Q.
Statement I For the following frequency distribution, the standard deviation is $1.38$.
x
2
3
4
5
6
7
f
4
9
16
14
11
6
Statement II The mean life of a sample of 60 bulbs was $650 h$ and the standard deviation was $8 h$. A second sample of 80 bulbs has a mean life of $660 h$ and standard deviation $7 h$. Then, the overall standard deviation is $8.9$.
| x | 2 | 3 | 4 | 5 | 6 | 7 |
| f | 4 | 9 | 16 | 14 | 11 | 6 |
Statistics
Solution:
$x_i$
$f_i$
$d_i = x_i - 4$
$f_id_i$
$f_id_i^2$
2
4
-2
-8
16
3
9
-1
-9
9
4
16
0
0
0
5
14
1
14
14
6
11
2
22
44
7
6
3
18
54
$\Sigma f_i = 60$
$\Sigma f_i d_i = 37$
$\Sigma f_i d_i^2 = 137$
Standard deviation $(\sigma) =\sqrt{\frac{\Sigma f d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f d_i}{\Sigma f_i}\right)^2} $
$ =\sqrt{\frac{137}{60}-\left(\frac{37}{60}\right)^2} $
$=\sqrt{2.2833-(0.616)^2} $
$ =\sqrt{2.2833-0.3794} $
$ =\sqrt{1.9038}=1.38$
II. We have, $n_1=60, \bar{x}_1=650, \sigma_1=8$
$n_2=80, \bar{x}_2=660, \sigma_2=7$
Combined standard deviation ( $\sigma$ )
$ =\sqrt{\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}}$
$=\sqrt{\frac{60 \times 64+80 \times 49}{60+80}+\frac{60 \times 80 \times(660-650)^2}{(60+80)^2}}$
$=\sqrt{\frac{6 \times 64+8 \times 49}{14}+\frac{60 \times 80 \times 100}{140 \times 140}} $
$=\sqrt{\frac{3 \times 64+4 \times 49}{7}+\frac{6 \times 8 \times 100}{14 \times 14}} $
$=\sqrt{\frac{388}{7}+\frac{1200}{49}}=\frac{\sqrt{2716+1200}}{49} $
$=\frac{\sqrt{3916}}{7}=\frac{62.58}{7}=8.9$
| $x_i$ | $f_i$ | $d_i = x_i - 4$ | $f_id_i$ | $f_id_i^2$ |
|---|---|---|---|---|
| 2 | 4 | -2 | -8 | 16 |
| 3 | 9 | -1 | -9 | 9 |
| 4 | 16 | 0 | 0 | 0 |
| 5 | 14 | 1 | 14 | 14 |
| 6 | 11 | 2 | 22 | 44 |
| 7 | 6 | 3 | 18 | 54 |
| $\Sigma f_i = 60$ | $\Sigma f_i d_i = 37$ | $\Sigma f_i d_i^2 = 137$ |