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Student’s Solutions Manual and Study Guide: Chapter 3
Page 1
Chapter 3
Descriptive Statistics
LEARNING OBJECTIVES
The focus of Chapter 3 is the use of statistical techniques to describe data. Here
are the key questions we would like to answer:
1. What are the measures of central tendency, measures of variability, measures of shape,
and measures of association?
2. What are the meanings of mean, median, mode, quartile, percentile, and range?
3. How can we compute the mean, median, mode, percentile, quartile, range, variance,
standard deviation, and mean absolute deviation on ungrouped data?
4. What is the difference between sample and population variance and standard deviation?
5. What is the meaning of standard deviation as it is applied by using the empirical rule and
Chebyshev’s theorem?
6. How can we compute the mean, median, mode, standard deviation, and variance on
grouped data?
7. What are the defi nitions of skewness and kurtosis and why are they important? What are
box and whisker plots and how can we construct them?
Student’s Solutions Manual and Study Guide: Chapter 3
Page 2
CHAPTER OUTLINE
3.1 Measures of Central Tendency: Ungrouped Data
Mode
Median
Mean
Percentiles
Quartiles
3.2 Measures of Variability - Ungrouped Data
Range
Interquartile Range
Mean Absolute Deviation, Variance, and Standard Deviation
Mean Absolute Deviation
Variance
Standard Deviation
Meaning of Standard Deviation
Empirical Rule
Chebyshev’s Theorem
Population Versus Sample Variance and Standard Deviation
Computational Formulas for Variance and Standard Deviation
z Scores
Coefficient of Variation
3.3 Measures of Central Tendency and Variability - Grouped Data
Measures of Central Tendency
Mean
Median
Mode
Measures of Variability
3.4 Measures of Shape
Skewness
Skewness and the Relationship of the Mean, Median, and Mode
Coefficient of Skewness
Kurtosis
Box and Whisker Plot
3.5 Descriptive Statistics on the Computer
Student’s Solutions Manual and Study Guide: Chapter 3
Page 3
KEY TERMS
Arithmetic Mean
Bimodal
Box and Whisker Plot
Chebyshev’s Theorem
Coefficient of Skewness
Coefficient of Variation (CV)
Deviation from the Mean
Empirical Rule
Interquartile Range
Kurtosis
Leptokurtic
Mean Absolute Deviation (MAD)
Measures of Central Tendency
Measures of Shape
Measures of Variability
Median
Mesokurtic
Mode
Multimodal
Percentiles
Platykurtic
Quartiles
Range
Skewness
Standard Deviation
Sum of Squares of x
Variance
z Score
Student’s Solutions Manual and Study Guide: Chapter 3
Page 4
STUDY QUESTIONS
1.
Statistical measures used to yield information about the centre or middle part of a
group of numbers are called _______________.
2. The "average" is the _______________.
3. The value occurring most often in a group of numbers is called _______________.
4.
In a set of 110 numbers arranged in order, the median is located at the
_______________ position.
5.
If a set of data has an odd number of values arranged in ascending order, the median
is the ________________ value.
Consider the data: 5, 4, 6, 6, 4, 5, 3, 2, 6, 4, 6, 3, 5
Answer questions 6-8 using this data.
6. The mode is _______________.
7. The median is _______________.
8. The mean is _______________.
9.
If a set of values is a population, then the mean is denoted by _______________.
10. In computing a mean for grouped data, the _______________ is used to represent all
data in a given class interval.
11. The mean for the data given below is _______________.
Class Interval
50 - under 53
53 - under 56
56 - under 59
59 - under 62
62 - under 65
Frequency
14
17
29
31
18
12. Measures of variability describe the _______________ of a set of data.
Student’s Solutions Manual and Study Guide: Chapter 3
Page 5
Use the following population data for Questions 13-17:
27 65 28 61 34 91 61 37 58 31
43 47 44 20 48 50 49 43 19 52
13. The range of the data is _______________.
14. The value of Q1 is _____________, Q2 is ____________, and Q3 is ____________.
15. The interquartile range is _____________.
16. The value of the 34th percentile is ________________.
17. The value of the Pearsonian coefficient of skewness for these data is
________________.
18. The Mean Absolute Deviation is computed by averaging the _______________ of
deviations around the mean.
19.
Subtracting each value of a set of data from the mean produces _______________
from the mean.
20. The sum of the deviations from the mean is always _______________.
21. The variance is the _______________ of the standard deviation.
22. The population variance is computed by using _______________ in the
denominator. Whereas, the sample variance is computed by using
_______________ in the denominator.
23.
If the sample standard deviation is 9, then the sample variance is
_______________.
Consider the data below and answer questions 24-26 using the data:
2, 3, 6, 12
24. The mean absolute deviation for this data is _______________.
25. The sample variance for this data is _______________.
26. The population standard deviation for this data is _______________.
27.
In estimating what proportion of values fall within so many standard deviations of
the mean, a researcher should use _______________ if the shape of the distribution
of numbers is unknown.
Student’s Solutions Manual and Study Guide: Chapter 3
28.
Page 6
Suppose a distribution of numbers is mound shaped with a mean of 150 and a
variance of 225. Approximately _______________ percent of the values fall
between 120 and 180. Between _______________ and _______________ fall
99.7% of these values.
29. The shape of a distribution of numbers is unknown. The distribution has a mean of
275 and a standard deviation of 12. The value of k for 299 is _______________.
At least _______________ percent of the values fall between 251 and 299.
30.
Suppose data are normally distributed with a mean of 36 and a standard deviation of
4.8. The z score for 30 is _______________. The z score for 40 is
_______________.
31.
A normal distribution of values has a mean of 74 and a standard deviation of 21.
The coefficient of variation for this distribution is _______________?
Consider the data below and use the data to answer questions 32-35.
Class Interval
2- 4
4- 6
6- 8
8-10
10-12
12-14
Frequency
5
12
14
15
8
4
32. The sample variance for the data above is _______________.
33. The population standard deviation for the data above is _______________.
34. The mode of the data is _________________________.
35. The median of the data is _________________________.
36.
If a unimodal distribution has a mean of 50, a median of 48, and a mode of 47, the
distribution is skewed _______________.
37.
If the value of Sk is positive, then it may be said that the distribution is
________________________ skewed.
38. The peakedness of a distribution is called _________________________.
39.
If a distribution is flat and spread out, then it is referred to as
_______________________; if it is "normal" in shape, then it is referred to as
_________________________; if it is high and thin, then it is referred to as
__________________________.
Student’s Solutions Manual and Study Guide: Chapter 3
Page 7
40.
In a box plot, the inner fences are computed by ____________________ and
_____________________. The outer fences are computed by
___________________ and _____________________.
41.
Data values that lie outside the mainstream of values in a distribution are referred to
as __________________.
Student’s Solutions Manual and Study Guide: Chapter 3
Page 8
ANSWERS TO STUDY QUESTIONS
1. Measures of Central Tendency
22. N, n – 1
2. Mean
23. 81
3. Mode
24. 3.25
4. 55.5th
25. 20.25
5. Middle
26. 3.897
6. 6
27. Chebyshev’s Theorem
7. 5
28. 95, 105, and 195
8. 4.54
29. 2, 75
9. 
30. –1.25, 0.83
10. Class Midpoint
31. 28.38%
11. 58.11
32. 7.54
12. Spread or Dispersion
33. 2.72
13. 72
34. 9
14. Q1 = 32.5, Q2 = 45.5, Q3 = 55
35. 7.7143
15. IQR = 22.5
36. Right
16. P34 = 37
37. Positively
17. Sk = -0.018
38. Kurtosis
18. Absolute Value
39. Platykurtic, Mesokurtic,
Leptokurtic
40. Q1 - 1.5 IQR and Q3 + 1.5 IQR
19. Deviations
Q1 - 3.0 IQR and Q3 + 3.0 IQR
20. Zero
21. Square
41. Outliers
Student’s Solutions Manual and Study Guide: Chapter 3
Page 9
SOLUTIONS TO PROBLEMS IN CHAPTER 3
3.1
Mean
17.3
44.5
µ = x/N = (333.6)/8 =
41.7
31.6
40.0
52.8
x = x/n = (333.6)/8 = 41.7
38.8
30.1
78.5 (It is not stated in the problem whether the
x = 333.6
data represent as population or a sample).
3.3
Median for values in 3.1
Arrange in ascending order:
17.3 , 30.1 , 31.6 , 38.8 , 40 , 44.5 , 52.8 , 78.5
There are 8 terms. Since there is an even number of terms, the median is the average of
the two middle values, 38.8 and 40.
38.8  40 78.8

 39.4
The median =
2
2
n1 th
8 1
Using the formula, the median is located at the
term =
= 4.5th term. The
2
2
median is located halfway between the 4th and 5th terms or the average of 38.8 and 40. So,
the median is 39.4.
3.5
Mode
2, 2, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8, 8, 8, 9
The mode = 4
4 is the most frequently occurring value
3.7
Rearranging the data in ascending order:
80, 94, 97, 105, 107, 112, 116, 116, 118, 119, 120, 127,
128, 138, 138, 139, 142, 143, 144, 145, 150, 162, 171, 172
n = 24
For P20:
i
20
(24)  4.8
100
Student’s Solutions Manual and Study Guide: Chapter 3
Page 10
Thus, P20 is located at the 4 + 1 = 5th term and
For P47:
i
P20 = 107
47
(24)  11.28
100
Thus, P47 is located at the 11 + 1 = 12th term and
For P83:
i
P47 = 127
83
(24)  19.92
100
Thus, P83 is located at the 19 + 1 = 20th term and
P83 = 145
Q1 = P25
For P25:
i
25
(24)  6
100
Thus, Q1 is located at the 6.5th term and
Q1 = (112 + 116)/ 2 =
114
Q2 = Median
 24  1 
th

  12.5 term
2


th
The median is located at the:
Thus, Q2 = (127 + 128)/ 2 = 127.5
Q3 = P75
For P75:
i
75
(24)  18
100
Thus, Q3 is located at the 18.5th term and
Q3 = (143 + 144)/ 2 = 143.5
3.9
Arrange terms in ascending order:
2,538,624 2,596,316 2,617,725
3,457,385 3,911,814 6,247,506
2,669,040 2,679,451 3,431,398
6,267,891 8,534,690 9,349,818
 12  1 
th
The median is located at the 
  6.5 position.
2


th
Student’s Solutions Manual and Study Guide: Chapter 3
Page 11
The median = (3,431,398 + 3,457,385)/2 = 3,444,391.5
75
(12)  9
100
P75 is located halfway between the 9th and 10th terms.
Q3 = P75 = (6,247,506 + 6,267,891)/ 2 = 6,257,698.5
i
For Q3 = P75:
20
(12)  2.4
100
P20 is located at the 3rd term:
For P20:
i
P20 = 2,617,725
60
(12)  7.2
100
P60 is located at the 8th term:
P60 = 3,911,814
For P60:
i
80
(12)  9.6
100
P80 is located at the 10th term:
For P80:
i
P80 = 6,267,891
93
(12)  11.16
100
P93 is located at the 12th term:
P93 = 9,349,818
For P93:
i
3.11
x
6
2
4
9
1
3
5
x = 30
6-4.2857 =
x-µ =

 x -µ
1.7143
2.2857
0.2857
4.7143
3.2857
1.2857
0.7143
14.2857
x 30

 4.2857
N
7
a.) Range = 9 - 1 = 8
b.) M.A.D. =
c.) 2 =
x
N

14.2857
 2.0408
7
( x   ) 2 43.4284

= 6.2041
N
7
(x-µ)2
2.9388
5.2244
0.0816
22.2246
10.7958
1.6530
0.5102
(x -µ)2 = 43.4284
Student’s Solutions Manual and Study Guide: Chapter 3
d.)  =
Page 12
( x   ) 2
 6.2041 = 2.4908
N
e.) Arranging the data in order:
Q1 = P25
1, 2, 3, 4, 5, 6, 9
i =
25
(7) = 1.75
100
Q1 is located at the 2nd term, Q1 = 2
Q3 = P75:
i =
75
(7) = 5.25
100
Q3 is located at the 6th term, Q3 = 6
IQR = Q3 - Q1 = 6 - 2 = 4
f.)
z =
6  4.2857
= 0.69
2.4908
z =
2  4.2857
= -0.92
2.4908
z =
4  4.2857
= -0.11
2.4908
z =
9  4.2857
= 1.89
2.4908
z =
1  4.2857
= -1.32
2.4908
z =
3  4.2857
= -0.52
2.4908
z =
5  4.2857
= 0.29
2.4908
Student’s Solutions Manual and Study Guide: Chapter 3
3.13
Page 13
a.)
x
12
23
19
26
24
23
x = 127
 =
 =
(x -µ)2
84.034
3.360
4.696
23.358
8.026
3.360
2
(x -µ) = 126.834
(x-µ)
12-21.167= -9.167
1.833
-2.167
4.833
2.833
1.833
(x -µ) = -0.002
x 127

= 21.167
N
6
( x   ) 2
126.834

 21.139 = 4.598
N
6
ORIGINAL FORMULA
b.)
x
12
23
19
26
24
23
x = 127
x2
144
529
361
676
576
529
x2= 2815
=
(x) 2
N 
N
x 2 
= 4.598
(127) 2
6

6
2815 
2815  2688.17
126.83

 21.138
6
6
SHORT-CUT FORMULA
The short-cut formula is faster, but the original formula gives insight
into the meaning of a standard deviation.
Student’s Solutions Manual and Study Guide: Chapter 3
3.15
Page 14
2 = 58,631.295
 = 242.139
x = 6886
x2 = 3,901,664
n = 16
µ = 430.375
3.17
a)
1
1
1 3
 1    .75
2
2
4 4
.75
b)
1
1
1
 1
 .84
2
2.5
6.25
.84
c)
1
1
1
 1
 .609
2
1.6
2.56
.609
d)
1
1
1
 1
 .902
2
3.2
10.24
.902
Set 2:
2 
x 570

 142.5
N
4
2 
(x) 2
(570) 2
82,070 
N 
4
N
4
x 2 
CV1 =
14.2215
(100) = 21.71%
65.5
CV2 =
14.5344
(100) = 10.20%
142.5
= 14.5344
Student’s Solutions Manual and Study Guide: Chapter 3
3.19
x
Page 15
x
xx
7
5
10
12
9
8
14
3
11
13
8
6
106
1.833
3.833
1.167
3.167
0.167
0.833
5.167
5.833
2.167
4.167
0.833
2.833
32.000
( x  x) 2
3.361
14.694
1.361
10.028
0.028
0.694
26.694
34.028
4.694
17.361
0.694
8.028
121.665
x 106

= 8.833
n
12
xx
a) MAD =
n

32
= 2.667
12
b) s2 =
( x  x) 2 121.665
= 11.06

n 1
11
c) s =
s 2  11.06 = 3.326
d) Rearranging terms in order:
3 5 6 7 8 8 9 10 11 12 13 14
Q1 = P25: i = (.25)(12) = 3
Q1 = the average of the 3rd and 4th terms:
Q1 = (6 + 7)/2 = 6.5
Q3 = P75: i = (.75)(12) = 9
Q3 = the average of the 9th and 10th terms:
IQR = Q3 - Q1 = 11.5 – 6.5 = 5
e.) z =
f.) CV =
6  8.833
= - 0.85
3.326
(3.326)(100)
= 37.65%
8.833
Q3 = (11 + 12)/2 = 11.5
Student’s Solutions Manual and Study Guide: Chapter 3
µ = 125
3.21
Page 16
 = 12
68% of the values fall within:
µ ± 1 = 125 ± 1(12) = 125 ± 12
between 113 and 137
95% of the values fall within:
µ ± 2 = 125 ± 2(12) = 125 ± 24
between 101 and 149
99.7% of the values fall within:
µ ± 3 = 125 ± 3(12) = 125 ± 36
between 89 and 161
3.23
1
= .80
k2
1
1 - .80 =
k2
1-
.20 =
k2 = 5
1
k2
and
and
.20k2 = 1
k = 2.236
2.236 standard deviations
3.25
=4
µ = 29
Between 21 and 37 days:
x1  

x2  


21  29  8
= -2 Standard Deviations

4
4

37  29 8
= 2 Standard Deviations

8
4
Student’s Solutions Manual and Study Guide: Chapter 3
Page 17
Since the distribution is normal, the empirical rule states that 95% of
the values fall within µ ± 2 .
Exceed 37 days:
Since 95% fall between 21 and 37 days, 5% fall outside this range. Since the
normal distribution is symmetrical, 2½% fall below 21 and above 37.
Thus, 2½% lie above the value of 37.
Exceed 41 days:
x


41  29 12

= 3 Standard deviations
4
4
The empirical rule states that 99.7% of the values fall within µ ± 3 = 29 ± 3(4)
= 29 ± 12. That is, 99.7% of the values will fall between 17 and 41 days.
0.3% will fall outside this range and half of this or .15% will lie above 41.
Less than 25: µ = 29
x


= 4
25  29  4

= -1 Standard Deviation
4
4
According to the empirical rule, µ ± 1 contains 68% of the values.
29 ± 1(4) = 29 ± 4
Therefore, between 25 and 33 days, 68% of the values lie and 32% lie outside this
range with ½(32%) = 16% less than 25.
3.27
Mean
Class
0- 2
2- 4
4- 6
6- 8
8 - 10
10 - 12
12 - 14
f
39
27
16
15
10
8
6
f = 121
M
1
3
5
7
9
11
13
fM
39
81
80
105
90
88
78
fM = 561
Student’s Solutions Manual and Study Guide: Chapter 3
Mean:  =
Page 18
fM 561

= 4.64
f
121
Median:
N   f  121
The observation 60.5, 50% of N = 121, is in the class corresponding to 2–under 4.
Therefore, l  2 , F  39 , f  27 and w  2 .
N

 121

 39 
 F

w  2 2
 2  2   21.5  2  3.59
median grouped  l   2
 f 
 27 
 27 








Mode: The modal class is 0 – 2.
The midpoint of the modal class = the mode = 1
3.29
Class
f
M
fM
20-30
30-40
40-50
50-60
60-70
70-80
Total
7
11
18
13
6
4
59
25
35
45
55
65
75
175
385
810
715
390
300
2775
 =
fM 2775

= 47.034
f
59
M-µ
-22.0339
-12.0339
- 2.0339
7.9661
17.9661
27.9661
2 =
 =
(M - µ)2
f(M - µ)2
485.4927
3398.449
144.8147
1592.962
4.1367
74.462
63.4588
824.964
322.7808
1936.685
782.1028
3128.411
Total 10,955.933
f ( M   ) 2 10,955.93

= 185.694
f
59
185.694
= 13.627
Student’s Solutions Manual and Study Guide: Chapter 3
3.31
Class
18 - 24
24 - 30
30 - 36
36 - 42
42 - 48
48 - 54
54 - 60
60 - 66
66 - 72
f
17
22
26
35
33
30
32
21
15
f= 231
a.) Mean: x 
Page 19
M
21
27
33
39
45
51
57
63
69
fM
357
594
858
1,365
1,485
1,530
1,824
1,323
1,035
fM= 10,371
fM2
7,497
16,038
28,314
53,235
66,825
78,030
103,968
83,349
71,415
fM2= 508,671
fM fM 10,371
= 44.896


n
f
231
b.) Median:
N   f  231
The observation 115.5, 50% of N = 231, is in the class corresponding to 42 –under 48.
Therefore, l  42 , F  17  22  26  35  100 , f  33 and w  6 .
N

 231

 100 
 F

 w  42   2
6  42   15.5  6  44.818
median grouped  l   2
33
 f 


 33 








c.) Mode. The Modal Class = 36-42. The mode is the class midpoint = 39
(fM ) 2
(10,371) 2
fM 
508,671 
n
231  43,053.5065 = 187.189
d.) s2 =

n 1
230
230
2
e.) s =
187.2 = 13.682
Student’s Solutions Manual and Study Guide: Chapter 3
Page 20
3.33
f
M
fM
fM2
0 - 50,000
18
25,000
450,000
11,250,000,000
50,000 - 100,000
4
75,000
300,000
22,500,000,000
100,000 - 150,000
7
125,000
875,000
109,375,000,000
150,000 - 200,000
7
175,000
1,225,000
214,375,000,000
200,000 - 250,000
2
225,000
450,000
101,250,000,000
250,000 - 300,000
0
275,000
0
0
300,000 - 350,000
1
325,000
325,000
105,625,000,000
350,000 - 400,000
0
375,000
0
0
400,000 - 450,000
1
 f  40
425,000
425,000
 f M  4,050,000
180,625,000,000
 f M 2  745,000,000,000
Class
a.) Mean:
fM 4,050,000
 =
= 101,250

f
40
b.) Median:
N   f  40
The observation 20, 50% of N = 40, is in the class corresponding to 50,000 – under
100,000.
Therefore, l  50,000 , F  18 , f  4 and w  50,000 .
N

 40

 18 
 F

 w  50,000   2
50,000 
median grouped  l   2
 f 
 4 








2
 50,000    50,000  75,000
4
c.) Mode.
The Modal Class = 0 - 50,000
The mode is the midpoint of this class = 25,000.
d.) Variance:
2 =
(fM ) 2
(4,050,000) 2
745,000,000,000 
N
40
= 8,373,437,500

N
40
fM 2 
e.) Standard Deviation:
 =
 2  8,373,437,500 = 91,506.49
Student’s Solutions Manual and Study Guide: Chapter 3
3.35
Page 21
mean = $35
median = $33
mode = $21
The stock prices are skewed to the right. While many of the stock prices are at the
cheaper end, a few extreme prices at the higher end pull the mean.
3.37
3(  M d )
3(551  319)
= 0.726

959
Because the value of Sk is positive, the distribution is positively skewed.
Sk =

3.39
379
300
500
400
Q1 = 500.
558.5
589
500
Median = 558.5.
690
600
700
Q3 = 589.
IQR = 589 - 500 = 89
Inner Fences:
Q1 - 1.5 IQR = 500 - 1.5 (89) = 366.5
and Q3 + 1.5 IQR = 589 + 1.5 (89) = 722.5
Outer Fences: Q1 - 3.0 IQR = 500 - 3 (89) = 233
and Q3 + 3.0 IQR = 589 + 3 (89) = 856
The distribution is negatively skewed. There are no mild or extreme outliers.
Student’s Solutions Manual and Study Guide: Chapter 3
3.41
Page 22
Arranging the values in an ordered array:
1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
3, 3, 3, 3, 3, 3, 4, 4, 5, 6, 8
Mean:
x
x 75

= 2.5
n 30
Mode = 2 (There are eleven 2’s)
Median: There are n = 30 terms.
n 1
30  1 31


= 15.5th position.
2
2
2
th
The median is located at
Median is the average of the 15th and 16th value.
However, since these are both 2, the median is 2.
Range = 8 - 1 = 7
Q1 = P25:
Q1 is the 8th term =
Q3 = P75:
25
(30) = 7.5
100
i =
1
i =
75
(30) = 22.5
100
Q3 is the 23rd term = 3
IQR = Q3 - Q1 = 3 - 1 = 2
Student’s Solutions Manual and Study Guide: Chapter 3
3.43
µ =
Page 23
x 126,904

= 6345.2
N
20
The median is located at the
n 1
th value = 21/2 = 10.5th value
2
The median is the average of 5414 and 5563 = 5488.5
P30:
i = (.30)(20) = 6
P30 is located at the average of the 6th and 7th terms
P30 = (4507+4541)/2 = 4524
P60:
i = (.60)(20) = 12
P60 is located at the average of the 12th and 13th terms
P60 = (6101+6498)/2 = 6299.5
P90:
i = (.90)(20) = 18
P90 is located at the average of the 18th and 19th terms
P90 = (9863+11,019)/2 = 10,441
Q1 = P25:
i = (.25)(20) = 5
Q1 is located at the average of the 5th and 6th terms
Q1 = (4464+4507)/2 = 4485.5
Q3 = P75:
i = (.75)(20) = 15
Q3 is located at the average of the 15th and 16th terms
Q3 = (6796+8687)/2 = 7741.5
Range = 11,388 - 3619 = 7769
IQR = Q3 - Q1 = 7741.5 - 4485.5 = 3256
Student’s Solutions Manual and Study Guide: Chapter 3
3.45
a.)
µ=
Page 24
x
= 26,675/11 = 2425
N
Median = 1965
b.)
Range = 6300 - 1092 = 5208
Q3 = 2867
c.)
Q1 = 1532
IQR = Q3 - Q1 = 1335
Variance:
2 =
(x) 2
(26,675) 2
86,942,873 
N 
11
= 2,023,272.55
N
11
x 2 
Standard Deviation:
 =
d.)
 2  2,023,272.55 = 1422.42
Texaco:
z =
x


1532  2425
= -0.63
1422.42
ExxonMobil:
z =
e.)
x


6300  2425
= 2.72
1422.42
Skewness:
Sk =
3(   M d )


3(2425  1965)
= 0.97
1422.42
Student’s Solutions Manual and Study Guide: Chapter 3
Page 25
3.47
f
15-20
9
20-25
16
25-30
27
30-35
44
35-40
42
40-45
23
45-50
7
50-55
2
f = 170
M
fM
fM2
17.5
157.5
2756.25
22.5
360.0
8100.00
27.5
742.5
20418.75
32.5
1430.0
46475.00
37.5
1575.0
59062.50
42.5
977.5
41543.75
47.5
332.5
15793.75
52.5
105.0
5512.50
2
fM = 5680.0 fM = 199662.50
fM 5680

= 33.412
f
170
a.) Mean:  =
Median:
N   f  170
The observation 85, 50% of N = 170, is in the class corresponding to 30 – under 35.
Therefore, l  30 , F  9  16  27  52 , f  44 and w  5 .
N

 170

 52 
 F

 w  30   2
5  30   33  5  33.75
median grouped  l   2
 f 
 44 
 44 








Mode: The Modal Class is 30-35. The class midpoint is the mode = 32.5.
b.) Variance:
(fM ) 2
(5680.0) 2
199,662.50 
n
170
= 58.483

n 1
169
fM 2 
s2 =
Standard Deviation:
s =
3.49
s 2  58.483 = 7.647
CVx =
x
3.45
(100%) 
(100%) = 10.78%
x
32
CVY =
y
5.40
(100%) 
(100%) = 6.43%
y
84
Stock X has a greater relative variability.
Student’s Solutions Manual and Study Guide: Chapter 3
3.51
Page 26
µ = 419,  = 27
a.) 68%:
µ + 1
419 + 27
95%:
µ + 2
419 + 2(27) 365 to 473
99.7%: µ + 3
419 + 3(27) 338 to 500
392 to 446
b.) Use Chebyshev’s:
Each of the points, 359 and 479 is a distance of 60 from the mean, µ = 419.
k = (distance from the mean)/ = 60/27 = 2.22
Proportion = 1 - 1/k2 = 1 - 1/(2.22)2 = .797 = 79.7%
400  419
= -0.704. This worker is in the lower half of
27
workers but within one standard deviation of the mean.
c.) Since x = 400, z =
3.53
Mean
Median
Mode
$30,468
$26,721
$25,000
Since these three measures are not equal, the distribution is skewed. The distribution is
skewed to the right because the mean is greater than the median. Often, the median is
preferred in reporting income data because it yields information about the middle of the
data while ignoring extremes.
3.55
Paris:
Since 1 - 1/k2 = .53, solving for k:
k = 1.459
The distance from µ = 349 to x = 381 is 32
1.459 = 32
 = 21.93
Moscow:
Since 1 - 1/k2 = .83, solving for k:
The distance from µ = 415 to x = 459 is 44
2.425 = 44
 = 18.14
k = 2.425
Student’s Solutions Manual and Study Guide: Chapter 3
Page 27
3.57
First of all, the graph indicates that the data are very skewed to the right. This is
supported by the output which yields a skewness value of 3.6214. This is a very
high positive skewness number telling us that there are a few advertisers in the
Hispanic market who are spending extremely large numbers on advertising while
most of the other companies are spending moderate to less amounts. In fact,
looking at the graph, there is a very high modal group of advertisers centering
around the $ 5 million mark. The output reveals that the mean amount being
spent by Hispanic advertisers is $ 7.8560 million while the median is $ 5.75
million. The gap between the mean and the median supports the notion that the
data are positively skewed. The standard deviation of the advertising data is
5.8860. If we apply the empirical rule to these data using the mean and the
standard deviation, we can see that there are less than one standard deviation
between the mean and the minimum value reinforcing the notion that the data are
not normally distributed (because there should be at least three standard
deviations of data on either side of the mean if the data are approximately
normally distributed) and are likely to be positively skewed. The “N” indicates
that there were 50 companies in this data analysis. The minimum amount spent
on advertising cultivating the Hispanic market by one of these companies is $
3.25 million and the maximum spent is $ 40.00 million. Also, presented in the
data are the first and third quartiles. Observe the gap between the third quartile
and the maximum amount. The “range” of this top 25 percent is over $31 million
(8.625 – 40.00) and this is where the skewness is occurring.
3.59
The N indicates that there are 25 companies in this database. The mean
advertising expenditure is $ 772,702 thousands while the median is $ 613,823
thousands. The gap between the mean and the median indicates that there is
positive skewness. While no skewness figure is given in the MINITAB output,
the boxplot shows extreme outliers on the right or positive side of the data
supporting the conclusion that there is positive skewness. Since the median is
located on the left side of the “box”, the data are likely to be skewed to the right.
The standard deviation is $ 436,067 thousand. Using this figure, it is possible to
see that there is less than one standard deviation between the mean and the
minimum ($ 445,958). If we apply the empirical rule to these data using the mean
and the standard deviation, we can see that the data are not normally distributed
(because there should be at least three standard deviations of data on either side of
the mean if the data are approximately normally distributed) and are likely to be
positively skewed. The values of the first quartile ($ 484,600 thousand) and the
third quartile ($788,256 thousand) are also given and can be used to compute the
interquartile deviation (middle 50% of the data), construct the boxplot, and
indicate advertising expenditures for both the bottom and top 25 percent of
companies.