Measures of Dispersion
BAN 530
1. For the following data set find the range, mean absolute deviation, variance and standard
deviation.
5, 12, 3, 4, 4
2. For the following data set find the range, mean absolute deviation, variance and standard
deviation.
-1, 3, 2, 6, 2, 1, 0, 0
3. A sample of twelve cars on IH 45 was selected and the speed of each was measured by
radar. The speeds (in miles per hour) are:
63 65 67 64 66 72 78 80 66 82 78 83
a) Find the mean.
b) Find the median.
c) Find the range.
d) Find the variance.
e) Find the standard deviation.
f) Find the mean absolute deviation.
g) Find the Z-score for a speed of 74 mph. Interpret.
4. A bank examiner recently audited eight banks. The return on investments (%) for each of
the banks is listed below.
-1 5 8 2 6 12 7 6
a)
b)
c)
d)
Find the mean.
Find the median.
Find the standard deviation.
Find the Z-score for a bank with a return on investment of 8%. Interpret.
5. Many undergraduate students change their major several times before they graduate. A
sample of eight recent SHSU graduates was obtained. Each graduate was asked how many
times he/she changed majors. The data are listed below.
a)
b)
c)
d)
5, 8, 2, 0, 0, 1, 6, 6
Find the mean.
Find the median.
Find the standard deviation.
Find the z-score for a graduate who changed majors 4 times. Interpret.
Measures of Dispersion
BAN 530
6.
A sample of eight runs from a production run of 1000 items yielded the following number
of defective items: 6, 12, 14, 8, 10, 12, 7, 7. Find the
a) mean
b) variance
c) standard deviation
d) z-score for a production run with 10 defective items. Interpret.
7.
A sample of six new graduates produced the following starting salaries (in $1,000s): 38.5,
42.1, 29.5, 30.1, 45.7, 27.3. Find the
a) mean
b) variance
c) standard deviation
d) z-score for a starting salary of $32,000. Interpret.
8.
A sample of returns for ten portfolios managed by a particular financial manager produced
the following returns (in %): 6.5, 12.2, 3.5, 15.6, 10.4, 8.5, 7.8, 9.3, 11.1, 17.3. Find the
a) mean
b) variance
c) standard deviation
d) z-score for a portfolio with a return of 8.4%. Interpret.
9.
Records revealed the below listed probability distribution for the number of defective items
produced per day. Compute the expected number of defective items produced per day and
the standard deviation.
X, No. of defective items
P(X = x)
0
0.30
1
0.25
2
0.20
3
0.15
4
0.10
Measures of Dispersion
BAN 530
10. Suppose that a car salesman uses past records to determine the probability distribution for
weekly sales of luxury cars. The distribution is tabled below. How many luxury cars does
the salesman expect to sell per week. What is the standard deviation?
X, Sales per Week
P(X)
0
0.05
1
0.20
2
0.35
3
0.20
4
0.15
5
0.05
Measures of Dispersion
BAN 530
SOLUTIONS
1. For the following data set find the range, mean absolute deviation, variance and standard deviation.
5, 12, 3, 4, 4
a) Range = H – L = 12 – 3 = 9
b) MAD 
x
2
c) s 
2
s 
 x  x 12.8

 2.56
n
5
x
xx
x x
5
12
3
4
4
28
-0.6
6.4
-2.6
-1.6
-1.6
0.0
0.6
6.4
2.6
1.6
1.6
12.8
x
xx
(x  x )
5
12
3
4
4
28
-0.6
6.4
-2.6
-1.6
-1.6
0.0
0.36
40.96
6.76
2.56
2.56
53.20
 x 28

 5.6
n
5
(x  x )2 53.2

 13.30
n 1
4
2
1
1
2
( x) 2 210  (28)
210  156.8 53.2
n
5



 13.30
n 1
5 1
4
4
 x2 
x
5
12
3
4
4
28
2
(x  x )
25
144
9
16
16
210
Measures of Dispersion
BAN 530
2
d) s  s  13.3  3.65
2. For the following data set find the range, mean absolute deviation, variance and standard deviation.
-1, 3, 2, 6, 2, 1, 0, 0
a. Range = H – L = 6 – (-1) = 7
b. MAD 
 x  x 13.00

 1.625
n
8
x
-1
3
2
6
2
1
0
0
13
x
c.
s2 
xx
x x
(x  x )
-2.625
1.375
0.375
4.375
0.375
-0.625
-1.625
-1.625
0.000
2.625
1.375
0.375
4.375
0.375
0.625
1.625
1.625
13.000
6.890625
1.890625
0.140625
19.140625
0.140625
0.390625
2.640625
2.640625
33.875000
2
 x 13

 1.625
n
8
(x  x )2 33.875

 4.839
n 1
7
2
s 
1
1
2
( x) 2 55  (13)
55  21.125 33.875
n
8



 4.839
n 1
8 1
7
7
 x2 
x
-1
3
2
6
2
1
0
0
13
2
d. s  s  4.839  2.200
x2
1
9
4
36
4
1
0
0
55
Measures of Dispersion
BAN 530
3. A sample of twelve cars on IH 45 was selected and the speed of each was measured by
radar. The speeds (in miles per hour) are:
63 65 67 64 66 72 78 80 66 82 78 83
a) Find the mean.
y
y 63 65  67  64  66  72  78  80  66  82  78  83 864


 72
n
12
12
b) Find the median.
63, 64, 65, 66, 66, 67, 72, 78, 78, 80, 82, 83
i
n 1 12 1

 6.5
2
2
y˜ 
67  72
 69.5
2
c) Find the range.
R = H – L = 83 – 63 = 20
d) Find the variance.
s2 
(y  y) 2 648

 58.909
n 1
11
y
63
64
65
66
66
67
72
78
78
80
82
83
864
yy
-9
-8
-7
-6
-6
-5
0
6
6
8
10
11
0
e) Find the standard deviation.
2
s  s  58.909  7.68
f) Find the mean absolute deviation.
(y  y)
81
64
49
36
36
25
0
36
36
64
100
121
648
2
Measures of Dispersion
BAN 530
MAD 
yy
n

82
 6.83
12
y
63
64
65
66
66
67
72
78
78
80
82
83
864
yy
(y  y)
-9
-8
-7
-6
-6
-5
0
6
6
8
10
11
0
9
8
7
6
6
5
0
6
6
8
10
11
82
g) Find the Z-score for a speed of 74 mph. Interpret.
y  y 74  72

 0.26 A car traveling 74 mph is traveling .26 standard
s
7.68
deviations faster than the mean.
z
4.
A bank examiner recently audited eight banks. The return on investments (%) for
each of the banks is listed below.
-1 5 8 2 6 12 7 6
a) Find the mean.
x
x 1 5  8  2  6 12  7  6 45


 5.625
n
8
8
b) Find the median.
-1, 2, 5, 6, 6, 7, 8, 12
i
n 1 8 1

 4.5
2
2
˜x 
66
6
2
Measures of Dispersion
BAN 530
c) Find the standard deviation.
1
1
x 2  (x) 2 359  (45) 2
359  253.125 105.875
2
n
8
s 



 15.125
n 1
8 1
7
7
2
s  s  15.125  3.89
x
-1
2
5
6
6
7
8
12
45
x2
1
4
25
36
36
49
64
144
359
d) Find the Z-score for a bank with a return on investment of 8%. Interpret.
x  x 8 5.625

 0.61 A bank with a return on investment of 8% is .61
s
3.89
standard deviations above the mean return.
z
5.
Many undergraduate students change their major several times before they graduate. A
sample of eight recent SHSU graduates was obtained. Each graduate was asked how
many times he/she changed majors. The data are listed below.
5, 8, 2, 0, 0, 1, 6, 6
a) Find the mean.
y
y 5  8  2  0  0 1 6  6 28


 3.5
n
8
8
b) Find the median.
0, 0, 1, 2, 5, 6, 6, 8
i
n 1 8 1

 4.5
2
2
˜y 
25
 3.5
2
Measures of Dispersion
BAN 530
c) Find the standard deviation.
1
1
y 2  (y) 2 166  (28) 2
166  98 68
2
n
8
s 



 9.714
n 1
8 1
7
7
2
s  s  9.714  3.12
y
0
0
1
2
5
6
6
8
28
y2
0
0
1
4
25
36
36
64
166
d) Find the z-score for a graduate who changed majors 4 times. Interpret.
y  y 4  3.5

 0.16 A graduate who changed majors 4 times changed .16
s
3.12
standard deviations more than the mean.
z
6.
A sample of eight runs from a production run of 1000 items yielded the following number
of defective items: 6, 12, 14, 8, 10, 12, 7, 7. Find the
a)
mean
x
b)
 x  6  12  14  8 10  12  7  7  76  9.5
n
8
8
variance
x
x2
6
12
14
36
144
196
x  x
x x
-3.5
2.5
4.5
12.25
6.25
20.25
2
Measures of Dispersion
BAN 530
8
10
12
7
7
64
100
144
49
49
-1.5
0.5
2.5
-2.5
-2.5
2.25
0.25
6.25
6.25
6.25
76
782
0.0
60.00
Using the definitional formula the sample variance is:
 x x

2
s
2
n 1

60
60

 8.57
8 1 7
Using the computational formula the sample variance is:
2
s 
x
c)
2
 x
1
n
n 1

2
1
5776
2
782 76
782 
8
8  782  722  60  8.57


8 1
7
7
7
standard deviation
The sample standard deviation is: s  s 2  8.57  2.93
d)
z-score for a production run with 10 defective items. Interpret.
z
7.
x  x 10  9.5

 0.17
s
2.93
A production run with 10 defective items is .17 standard deviations above the mean.
A sample of six new graduates produced the following starting salaries (in $1,000s): 38.5,
42.1, 29.5, 30.1, 45.7, 27.3. Find the
a)
x
b)
mean
x
n

38.5  42.1 29.5  30.1  45.7  27.3 213.2

 35.53
6
6
variance
x
x2
x  x
38.5
42.1
29.5
1482.25
1772.41
870.25
2.97
6.57
-6.03
x x
2
8.8209
43.1649
36.3609
Measures of Dispersion
BAN 530
30.1
45.7
27.3
906.01
2088.49
745.29
-5.43
10.17
-8.23
29.4849
103.4289
67.7329
213.2
7864.70
0.02
288.9934
Using the definitional formula the sample variance is:
 x x

2
s
2
n 1

288.9934 288.9934

 57.80
6 1
5
Using the computational formula the sample variance is:
s2 
x
2
 x
1
n
n 1

c)
2
1
45454.24
2
7864.70  213.2
7864.70 
7864.70  7575.71 288.99
6
6




 57.80
6 1
5
5
5
standard deviation
The sample standard deviation is: s  s 2  57.80  7.60
d)
z-score for a starting salary of $32,000. Interpret.
z
x  x 32  35.53

 0.46
s
7.60
A starting salary of $32,000 is .46 standard deviations below the mean starting
salary.
Measures of Dispersion
BAN 530
8.
A sample of returns for ten portfolios managed by a particular financial manager produced
the following returns (in %): 6.5, 12.2, 3.5, 15.6, 10.4, 8.5, 7.8, 9.3, 11.1, 17.3. Find the
a)
mean
x
x
n

b)
6.5  12.2  3.5  15.6  10.4  8.5  7.8  9.3  11.1  17.3 102.2

 10.22
10
10
variance
x  x
x x
2
x
x2
6.5
12.2
3.5
15.6
10.4
8.5
7.8
9.3
11.1
17.3
42.25
148.84
12.25
243.36
108.16
72.25
60.84
86.49
123.21
299.29
-3.72
1.98
-6.72
5.38
0.18
-1.72
-2.42
-0.92
0.88
7.08
13.8384
3.9204
45.1584
28.9444
0.0324
2.9584
5.8564
0.8464
0.7744
50.1264
102.2
1196.94
0.0
152.456
Using the definitional formula the sample variance is:
 x x

2
s
2
n 1

152.456 152.456

 16.94
10  1
9
Using the computational formula the sample variance is:
s2 
x
2
 x
1
n
n 1

c)
2

1
10444.84
102.22 1196.94 
1196.94  1044.484 152.456
10
10



 16.94
10  1
9
9
9
1196.94 
standard deviation
The sample standard deviation is: s  s 2  16.94  4.12
d)
z-score for a portfolio with a return of 8.4%. Interpret.
z
x  x 8.4 10.22

 0.44
s
4.12
A return of 8.4% is 0.44 standard deviations less than the average return.
Measures of Dispersion
BAN 530
9.
Records revealed the below listed probability distribution for the number of defective items
produced per day. Compute the expected number of defective items produced per day and
the standard deviation.
X, No. of defective items
P(X = x)
xP(x)
x2P(x)
0
0.30
0
0
1
0.25
0.25
0.25
2
0.20
0.40
0.80
3
0.15
0.45
1.35
4
0.10
0.40
1.60
1.0
1.5
4.0
  x2P(x)  2  4.0 1.52  1.75 1.323
E(X)  xP(x) 1.5
10. Suppose that a car salesman uses past records to determine the probability distribution for
weekly sales of luxury cars. The distribution is tabled below. How many luxury cars does
expect to sell per week. What is the standard deviation?
the salesman
X, Sales per Week
P(X)
xP(x)
x2P(x)
0
0.05
0
0
1
0.20
0.20
0.20
2
0.35
0.70
1.40
3
0.20
0.60
1.80
4
0.15
0.60
2.40
5
0.05
0.25
1.25
2.35
7.05
E(X)  xP(x)  2.35
  x P(x)    7.05  2.35  1.5275  1.236
2
2
2