Chapter 3: Descriptive Statistics
1
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.
8.
Understand box and whisker plots, skewness, and kurtosis.
Compute a coefficient of correlation and interpret it.
CHAPTER TEACHING STRATEGY
In chapter 2, the student learned how to summarize data by constructing
frequency distributions (grouping data) and by using graphical depictions. Much of the
time, statisticians need to describe data by using single numerical measures. Chapter 3
presents a cadre of statistical measures for describing numerically sets of data.
It can be emphasized in this chapter that there are at least two major dimensions
along which data can be described. One is the measure of central tendency with which
statisticians attempt to describe the more central portions of the data. Included here are
the mean, median, mode, percentiles, and quartiles. It is important to establish that the
Chapter 3: Descriptive Statistics
2
median is a useful device for reporting some business data such as income and housing
costs because it tends to ignore the extremes. On the other hand, the mean utilizes every
number of a data set in its computation. This makes the mean an attractive tool in
statistical analysis.
A second major category of descriptive statistical techniques are the measures of
variability. Students can understand that a measure of central tendency is often not
enough to fully describe data. The measure of variability helps the researcher get a
handle on the spread of the data. An attempt is made in this text to communicate to the
student that through the use of the empirical rule and through Chebyshev’s Theorem,
students can better understand what a standard deviation means. The empirical rule will
be referred to quite often throughout the course; and therefore, it is important to
emphasize it as a rule of thumb. For example, in discussing control charts in chapter 17,
the upper and lower control limits are established by using the range of + 3 standard
deviations of the statistic as limits within which virtually all points should fall if the
process is in control. z scores are presented mainly to help bridge the gap between the
discussion of means and standard deviations in chapter 3 and the normal curve of chapter
6.
It should be emphasized that the calculation of measures of central tendency and
variability for grouped data is different than for ungrouped or raw data. While the
principles are the same for the two types of data, the implementation of the formulas are
different. Grouped data are represented by the class midpoints rather than computation
based on raw values. For this reason, the student should be cautioned that group statistics
are often just approximations.
Measures of shape are useful in helping the researcher describe a distribution of
data. The Pearsonian coefficient of skewness is a handy tool for ascertaining the degree
of skewness in the distribution. Box and Whisker plots can be used to determine the
presence of skewness in a distribution and to locate outliers. The coefficient of
correlation is introduced here instead of chapter 13 (regression chapter) so that the
student can begin to think about two variable relationships and analyses. Since the
coefficient of correlation is a stand-alone statistic, it is appropriate to present it here. In
addition, when the student studies simple regression in chapter 13, there will be a
foundation upon which to build. All in all, chapter 3 is quite important because it
presents some of the building blocks for many of the later chapters.
CHAPTER OUTLINE
3.1 Measures of Central Tendency: Ungrouped Data
Mode
Median
Mean
Percentiles
Chapter 3: Descriptive Statistics
3
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
Kurtosis
Box and Whisker Plot
3.5 Measures of Association
Correlation
3.6 Descriptive Statistics on the Computer
KEY TERMS
Arithmetic Mean
Bimodal
Box and Whisker Plot
Chebyshev’s Theorem
Coefficient of Correlation (r)
Coefficient of Skewness
Measures of Variability
Median
Mesokurtic
Mode
Multimodal
Percentiles
Chapter 3: Descriptive Statistics
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
4
Platykurtic
Quartiles
Range
Skewness
Standard Deviation
Sum of Squares of x
Variance
z Score
SOLUTIONS TO 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
3.2
Median for values in 3.1
Arrange in ascending order:
2, 2, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8, 8, 8, 9
There are 15 terms.
Since there are an odd number of terms, the median is the middle number.
The median = 4
Using the formula, the median is located
at the n + 1 th term = 15+1 = 16 = 8th term
2
2
2
The 8th term = 4
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
Chapter 3: Descriptive Statistics
5
Using the formula, the median is located at the n + 1 th term.
2
n = 10 therefore 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
3.4
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).
3.5
Mean
7
-2
5
9
0
-3
-6
-7
-4
-5
2
-8
x = -12
3.6
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 into ascending order:
11, 13, 16, 17, 18, 19, 20, 25, 27, 28, 29, 30, 32, 33, 34
i
35
(15)  5.25
100
P35 is located at the 5 + 1 = 6th term
Chapter 3: Descriptive Statistics
6
P35 = 19
i
55
(15)  8.25
100
P55 is located at the 8 + 1 = 9th term
P55 = 27
Q1 = P25
i
25
(15)  3.75
100
Q1 = P25 is located at the 3 + 1 = 4th term
Q1 = 17
Q2 = Median
 15  1 
th
The median is located at the 
  8 term
 2 
th
Q2 = 25
Q3 = P75
i
75
(15)  11.25
100
Q3 = P75 is located at the 11 + 1 = 12th term
Q3 = 30
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
Chapter 3: Descriptive Statistics
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
7
Chapter 3: Descriptive Statistics
8
Q3 is located at the 18.5th term
Q3 = (143 + 144)/ 2 = 143.5
3.8
Mean =
 x  43,063  2,870.87
N
15
 15  1 
The median is located at the 
 = 8th term
 2 
Median = 2,264
th
Q2 = Median = 2,264
i
63
(15)  9.45
100
P63 is located at the 9 + 1 = 10th term
P63 = 2,646
 10  1 
th
The median is located at the 
  5.5 position
 2 
th
3.9
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
Chapter 3: Descriptive Statistics
9
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
3.10
n = 17
Mean =
 x  61  3.59
N
17
 17  1 
The median is located at the 
 = 9th term
 2 
th
Median = 4
Mode = 4
Q3 = P75:
i=
75
(17)  12.75
100
Q3 is located at the 13th term
Q3 = 4
P11:
i=
11
(17)  1.87
100
Chapter 3: Descriptive Statistics
10
P11 is located at the 2nd term
P11 = 1
P35:
i=
35
(17)  5.95
100
P35 is located at the 6th term
P35 = 3
P58:
i=
58
(17)  9.86
100
P58 is located at the 10th term
P58 = 4
P67:
i=
67
(17)  11.39
100
P67 is located at the 12th term
P67 = 4
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. =
x
N

14.2857
 2.041
7
(x-µ)2
2.9388
5.2244
.0816
22.2246
10.7958
1.6530
.5102
2
(x -µ) = 43.4284
Chapter 3: Descriptive Statistics
c.) 2 =
( x   ) 2 43.4284
= 6.204

N
7
( x   ) 2
 6.204 = 2.491
N
 =
d.)
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.)
11
z =
6  4.2857
= 0.69
2.491
z =
2  4.2857
= -0.92
2.491
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
Chapter 3: Descriptive Statistics
5  4.2857
= 0.29
2.491
z =
3.12
x
xx
4
3
0
5
2
9
4
5
0
1
4
1
2
5
0
1
x = 32
x
12
 x  x = 14
( x  x) 2
0
1
16
1
4
25
0
1
( x  x ) 2 = 48
x 32

=4
n
8
a) Range = 9 - 0 = 9
xx
b) M.A.D. =
n

14
= 1.75
8
( x  x) 2 48

c) s =
= 6.857
n 1
7
2
d) s =
( x  x ) 2
 6.857 = 2.619
n 1
e) Numbers in order: 0, 2, 3, 4, 4, 5, 5, 9
Q1 = P25
i =
25
(8) = 2
100
Q1 is located at the average of the 2nd and 3rd terms,
2.5th term
Q1 = 2.5
Q1 = (2 + 3)/2 =
Chapter 3: Descriptive Statistics
13
Q3 = P75
i =
75
(8) = 6
100
Q3 is located at the average of the 6th and 7th terms
Q3 = (5 + 5)/2 = 5
IQR = Q3 - Q1 = 5 - 2.5 = 2.5
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 127

= 21.167
N
6
( x   ) 2
126.834

 21.139 = 4.598
N
6
b.)
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
x2
144
529
361
676
576
529
2
x = 2815
ORIGINAL FORMULA
Chapter 3: Descriptive Statistics
14
 =
(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.
3.14
s2 = 433.9267
s = 20.8309
x = 1387
x2 = 87,365
n = 25
x = 55.48
3.15
2 = 58,631.359
 = 242.139
x = 6886
x2 = 3,901,664
n = 16
µ = 430.375
3.16
14, 15, 18, 19, 23, 24, 25, 27, 35, 37, 38, 39, 39, 40, 44,
46, 58, 59, 59, 70, 71, 73, 82, 84, 90
Q1 = P25
i =
25
( 25) = 6.25
100
P25 is located at the 6 + 1 = 7th term
Chapter 3: Descriptive Statistics
15
Q1 = 25
Q3 = P75
i =
75
( 25) = 18.75
100
P75 is located at the 18 + 1 = 19th term
Q3 = 59
IQR = Q3 - Q1 = 59 - 25 = 34
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
3.18
Set 1:
1 
1 
x 262

 65.5
N
4
(x) 2
(262) 2
17,970 
N 
4
N
4
x 2 
Set 2:
2 
x 570

 142.5
N
4
= 14.2215
Chapter 3: Descriptive Statistics
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
3.19
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 106

= 8.833
n
12
xx
a) MAD =
16
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
= 14.5344
( 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
Chapter 3: Descriptive Statistics
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 =
6  8.833
= - 0.85
3.326
f.) CV =
(3.326)(100)
= 37.65%
8.833
3.20 n = 11
x
768
429
323
306
286
262
215
172
162
148
145
x = 3216
x-
475.64
136.64
30.64
13.64
6.36
30.36
77.36
120.36
130.36
144.36
147.36
x-µ = 1313.08
µ = 292.36 x = 3216 x2 = 1,267,252
a.) range = 768 - 145 = 623
b.) MAD =
x
N

1313.08
= 119.37
11
17
Chapter 3: Descriptive Statistics
(x) 2
(3216) 2
x 
1,267,252 
N 
11
c.) 2 =
N
11
18
2
d.)  =
= 29,728.23
29,728.23 = 172.42
e.) Q1 = P25:
i = .25(11) = 2.75
Q1 is located at the 2 + 1 = 3rd term
Q1 = 162
Q3 = P75:
i = .75(11) = 8.25
Q3 is located at the 8 + 1 = 9th term
Q3 = 323
IQR = Q3 - Q1 = 323 - 162 = 161
f.) xnestle = 172
Z =
g.) CV =
3.21
x


172  292.36
= -0.70
172.42

172.42
(100) 
(100) = 58.98%

292.36
µ = 125
 = 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:
Chapter 3: Descriptive Statistics
19
µ ± 3 = 125 ± 3(12) = 125 ± 36
between 89 and 161
3.22
=6
µ = 38
between 26 and 50:
x1 - µ = 50 - 38 = 12
x2 - µ = 26 - 38 = -12
x1  

12
= 2
6
x2  

 12
= -2
6


K = 2, and since the distribution is not normal, use Chebyshev’s theorem:
1
1
1
1 3
 1 2  1 
= .75
2
K
2
4 4
at least 75% of the values will fall between 26 and 50
between 14 and 62?
µ = 38
 =6
x1 - µ = 62 - 38 = 24
x2 - µ = 14 - 38 = -24
x1  

24
= 4
6
x2  

 24
= -4
6


K=4
1
1
1
1 15
 1 2  1

= .9375
2
K
4
16 16
at least 93.75% of the values fall between 14 and 62
between what 2 values do at least 89% of the values fall?
Chapter 3: Descriptive Statistics
1-
20
1
=.89
K2
.11 =
1
K2
.11 K2 = 1
K2 =
1
.11
K2 = 9.09
K
= 3.015
With µ = 38,  = 6 and K = 3.015 at least 89% of the values fall within:
µ ± 3.015
38 ± 3.015 (6)
38 ± 18.09
19.91 and 56.09
3.23
1-
1
= .80
K2
1 - .80 =
1
K2
.20 =
1
K2
.20K2
K2 = 5
=1
and
K = 2.236
2.236 standard deviations
3.24
µ = 43. within 68% of the values lie µ + 1
1 = 46 - 43 = 3
Chapter 3: Descriptive Statistics
21
within 99.7% of the values lie µ + 3
3 = 51 - 43 = 8
 = 2.67
µ = 28 and 77% of the values lie between 24 and 32 or + 4 from the mean:
1-
1
= .77
K2
k2 = 4.3478
k = 2.085
2.085 = 4
 =
3.25
4
= 1.918
2.085
=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:
Chapter 3: Descriptive Statistics
x


22
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.
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.
3.26
x
97
109
111
118
120
130
132
133
137
137
x = 1224
x2 = 151,486
Bordeaux: x = 137
z =
137  122.4
= 1.07
13.615
n = 10
x = 122.4
s = 13.615
Chapter 3: Descriptive Statistics
23
Montreal: x = 130
z =
130  122.4
= 0.56
13.615
Edmonton: x = 111
z =
111  122.4
= -0.84
13.615
Hamilton: x = 97
z =
3.27
97  122.4
= -1.87
13.615
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
fM 561

= 4.64
f
121
Mode: The modal class is 0 – 2.
The midpoint of the modal class = the mode = 1
3.28
Class
1.2 - 1.6
1.6 - 2.0
2.0 - 2.4
2.4 - 2.8
2.8 - 3.2
Mean:
f
220
150
90
110
280
f=850
 =
fM 1902

= 2.24
f
850
M
1.4
1.8
2.2
2.6
3.0
fM
308
270
198
286
840
fM=1902
Chapter 3: Descriptive Statistics
Mode:
3.29
The modal class is 2.8 – 3.2.
The midpoint of the modal class is the mode = 3.0
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
 =
M-µ
-22.0339
-12.0339
- 2.0339
7.9661
17.9661
27.9661
24
2775
fM 2775

= 47.034
f
59
(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
2
3.30
 =
185.694
Class
5- 9
9 - 13
13 - 17
17 - 21
21 - 25
f
20
18
8
6
2
f=54
= 13.627
M
7
11
15
19
23
fM
140
198
120
114
46
fM= 618
fM2
980
2,178
1,800
2,166
1,058
fm2= 8,182
(fM ) 2
(618) 2
fM 
8182 
n
54  8182  7071.7 = 20.9
s2 =

n 1
53
53
2
Chapter 3: Descriptive Statistics
S 2  20.9 = 4.57
s =
3.31
25
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 
M
21
27
33
39
45
51
57
63
69
fM
fM2
357
7,497
594
16,038
858
28,314
1,365
53,235
1,485
66,825
1,530
78,030
1,824
103,968
1,323
83,349
1,035
71,415
fM= 10,371 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 =
187.2 = 13.7
3.32
a.) Mean
Class
0-1
1-2
2-3
3-4
4-5
5-6
 =
f
31
57
26
14
6
3
f=137
fM 258.5

= 1.89
f
137
M
fM
fM2
0.5
1.5
2.5
3.5
4.5
5.5
15.5
85.5
65.0
49.0
27.0
16.5
fM=258.5
7.75
128.25
162.50
171.50
121.50
90.75
2
fM =682.25
Chapter 3: Descriptive Statistics
b.) Mode:
26
Modal Class = 1-2. Mode = 1.5
c.) Variance:
2 =
(fM ) 2
(258.5) 2
682.25 
N
137

N
137
fM 2 
= 1.4197
d.) standard Deviation:
 =
3.33
f
20-30
8
30-40
7
40-50
1
50-60
0
60-70
3
70-80
1
f = 20
M
 2  1.4197 = 1.1915
fM2
fM
25
200
5000
35
245
8575
45
45
2025
55
0
0
65
195
12675
75
75
5625
2
fM = 760 fM = 33900
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:
 =
 2  251 = 15.843
Chapter 3: Descriptive Statistics
3.34
No. of Farms
f
0 - 20,000
20,000 - 40,000
40,000 - 60,000
60,000 - 80,000
80,000 - 100,000
100,000 - 120,000
16
11
10
6
5
1
f = 49
 =
M
10,000
30,000
50,000
70,000
90,000
110,000
27
fM
160,000
330,000
500,000
420,000
450,000
110,000
fM2 = 1,970,000
fM 1,970,000

= 40,204
f
49
The actual mean for the ungrouped data is 37,816. This computed group
mean, 40,204, is really just an approximation based on using the class
midpoints in the calculation. Apparently, the actual numbers of farms per
state in some categories do not average to the class midpoint and in fact
might be less than the class midpoint since the actual mean is less than the
grouped data mean.
fM2 = 1.185x1011
2 =
(fm) 2
(1.97 x106 ) 2
1.18 x1011 
N 
49
= 801,999,167
N
49
fM 2 
 = 28,319.59
The actual standard deviation was 29,341. The difference again is due to
the grouping of the data and the use of class midpoints to represent the
data. The class midpoints due not accurately reflect the raw data.
3.35
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.
3.36
mean = 51
median = 54
Chapter 3: Descriptive Statistics
28
mode = 59
The distribution is skewed to the left. More people are older but the most
extreme ages are younger ages.
3.37
Sk =
3.38
n = 25
3(   M d )

3(5.51  3.19)
9.59
= 0.726
x = 600
x2 = 15,462
x = 24
Sk =

s = 6.6521
3( x  M d ) 3(24  23)

= 0.451
s
6.6521
There is a slight skewness to the right
3.39
Q1 = 500.
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.
3.40
n = 18
i =
Q1 = P25:
25
(18) = 4.5
100
Q1 = 5th term =
66
Q3 = P75:
i =
75
(18) = 13.5
100
Chapter 3: Descriptive Statistics
Q3 = 14th term =
29
90
(n  1)th (18  1)th 19


= 9.5th term
2
2
2
th
Median:
Median = 74
IQR = Q3 - Q1 = 90 - 66 = 24
Inner Fences: Q1 - 1.5 IQR = 66 - 1.5 (24) = 30
Q3 + 1.5 IQR = 90 + 1.5 (24) = 126
Outer Fences: Q1 - 3.0 IQR = 66 - 3.0 (24) = -6
Q3 + 3.0 IQR = 90 + 3.0 (24) = 162
There are no extreme outliers. The only mild outlier is 21. The
distribution is positively skewed since the median is nearer to Q1 than Q3.
3.41
x = 80
y2 = 815
r =
x2 = 1,148
xy = 624
 xy 

( x)
2
 x 
n

2
 x y
n

( y )
2
  y 
n
 
(80)(69)
7
=
2

(80)  
(69) 2 
1,148 
 815 

7 
7 

624 
r =
r =
y = 69
n=7
 164.571
= -0.927
177.533
2



=
 164.571
=
(233.714)(134.857)
Chapter 3: Descriptive Statistics
3.42
x = 1,087
y2 = 878,686
r =
30
x2 = 322,345
xy= 507,509
 xy 
y = 2,032
n=5
 x y
n

( x)  
( y ) 2 
2
2
 x 
  y 

n  
n 

=
2
(1,087)( 2,032)
5
=
2
2




(1,087)
(2,032)
322,345 
 878,686 

5 
5


507,509 
r =
r =
3.43
65,752.2
65,752.2
=
= .975
67,449.5
(86,031.2)(52,881.2)
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 =
 xy 
y = 96.5
xy = 4,870.11
2
y = 1,557.91
 x y
n

( x)  
( y ) 2 
2
2
 x 
  y 

n  
n 

2
=
Chapter 3: Descriptive Statistics
31
(302.2)(96.5)
6
= .6445
2

(302.2)  
(96.5) 2 
15,259.62 
 1,557.91 

6
6 


4,870.11 
r =
3.44
x = 6,087
y = 1,050
xy = 1,130,483
r =
x2 = 6,796,149
y2 = 194,526
n=9
 xy 

( x)
2
 x 
n

2
 x y
n

( y )
2
  y 
n
 
2



=
(6,087)(1,050)
9
=
2

(6,087)  
(1,050)2 
6,796,149 
 194,526 

9
9 


1,130,483 
r =
r =
3.45
420,333
420,333
=
= .957
439,294.705
(2,679,308)(72,026)
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) 2  
( y ) 2 
2
2
x

y


 

n  
n 

=
Chapter 3: Descriptive Statistics
32
(17.09)(15.12)
8
=
2

(17.09)  
(15.12) 2 
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)  
( y ) 2 
2
2
 x 
  y 

n  
n 

=
2
(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 y
n

( x)  
( y ) 2 
2
2
 x 
  y 

n  
n 

2
=
Chapter 3: Descriptive Statistics
33
(17.09)(15.86)
8
2

(17.09)  
(15.86) 2 
58
.
7911

42
.
0396




8 
8 

48.5827 
r =
r =
14.702
14.702
=
= .957
15.367
(2.2826)(10.5972)
The years 2 and 3 are the most correlated with r = .985.
3.46
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
x 75

= 2.5
n 30
Mode = 2 (There are eleven 2’s)
Mean:
x
Median: There are n = 30 terms.
n 1
30  1 21


The median is located at
= 15.5th position.
2
2
2
th
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:
25
(30) = 7.5
100
i =
Q1 is the 8th term =
1
Q3 = P75:
i =
75
(30) = 22.5
100
Chapter 3: Descriptive Statistics
Q3 is the 23rd term = 3
IQR = Q3 - Q1 = 3 - 1 = 2
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
34
Chapter 3: Descriptive Statistics
3.48
µ =
35
x 126,904

= 6345.2
N
20
The median is located at the (n+1)/2 th value = 21/2 = 10.5th value
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
Chapter 3: Descriptive Statistics
3.49
x = 9,332,908
n = 10
36
x2 = 1.06357x1013
µ = (x)/N = 9,332,908/10 = 933,290.8
(x) 2
(9332908) 2
13
x 
1.06357 x10 
N 
10
N
10
2
 =
= 438,789.2
3.50
a.)
µ=
x
= 26,675/11 = 2425
N
Median = 1965
b.)
Range = 6300 - 1092 = 5208
Q3 = 2867
Q1 = 1532
c.)
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 =
x


6300  2425
= 2.72
1422.42
Chapter 3: Descriptive Statistics
e.)
37
Skewness:
Sk =
3(   M d )


3(2425  1965)
= 0.97
1422.42
3.51 a.)
x 20,310

= 2,031
n
10
Mean:

Median:
1670  1920
= 1795
2
Mode:
No Mode
b.) Range:
3350 – 1320 = 2030
Q1:
1
(10)  2.5
4
Located at the 3rd term. Q1 = 1570
Q3:
3
(10)  7.5
4
Located at the 8th term.
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

2
5252
= 525.2
10
Q3 = 2550
Chapter 3: Descriptive Statistics
s =
 x  x 
s=
s 2  441,387.78 = 664.37
38
2
2
n 1

3,972,490
= 441,387.78
9
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:
Inner Fences:
1320 and 3350
1570 – 1.5(980) = 100
2550 + 1.5(980) = 4020
Outer Fences: 1570 + 3.0(980) = -1370
2550+ 3.0(980) = 5490
No apparent outliers
1500
2500
$ Millions
3500
Chapter 3: Descriptive Statistics
3.52
15-20
20-25
25-30
30-35
35-40
40-45
45-50
50-55
39
f
M
fM
fM2
9
16
27
44
42
23
7
2
17.5
22.5
27.5
32.5
37.5
42.5
47.5
52.5
157.5
360.0
742.5
1430.0
1575.0
977.5
332.5
105.0
2756.25
8100.00
20418.75
46475.00
59062.50
41543.75
15793.75
5512.50
fM = 5680.0
fM2 = 199662.50
f = 170
a.) Mean:
 =
fM 5680

= 33.412
f
170
Mode: The Modal Class is 30-35. The class midpoint is the mode = 32.5.
b.) Variance:
2 =
(fM ) 2
(5680) 2
199,662.5 
N
170
= 58.139

N
170
fM 2 
standard Deviation:
 =
3.53
Class
0 - 20
20 - 40
40 - 60
60 - 80
80 - 100
f
32
16
13
10
19
f = 90
 2  58.139 = 7.625
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
Chapter 3: Descriptive Statistics
40
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:
(fM ) 2
n

n 1
fM 2 
s =

253,000 
89
(3860) 2
90

87,448.9
 982.571
89
= 31.346
3.54
x = 36
y = 44
xy = 188
r =
x2 = 256
y2 = 300
n =7
 xy 
 x y
n

( x)  
( y ) 2 
2
2
 x 
  y 

n  
n 

=
2
(36)( 44)
7
=
2
2




(36)
(44)
256 
 300 

7 
7 

188 
r =
r =
 38.2857
 38.2857
=
= -.940
40.7441
(70.8571)( 23.42857)
253,000  165,551.1
89
Chapter 3: Descriptive Statistics
3.55
41
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.
3.56
µ = 7.5
From the Empirical Rule: 99.7% of the values lie in µ + 3 = 7.5 + 3
3 = 14 - 7.5 = 6.5
 = 2.167
suppose that µ = 7.5,  = 1.7:
95% lie within µ + 2 = 7.5 + 2(1.7) = 7.5 + 3.4
Between 4.1 and 10.9
3.57
µ = 419,  = 27
a.) 68%:
µ + 1
419 + 27
392 to 446
95%:
µ + 2
419 + 2(27) 365 to 473
99.7%: µ + 3
419 + 3(27) 338 to 500
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%
Chapter 3: Descriptive Statistics
42
400  419
= -0.704. This worker is in the lower half of
27
workers but within one standard deviation of the mean.
c.)
x = 400. z =
3.58
x
x2
Albania
1,650
2,722,500
Bulgaria
4,300
18,490,000
Croatia
5,100
26,010,000
Germany 22,700
515,290,000
x=33,750 x2= 562,512,500
 =
 =
x 33,750

= 8,437.50
N
4
(x) 2
(33,750) 2
562,512,500 
N 
4
N
4
x 2 
= 8332.87
b.)
x
x2
Hungary
7,800
60,840,000
Poland
7,200
51,840,000
Romania
3,900
15,210,000
Bosnia/Herz 1,770
3,132,900
2
x=20,670 x =131,022,900
 =
 =
x 20,670

= 5,167.50
N
4
(x) 2
(20,670) 2
131,022,900 
N 
4
N
4
x 2 
= 2,460.22
c.)
CV1 =
1
8,332.87
(100) 
(100) = 98.76%
1
8,437.50
CV2 =
2
2,460.22
(100) 
(100) = 47.61%
2
5,167.50
The first group has a much larger coefficient of variation
Chapter 3: Descriptive Statistics
3.59
Mean
Median
Mode
43
$35,748
$31,369
$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.
3.60
x = 36.62
y = 57.23
xy = 314.9091
r =
x2 = 217.137
y2 = 479.3231
n =8
 xy 

( x)
2
 x 
n

2
 x y
n

( y ) 2 
2
y

 

n 
 
=
(36.62)(57.23)
8
=
2

(36.62)  
(57.23) 2 
217.137 
 479.3231 

8
8 


314.9091 
r =
r =
52.938775
= .90
(49.50895)(69.91399)
There is a strong positive relationship between the inflation rate and
the thirty-year treasury yield.
3.61
a.)
Q1 = P25:
i =
25
( 20) = 5
100
Q1 = 5.5th term = (42.3 + 45.4)/2 = 43.85
Q3 = P75:
Chapter 3: Descriptive Statistics
i =
44
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.
Chapter 3: Descriptive Statistics
20
120
220
Tonnage
3.62
Paris:
1 - 1/k2 = .53
k = 1.459
The distance from µ = 349 to x = 381 is 32
1.459 = 32
 = 21.93
Moscow:
1 - 1/k2 = .83
k = 2.425
The distance from µ = 415 to x = 459 is 44
2.425 = 44
 = 18.14
45