The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.
Here, $n_{1}=60, \bar{x}_{1}=650, s_{1}=8$ and $n_{2}=80, \bar{x}_{2}=660, s_{2}=7$
$\therefore \quad \sigma=\sqrt{\frac{n_{1} s_{1}^{2}+n_{2} s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$
$=\sqrt{\frac{60 \times(8)^{2}+80 \times(7)^{2}}{60+80}+\frac{60 \times 80(650-660)^{2}}{(60+80)^{2}}}$
$=\sqrt{\frac{6 \times 64+8 \times 49}{14}+\frac{60 \times 80 \times 100}{140 \times 140}}$
$=\sqrt{\frac{192+196}{7}+\frac{1200}{49}=\sqrt{\frac{388}{7}+\frac{1200}{49}}}{\sqrt{\frac{2716+1200}{49}}}$
$=\sqrt{\frac{3915}{49}}=\sqrt{79.9}=8.9$
The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking it was found that an observation was read as $10$ in place of $12$ . If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^2+\sigma^2\right)$ is equal to$...................$
Find the variance and standard deviation for the following data:
${x_i}$ | $4$ | $8$ | $11$ | $17$ | $20$ | $24$ | $32$ |
${f_i}$ | $3$ | $5$ | $9$ | $5$ | $4$ | $3$ | $1$ |
If the variance of observations ${x_1},\,{x_2},\,......{x_n}$ is ${\sigma ^2}$, then the variance of $a{x_1},\,a{x_2}.......,\,a{x_n}$, $\alpha \ne 0$ is
The $S.D$ of $15$ items is $6$ and if each item is decreased or increased by $1$, then standard deviation will be