Q.
The variance of the following continuous frequency distribution is
Class Interval 0-10 10-20 20-30 30-40 Frequency 2 3 4 1
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 |
| Frequency | 2 | 3 | 4 | 1 |
AP EAMCETAP EAMCET 2019
Solution:
Given
Class Interval
Frequency $(f_{i})$
$x_{i}$
$x_{i}f_{i}$
$\left(\bar{x}-x_{i}\right)^{2}$
$f_{i}\left(\bar{x}-x_{i}\right)^{2}$
0-10
2
5
10
196
392
10-20
3
15
45
16
48
20-30
4
25
100
36
144
30-40
1
35
35
256
256
$N=\Sigma f_{i}=10$
$\Sigma x_{i} f_{i}=190$
$\Sigma f_{i}\left(\bar{x}-x_{i}\right)^{2}=840$
$\because \bar{x}=\frac{\sum x_{i} f_{i}}{N}=\frac{190}{10}=19$
$\therefore $ Variance $(\sigma)^{2}=\frac{1}{N} \Sigma f_{i}\left(\bar{x}-x_{i}\right)^{2}$
$=\frac{1}{10}(840)=84$
| Class Interval | Frequency $(f_{i})$ | $x_{i}$ | $x_{i}f_{i}$ | $\left(\bar{x}-x_{i}\right)^{2}$ | $f_{i}\left(\bar{x}-x_{i}\right)^{2}$ |
|---|---|---|---|---|---|
| 0-10 | 2 | 5 | 10 | 196 | 392 |
| 10-20 | 3 | 15 | 45 | 16 | 48 |
| 20-30 | 4 | 25 | 100 | 36 | 144 |
| 30-40 | 1 | 35 | 35 | 256 | 256 |
| $N=\Sigma f_{i}=10$ | $\Sigma x_{i} f_{i}=190$ | $\Sigma f_{i}\left(\bar{x}-x_{i}\right)^{2}=840$ |