
7.6
12.1
4.7
6.2
Reorder: 1.7 1.9 2.0 2.8
Low = 1.7, Q1  2.0 , median =
1.9
3.9
3.9
2.8
4.7 6.2
3.9  4.7
 4.3 ,
2
2.0
1.7
7.6 12.1
29.3
Q3  7.6
High = 29.3
13. Vacation Days construct a box plot for the following average number of vacation days in selected countries.
13 25 25 26 28 34 35 37 42
Low = 13,
Q1 
25  25
 25 ,
2
median = 28 ,
Q3 
35  37
 36
2
High = 42
15) Inches of snow these data are the number of inches of snow reported in randomly selected U. S. cities for
September 1 through January 10. Construct a box plot and comment on the skewness of the data.
3.2 3.4 4.4 8.0 9.8 11.7 13.9 15.9 17.6 21.7 24.8 34.1
Low = 3.2, Q1 
High = 34.1
4.4  8.0
 6.2 ,
2
median =
11.7  13.9
 12.8 ,
2
Q3 
17.6  21.7
 19.65 ,
2