Q.
The coefficient of variation from the given data
class interval
0 -10
10- 20
20 -30
30- 40
40 -50
frequency
2
10
8
4
6
is
| class interval | 0 -10 | 10- 20 | 20 -30 | 30- 40 | 40 -50 |
| frequency | 2 | 10 | 8 | 4 | 6 |
Statistics
Solution:
Class
mid value (x)
f
f x
d = x -M
$d^2$
$fd^2$
0 - 10
5
2
10
-20.7
428.49
856.98
10 - 20
15
10
150
-10.7
114.49
1144.9
20 - 30
25
8
200
-0.7
0.49
3.92
30 - 40
35
4
140
9.3
86.49
345.96
40 - 50
45
6
270
19.3
372.49
2234.94
$\sum f = 30$
$\sum fx = 770$
$ \sum fd^2 = 4586 .7$
Now, M (A.M) = $\frac{\sum fx}{\sum f} = \frac{770}{30} = 25.7$
Now, standard deviation (S.D)
$ = \sqrt{\frac{\sum fd^2}{\sum f}} = \sqrt{\frac{4586.70}{30}} = 12.36$
$\therefore $ Coeff of SD = $\frac{SD}{M} = \frac{12.36}{25.7} = 0.480$
$\therefore $ Coeff of variation = Coeff of S.D $\times$ 100
= 0.480 $\times$ 100 = 48.
| Class | mid value (x) | f | f x | d = x -M | $d^2$ | $fd^2$ |
| 0 - 10 | 5 | 2 | 10 | -20.7 | 428.49 | 856.98 |
| 10 - 20 | 15 | 10 | 150 | -10.7 | 114.49 | 1144.9 |
| 20 - 30 | 25 | 8 | 200 | -0.7 | 0.49 | 3.92 |
| 30 - 40 | 35 | 4 | 140 | 9.3 | 86.49 | 345.96 |
| 40 - 50 | 45 | 6 | 270 | 19.3 | 372.49 | 2234.94 |
| $\sum f = 30$ | $\sum fx = 770$ | $ \sum fd^2 = 4586 .7$ |