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G. Measures of Dispersion and Asymmetry.
1. Range
Downing & Clark, problem 7 above (Use data to find IQR). Review solutions and terms on page 41 (36 in 3 rd ed.) of Downing &
Clark.
2. The Variance and Standard Deviation of Ungrouped Data.
Text exercises 3.1b, 3.2b, 3.6, 3.37, 3.24 [3.1b, 3.2b, 3.7, 3.37, 3.23] (3.1b, 3.2b, 3.7, 3.23, 3.33)
3. The Variance and Standard Deviation of Grouped Data.
Text exercises 3.28, 3.30 (3.68, 3.70) (work 3.30 in thousands), Downing & Clark pg 42 or 37, problems 6,7 (Find sample standard
deviation – hint: run problem 6 in hundreds) (Note that you can use the Excel or Minitab techniques in the graded assignment to
compute and sum the
fx
and
fx 2
columns in problems 6 and 7. ), Problems G1, G2. Graded Assignment 1
4. Skewness and Kurtosis.
Find the standard deviation, coefficient of variation and measures of skewness in Text problem 3.1, 3.2. Problems G3A, G4 (See
251wrksht).
5. Review
a. Grouped Data.
b. Ungrouped Data.
Section 3 is in this document.
----------------------------------------------------------------------------------------------------------------------------.
Exercise 3.28 (Formerly 3.68): The following data is given. Find a) the mean and b) the variance.
Class Intervals
f
0 - under 10
10
10 - under 20
20
20 - under 30
40
30 - under 40
20
40 - under 50
10
Total
100
Solution: Of course, it is unnecessary to do everything in the table below, but you should know how to do
the problem using both computational and definitional formulas.
class
0-10
10-20
20-30
30-40
40-50
f
10
20
40
20
10
100
x
5
15
25
35
45
fx
50
300
1000
700
450
2500
x  x 
fx 2
250
4500
25000
24500
20250
74500
-20
-10
0
10
20
f x  x 
-200
-200
0
200
200
0
 f  n  100 ,  fx  2500 ,  fx  74500 ,
 fx  2500  25 and, using the computational formula,
a) x 
2
So
s2 

n
100
2
fx  nx 2 74600  100

n 1
100  1
25 2

12000
 121 .21212 .
99
f x  x 
4000
2000
0
2000
4000
12000
2
251solnG2 1/31/08 (Open this document in 'Page Layout' view!)
 f x  x 
2
Or using the definitional formula s 2 

2
12000
 121 .21212 . So
99
n 1
s 11.0096
 0.4404 .
s  121.21212  11.0096 and C  
25
x
Exercise 3.30 (Formerly 3.70): The following sample data is given. f M is frequency in March, f A in
April. Find a) the means and b) the standard deviations. c) Have these statistics changed substantially
between the months?
Class Intervals
fM
fA
$0 - under $2000
6
10
$2000 - under $4000
13
14
$4000 - under $5000
17
13
$5000 - under $8000
10
10
$8000 - under $10000
4
0
$10000 - under $12000
0
3
Total
50
50
Solution: Note that the intervals are not even. Only computational formulas are used below. Numbers are in
thousands.
class
fM
fA
x
x
fM x2
f Ax2
fM x
f Ax
0- 2
6
1
6.0
6.00
10
1.0
10.0
10.00
2- 4
13
3
39.0
117.00
14
3.0
42.0
126.00
4- 5
17
4.5
76.5
344.25
13
4.5
58.5
263.25
5- 8
10
6.5
65.0
422.50
10
6.5
65.0
422.50
8-10
4
9
36.0
324.00
0
9.0
0.0
0.00
10-12
0
11
0.0
0.00
3
11.0
33.0
363.00
50
222.5 1213.75
50
208.5 1184.75
 f  n  50 ,  f x  222 .5 ,  f x  1213 .75 ,  f
 f x  222 .5  4.45 x   f x  208 .5  4.17
a) x 
2
So
M
M
M
n
50
b ) Using the computational formula,
f
f

M
 208 .5 and
f
Ax
2
 1184 .75 .
A
M
2
sM

Ax
x 2  nx M2
n 1

A
1213 .75  50
50  1
n
4.45 2
50

223 .625
 4.56378 . So s M  4.56378  2.1363 .
49
4.17 2  315 .305  6.43480 . So s  4.56378  2.5367
11184 .75  50
A
n 1
50  1
49
Though the mean has fallen somewhat (6%), there are more items in both the largest and smallest
categories, making the standard deviation larger (by 19%).
s A2
Ax
2
 nx A2

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Downing and Clark, pg. 37, Application 6: Find sample standard deviation of the data below– hint: run
problem 6 in hundreds). The given data is group number, (group), frequency: 1, ( 0-1000), 12; 2, (10002000), 15; 3, (2000-3000), 19; 4, (3000-4000), 22; 5, (4000-5000), 30; 6, (5000-6000), 56; 7, (6000-7000),
48; 8, (7000-8000), 40; 9, (8000-9000), 30; 10, (9000-10000), 16; 11, (10000-11000), 4; 12, (1100012000), 2.
Solution: To use the computational method, construct the following table. Note that we have converted the
data into hundreds so that the group (2000-3000) is treated as (20-30). This means that the fx column is in
hundreds, the fx 2 column is in 100 2  ten thousands and the fx 3 column is in 100 3  millions.
fx 2
fx 3
1
300
1500
2
3375
50625
3
11875
296875
4
26950
943250
5
60750
2733750
6
169400
9317000
7
202800
13182000
8
225000
16875000
9
216750
18423750
10
144400
13718000
11
44100
4630500
12
26450
30417500
1132150
fx 16800
fx  16800 ,
f  n  294 ,
fx 2  1132150 , x 

 57 .142857 (in hundreds
So
n
294
– so actually 5714.2857) and, using the computational formula to find the sample variance ,
fx 2  nx 2 1132150  294 57 .142857 2 172150
s2 


 587 .54266 (in 100 2  ten thousands). So
n 1
294  1
293
s
s  587.54266  24.2393 (in hundreds – so actually 2423.93) and C   24.2393  0.4242 .
x 57.142857
Downing and Clark, pg. 37, Application 7: Find sample standard deviation of the data given in the first
three columns below.
Row
group
x
fx
fx 2
fx 3
f
1
0-10
122
5
610
3050
15250
2
10-20
180
15
2700
40500
607500
3
20-30
256
25
6400
160000
4000000
4
30-40
350
35
12250
428750
15006250
5
40-50
311
45
13995
629775
28339876
6
50-60
278
55
15290
840950
46252248
7
60-70
250
65
16250
1056250
68656248
8
70-80
211
75
15825
1186875
89015624
9
80-90
180
85
15300
1300500
110542496
10
90-100
175
95
16625
1579375
150040624
11
100-110
143
105
15015
1576575
165540368
12
110-120
120
115
13800
1587000
182504992
13
120-130
106
125
13250
1656250
207031248
14
130-140
99
135
13365
1804275
243577120
15
140-150
97
145
14065
2039425
295716640
16
150-160
75
155
11625
1801875
279290624
Total
2953
196365 17691424 1886137088
f  n  2953 ,
fx 2  17691424 . This means
fx  196365 and
So
Row
group
0-1000
1000-2000
2000-3000
3000-4000
4000-5000
5000-6000
6000-7000
7000-8000
8000-9000
9000-10000
10000-11000
11000-12000

f
12
15
19
22
30
56
48
40
30
16
4
2
294

x
5
15
25
35
45
55
65
75
85
95
105
115

fx
60
225
475
770
1350
3080
3120
3000
2550
1520
420
230
16800




 fx  196365  66.496783
x
n
2953

and, using the computational formula to find the sample variance ,
251solnG2 1/31/08 (Open this document in 'Page Layout' view!)
s2 
 fx
2
 nx 2
n 1

17691424  2953 66 .496783 2 4633783 .2

 1569 .7098 . So
2953  1
2952
s  1569.7098  39.6196 .
Problem G1: If the mean return for an industry is 10% with a standard deviation of 6%, out of 100 firms
how many do you expect to have returns above 22%?
 1 
Solution: Use Chebyshef's Rule. For any number k greater than 1, at least 1   2  of the points will fall
k 
1
within k standard deviations of the mean (i. e. in the interval   k ) and no more than 2 of the points
k
will fall beyond k standard deviations of the mean . Note that 22% is above the mean by a number of
12 12
standard deviations. Since x    22 10  12 and

 2 , k  2. If the proportion in the tails is no

6
1
1
1 1
more than 2 , we can compute 2  2  and conclude that at most one quarter of the points are
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k
k
2
above 22%. This would be 25 firms.
Problem G2: If the mean is 5 and the standard deviation is 2, find an interval that must contain at least the
central two-thirds of the observations.
Solution: Again use Chebyshef's Rule. If we want a central interval with no less than 1/3 of the
1 1
observations, then the tails must contain at most 1/3. So we can say that 2  . So k 2  3 , which means
3
k
that k  3  1.732 . We know that   5 and   2 , so that our interval is   k
 5 1.732 2  5 3.464 , or 1.536 to 8.464.
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