Chapter 3
Descriptive Statistics
LEARNING OBJECTIVES
The focus of Chapter 3 is on the use of statistical techniques to describe data, thereby enabling you to:
1. Distinguish between measures of central tendency, measures of variability, and measures of shape.
2. Understand conceptually the meanings of mean, median, mode, quartile, percentile, and range.
3. Compute mean, median, mode, percentile, quartile, range, variance, standard deviation, and mean
absolute deviation on ungrouped data.
4. Differentiate between sample and population variance and standard deviation.
5. Understand the meaning of standard deviation as it is applied using the empirical rule and
Chebyshev’s theorem.
6. Compute the mean, median, standard deviation, and variance on grouped data.
7. Understand box and whisker plots, skewness, and kurtosis.
8. Compute a coefficient of correlation and interpret it.
CHAPTER OUTLINE
3.1
Measures of Central Tendency: Ungrouped Data
Mode
Median
Mean
Percentiles
Steps in Determining the Location of a Percentile 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
Mode
Measures of Variability
3.4
Measures of Shape
Skewness
Skewness and the Relationship of the Mean, Median, and Mode
Coefficient of Skewness
25
26 Solutions Manual and Study Guide
Kurtosis
Box and Whisker Plots
3.5
Measures of Association
Correlation
3.6
Descriptive Statistics on the Computer
KEY WORDS
arithmetic mean
bimodal
box and whisker plot
Chebyshev’s theorem
coefficient of correlation (r)
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
STUDY QUESTIONS
1.
Statistical measures used to yield information about the center 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 _______________.
Chapter 3: Descriptive Statistics 27
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
12.
Frequency
14
17
29
31
18
Measures of variability describe the _______________ of a set of data.
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 _______________.
28 Solutions Manual and Study Guide
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.
28.
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 median of the data is _________________________.
35. The mode 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 __________________________.
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
__________________.
Chapter 3: Descriptive Statistics 29
42.
_______________ is a measure of the degree of relatedness of two variables.
43.
The Pearson product-moment correlation coefficient is denoted by _______________.
44.
The value of r varies from _________________________.
45.
Perfect positive correlation results in an r value of _______________.
46.
The value of the coefficient of correlation from the following data is _______________.
x: 19, 20, 26, 31, 34, 45, 45, 51
y: 78, 100, 125, 120, 119, 130, 145, 143
47.
The value of r from the following data is __________.
x: –10, –6, 1, 4, 15
y: –26, –44, –36, –39, –43
30 Solutions Manual and Study Guide
ANSWERS TO STUDY QUESTIONS
1. Measures of Central Tendency
25. 20.25
2. Mean
26. 3.897
3. Mode
27. Chebyshev’s Theorem
4. 55.5th
28. 95, 105, and 195
5. Middle
29. 2, 75
6. 6
30. –1.25, 0.83
7. 5
31. 28.38%
8. 4.54
32. 7.54
9. 
33. 2.72
10. Class Midpoint
34. 7.7143
11. 58.11
35. 9
12. Spread or Dispersion
36. Right
13. 72
37. Positively
14. Q1 = 32.5, Q2 = 45.5, Q3 = 55
38. Kurtosis
15. IQR = 22.5
39. Platykurtic, Mesokurtic,
Leptokurtic
16. P34 = 37
17. Sk = -0.018
40. Q1 – 1.5 IQR and Q3 + 1.5 IQR
Q1 – 3.0 IQR and Q3 + 3.0 IQR
18. Absolute Value
41. Outliers
19. Deviations
42. Correlation
20. Zero
43. r
21. Square
44. –1 to 0 to +1
22. N, n – 1
45. +1
23. 81
46. .876
24. 3.25
47. –.581
Chapter 3: Descriptive Statistics 31
SOLUTIONS TO ODD-NUMBERED PROBLEMS IN CHAPTER 3
3.1
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
3.3
Median
Arrange terms in ascending order:
073, 167, 199, 213, 243, 345, 444, 524, 609, 682
There are 10 terms.
Since there are an even number of terms, the median is the average of the two middle terms:
Median =
243  345 588

= 294
2
2
Using the formula, the median is located at the
n = 10 therefore
n 1
th term.
2
10  1 11

= 5.5th term.
2
2
The median is located halfway between the 5th and 6th terms.
5th term = 243
6th term = 345
Halfway between 243 and 345 is the median = 294
3.5
Mean
7
–2
5
9
0
–3
–6
–7
–4
–5
2
–8
x = –12
3.7
µ = x/N = –12/12 = –1
x = x/n = –12/12 = –1
(It is not stated in the problem whether the
data represent a population or a sample).
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
32 Solutions Manual and Study Guide
i
20
(24)  4.8
100
P20 is located at the 4 + 1 = 5th term
P20 = 107
i
47
(24)  11.28
100
P47 is located at the 11 + 1 = 12th term
P47 = 127
i
83
(24)  19.92
100
P83 is located at the 19 + 1 = 20th term
P83 = 145
Q1 = P25
i
25
(24)  6
100
Q1 is located at the 6.5th term
Q1 = (112 + 116)/ 2 =
114
Q2 = Median
The median is located at the:
 24  1 
th

  12.5 term
 2 
th
Q2 = (127 + 128)/ 2 =
127.5
Q3 = P75
i
75
(24)  18
100
Q3 is located at the 18.5th term
Q3 = (143 + 144)/ 2 = 143.5
Chapter 3: Descriptive Statistics 33
 10  1 
th
3.9 The median is located at the 
  5.5 position
 2 
th
The median = (595 + 653)/2 = 624
Q3 = P75:
i
75
(10)  7.5
100
P75 is located at the 7+1 = 8th term
Q3 = 751
For P20:
i
20
(10)  2
100
P20 is located at the 2.5th term
P20 = (483 + 489)/2 = 486
For P60:
i
60
(10)  6
100
P60 is located at the 6.5th term
P60 = (653 + 701)/2 = 677
For P80:
i
80
(10)  8
100
P80 is located at the 8.5th term
P80 = (751 + 800)/2 = 775.5
For P93:
i
93
(10)  9.3
100
P93 is located at the 9+1 = 10th term
P93 = 1096
34 Solutions Manual and Study Guide
3.11
x
6
2
4
9
1
3
5
x = 30
x -µ
6-4.2857 = 1.7143
2.2857
0.2857
4.7143
3.2857
1.2857
0.7143
x-µ = 14.2857

x 30

 4.2857
N
7
a.) Range = 9 – 1 = 8
b.) M.A.D. =
c.) 2 =
x
N

14.2857
 2.041
7
( x   ) 2 43.4284

= 6.204
N
7
d.)  =
( x   ) 2
 6.204 = 2.491
N
e.) 1, 2, 3, 4, 5, 6, 9
Q1 = P25
i =
25
(7) = 1.75
100
Q1 is located at the 1 + 1 = 2th term, Q1 = 2
Q3 = P75:
i =
75
(7) = 5.25
100
Q3 is located at the 5 + 1 = 6th term, Q3 = 6
IQR = Q3 – Q1 = 6 – 2 = 4
f.)
z =
6  4.2857
= 0.69
2.491
z =
2  4.2857
= –0.92
2.491
(x-µ)2
2.9388
5.2244
.0816
22.2246
10.7958
1.6530
.5102
(x -µ)2 = 43.4284
Chapter 3: Descriptive Statistics 35
z =
4  4.2857
= –0.11
2.491
z =
9  4.2857
= 1.89
2.491
z =
1  4.2857
= –1.32
2.491
z =
3  4.2857
= –0.52
2.491
z =
5  4.2857
= 0.29
2.491
3.13
a.)
x
12
23
19
26
24
23
x = 127
 =
 =
(x – µ)
12-21.167 = –9.167
1.833
–2.167
4.833
2.833
1.833
(x – µ) = –0.002
(x – µ)2
84.034
3.360
4.696
23.358
8.026
3.360
2
(x – µ) = 126.834
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
x 
N 
N
2
= 4.598
(127) 2
2815 
6

6
SHORT-CUT FORMULA
2815  2688.17
126.83

 21.138
6
6
36 Solutions Manual and Study Guide
The short-cut formula is faster.
3.15
2 = 58,631.359
 = 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
b)
1
1
1
 1
 .84
2
2.5
6.25
c)
1
1
1
 1
 .609
2
1.6
2.56
d)
1
1
1
 1
 .902
2
3.2
10.24
3.19
x
x
xx
( x  x) 2
7
5
10
12
9
8
14
3
11
13
8
6
1.833
3.833
1.167
3.167
0.167
0.833
5.167
5.833
2.167
4.167
0.833
2.833
3.361
14.694
1.361
10.028
0.028
0.694
26.694
34.028
4.694
17.361
0.694
8.028
106
32.000
121.665
x 106

= 8.833
n
12
Chapter 3: Descriptive Statistics 37
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:
Q3 = (11 + 12)/2 = 11.5
IQR = Q3 – Q1 = 11.5 – 9 = 2.5
e.) z =
f.) CV =
3.21
µ = 125
6  8.833
= – 0.85
3.326
(3.326)(100)
= 37.65%
8.833
 = 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
38 Solutions Manual and Study Guide
between 89 and 161
1
= .80
K2
1–
3.23
1 – .80 =
1
K2
.20 =
1
K2
.20K2
K2 = 5
=1
and
K = 2.236
2.236 standard deviations
3.25
=4
µ = 29
x1 – µ = 21 – 29 = –8
x2 – µ = 37 – 29 = +8
x1  

8
= –2 Standard Deviations
4
x2  

8
= 2 Standard Deviations
4


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
between 17 and 41, 99.7% of the values will fall.
0.3% will fall outside this range.
Chapter 3: Descriptive Statistics 39
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
from 25 to 33.
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 561

= 4.64
f
121
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
7
11
18
13
6
4
25
35
45
55
65
75
175
385
810
715
390
300
Total
59
 =
2775
fM 2775

= 47.034
f
59
fM
39
81
80
105
90
88
78
fM=561
40 Solutions Manual and Study Guide
M–µ
–22.0339
–12.0339
– 2.0339
7.9661
17.9661
27.9661
2 =
(M – µ)2
485.4927
144.8147
4.1367
63.4588
322.7808
782.1028
f ( M   ) 2 10,955.93
= 185.694

f
59
 =
3.31
f(M – µ)2
3398.449
1592.962
74.462
824.964
1936.685
3128.411
Total 10,955.933
Class
18 - 24
24 - 30
30 - 36
36 - 42
42 - 48
48 - 54
54 - 60
60 - 66
66 - 72
185.694 = 13.627
f
17
22
26
35
33
30
32
21
15
f= 231
a.) Mean: x 
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
= 44.9


n
f
231
b.) Mode. The Modal Class = 36 – 42.
The mode is the class midpoint = 39
(fM ) 2
(10,371) 2
508,671 
n
231  43,053.5 = 187.2

n 1
230
230
fM 2 
c.) s2 =
d.) s =
3.33
20-30
30-40
40-50
50-60
60-70
70-80
187.2 = 13.7
f
M
8
7
1
0
3
1
f = 20
25
35
45
55
65
75
fM
200
245
45
0
195
75
fM = 760
fM2
5000
8575
2025
0
12675
5625
fM2 = 33900
Chapter 3: Descriptive Statistics 41
a.) Mean:
 =
fM 760

= 38
f
20
b.) Mode. The Modal Class = 20 - 30. The mode is the
midpoint of this class = 25.
c.) Variance:
2 =
(fM ) 2
(760) 2
33,900 
N
20
= 251

N
20
fM 2 
d.) Standard Deviation:
 =
3.35
 2  251 = 15.843
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
Sk =
3(   M d )

3.39 Q1 = 500.

3(5.51  3.19)
= 0.726
9.59
Median = 558.5.
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.
3.41
x = 80
y2 = 815
r =
x2 = 1,148
xy = 624
 xy 
 x y
n

( x)  
( y ) 2 
2
2
 x 
  y 

n  
n 

2
y = 69
n=7
=
42 Solutions Manual and Study Guide
(80)(69)
7
=
2

(80)  
(69) 2 
1,148 
 815 

7 
7 

624 
r =
r =
3.43
 164.571
= –0.927
177.533
Delta (x)
SW (y)
47.6
46.3
50.6
52.6
52.4
52.7
15.1
15.4
15.9
15.6
16.4
18.1
x = 302.2
x2 = 15,259.62
r =
 164.571
=
(233.714)(134.857)
 xy 
y = 96.5
y2 = 1,557.91
xy = 4,870.11
 x y
n

( x)  
( y ) 2 
2
2
 x 
  y 

n  
n 

=
2
(302.2)(96.5)
6
= .6445
2

(302.2)  
(96.5) 2 
15,259.62 
 1,557.91 

6 
6 

4,870.11 
r =
3.45
Correlation between Year 1 and Year 2:
x = 17.09
y = 15.12
xy = 48.97
r =
x2 = 58.7911
y2 = 41.7054
n=8
 xy 
 x y
n

( x)  
( y ) 2 
2
2
 x 
  y 

n  
n 

2
=
Chapter 3: Descriptive Statistics 43
(17.09)(15.12)
8
=
2
2

(17.09)  
(15.12) 
58.7911 
 41.7054 

8 
8 

48.97 
r =
r =
16.6699
16.6699
=
= .975
17.1038
(22.28259)(13.1286)
Correlation between Year 2 and Year 3:
x = 15.12
x2 = 41.7054
y = 15.86
y2 = 42.0396
xy = 41.5934
n=8
r =
 xy 
 x y
n

( x) 2  
( y ) 2 
2
2
x

y


 

n  
n 

=
(15.12)(15.86)
8
=
2
2

(15.12)  
(15.86) 
41.7054 
 42.0396 

8 
8 

41.5934 
r =
r =
11.618
11.618
=
= .985
11.795
(13.1286)(10.59715)
Correlation between Year 1 and Year 3:
x = 17.09
y = 15.86
xy = 48.5827
r =
x2 = 58.7911
y2 = 42.0396
n=8
 xy 

( x)
2
 x 
n

2
 x y
n

( y ) 2 
2
y

 

n 
 
(17.09)(15.86)
8
2

(17.09)  
(15.86) 2 
58
.
7911

42
.
0396




8 
8 

48.5827 
r =
=
44 Solutions Manual and Study Guide
r =
14.702
14.702
=
= .957
15.367
(2.2826)(10.5972)
The years 2 and 3 are the most correlated with r = .985.
3.47
P10:
i =
P10 = 4.5th term =
10
( 40) = 4
100
23
P80:
i =
P80 = 32.5th term =
80
( 40) = 32
100
49.5
Q1 = P25:
i =
P25 = 10.5th term =
25
( 40) = 10
100
27.5
Q3 = P75:
i =
P75 = 30.5th term =
75
( 40) = 30
100
47.5
IQR = Q3 – Q1 = 47.5 – 27.5 = 20
Range = 81 – 19 = 62
3.49
x = 9,332,908
n = 10
x2 = 1.06357×1013
µ = (x)/N = 9,332,908/10 = 933,290.8
(x) 2
(9332908) 2
1.06357 x1013 
N 
10
N
10
x 2 
 =
3.51 a.) Mean:
Median:

x 20,310

= 2,031
n
10
1670  1920
= 1795
2
= 438,789.2
Chapter 3: Descriptive Statistics 45
Mode:
No Mode
3350 – 1320 = 2030
b.) Range:
1
(10)  2.5
4
Q1:
3
(10)  7.5
4
Q3:
Located at the 3rd term. Q1 = 1570
Located at the 8th term.
Q3 = 2550
IQR = Q3 – Q1 = 2550 – 1570 = 980
x
xx
x  x 
3350
2800
2550
2050
1920
1670
1660
1570
1420
1320
1319
769
519
19
111
361
371
461
611
711
5252
1,739,761
591,361
269,361
361
12,321
130,321
137,641
212,521
373,321
505,521
3,972,490
xx
MAD =
n

 x  x 
=
2
5252
= 525.2
10
2
s
2
s=
n 1

3,972,490
= 441,387.78
9
s 2  441,387.78 = 664.37
c.) Pearson’s Coefficient of skewness:
Sk =
3( x  M d ) 3(2031  1795)

= 1.066
s
664.37
d.) Use Q1 = 1570, Q2 = 1795, Q3 = 2550, IQR = 980
Extreme Points:
1320 and 3350
Inner Fences: 1570 – 1.5(980) = 100
2550 + 1.5(980) = 4020
Outer Fences: 1570 + 3.0(980) = –1370
2550+ 3.0(980) = 5490
46 Solutions Manual and Study Guide
No apparent outliers
1500
2500
3500
$ Millions
3.53
Class
0 - 20
20 - 40
40 - 60
60 - 80
80 - 100
f
32
16
13
10
19
f = 90
M
10
30
50
70
90
fM
320
480
650
700
1,710
fM= 3,860
fM2
3,200
14,400
32,500
49,000
153,900
fM2= 253,000
a) Mean:
x
fM fm 3,860
= 42.89


n
f
90
Mode: The Modal Class is 0-20.
The midpoint of this class is the mode = 10.
b) Standard Deviation:
s =
(fM ) 2
n

n 1
fM 2 

253,000 
87,448.9
 982.571
89
= 31.346
89
(3860) 2
90

253,000  165,551.1
89
Chapter 3: Descriptive Statistics 47
3.55
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.
3.57
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:
The distance from 359 to 479 is 120
µ = 419 The distance from the mean to the limit is 60.
k = (distance from the mean)/ = 60/27 = 2.22
Proportion = 1 – 1/k2 = 1 – 1/(2.22)2 = .797 = 79.7%
c.)
x = 400. z =
400  419
= –0.704. This worker is in the lower half of workers but
27
within one standard deviation of the mean.
3.59
Mean
$35,748
Median $31,369
Mode
$29,500
Since these three measures are not equal, the distribution is skewed. The distribution is skewed to
the right. Often, the median is preferred in reporting income data because it yields information
about the middle of the data while ignoring extremes.
3.61
a.)
Q1 = P25:
i =
25
( 20) = 5
100
Q1 = 5.5th term = (42.3 + 45.4)/2 = 43.85
Q3 = P75:
48 Solutions Manual and Study Guide
i =
75
( 20) = 15
100
Q3 = 15.5th term = (69.4 +78.0)/2 = 73.7
Median:
n 1
20  1

= 10.5th term
2
2
th
th
Median = (52.9 + 53.4)/2 = 53.15
IQR = Q3 – Q1 = 73.7 – 43.85 = 29.85
1.5 IQR = 44.775;
3.0 IQR = 89.55
Inner Fences:
Q1 – 1.5 IQR = 43.85 – 44.775 = –0.925
Q3 + 1.5 IQR = 73.7 + 44.775 = 118.475
Outer Fences:
Q1 – 3.0 IQR = 43.85 – 89.55 = –45.70
Q3 + 3.0 IQR = 73.7 + 89.55 = 163.25
b.) and c.) There are no outliers in the lower end. There is one extreme outlier in the upper end
(214.2). There are two mile outliers at the upper end (133.7 and 158.8). since the median is
nearer to Q1, the distribution is positively skewed.
d.) There are three dominating, large ports
Displayed below is the MINITAB boxplot for this problem.
20
120
Tonnage
220