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MEASURES OF DISPERSION
EMPIRICAL RELATIONSHIPS
Mean Deviation = ( 0
...
Deviation
ABSOLUTE DISPERSION AND RELATIVE DISPERSION
Relative Dispersion = (Absolute Dispersion ) / Average
Here, Absolute Dispersion is the value obtained from
different measures of dispersions, like Mean deviation
or Standard deviation, etc
...
Question :
A manufacturer of Television Tubes produced
two types of tubes, which are Tube A and Tube
B
...
Which tube has greater relative dispersion ?
SOLUTION:
For Tube A :
Mean = XA = 1375 hours
Standard deviation = sA = 280 hours
So , Relative Dispersion = sA / XA
= 280 / 1375 = 0
...
For Tube B :
Mean = XB = 1525 hours
Standard deviation = sB = 335 hours
So , Relative Dispersion = sB / XB
= 335 / 1525 = 0
...
RESULT :
TUBE B has greater relative dispersion
...
From above result, which team is found to be more consistent ?
Number of Goals
Total (∑)
Number of Match played
(X)
Team A ( FA )
Team B ( FB )
0
22
16
1
19
15
2
12
12
3
8
10
4
3
6
5
4
3
FA X
FB X
Number of Goals
Number of Match played
(X)
Team A ( FA )
Team B ( FB )
FA X
FB X
0
22
16
0
0
1
19
15
19
15
2
12
12
24
24
3
8
10
24
30
4
3
6
12
24
5
4
3
20
15
68
62
99
108
Total (∑)
Arithmetic Means :
Team A : X’A = ∑ FAX / ∑FA
Team B :
= 99 / 68 = 1
...
742
Standard Deviations :
Team A
| X – X’A |
1
...
456
0
...
544
2
...
544
( X – x’A)2
FA( X – X’A)2
( X –X’A)2
FA( X – X’A)2
Team A ( FA )
| X – X’A |
22
1
...
1199
46
...
456
0
...
9501
12
0
...
2959
3
...
544
2
...
0712
3
2
...
4719
19
...
544
12
...
2396
68
SA = [ ∑ FA( X – X’A)2 / ∑ FA ]1/2 = [ 142
...
4494
=> SA = 1
...
8652
Team B ( FB )
16
15
12
10
6
3
62
| X – X’B |
( X –X’B)2
FA( X – X’B)2
Team B ( FB )
16
15
12
10
6
3
62
| X – X’B |
1
...
742
0
...
258
2
...
258
( X –X’B)2
3
...
55056
0
...
582564
5
...
614564
FA( X – X’B)2
48
...
25846
0
...
82564
30
...
843692
135
...
870968/62 ]1/2 = 1
...
Disp
...
4494 /1
...
9954
Rel
...
B = SB / X’B = 1
...
742 = 0
...
V
...
9954 x 100 = 99
...
V
...
8498 x 100 = 84
...