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 $n_{2}=80, \bar{x}_{2}=660, s_{2}=7$
$\therefore \quad \sigma=\sqrt{\frac{n_{1} s_{1}^{2}+n_{2} s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$
$=\sqrt{\frac{60 \times(8)^{2}+80 \times(7)^{2}}{60+80}+\frac{60 \times 80(650-660)^{2}}{(60+80)^{2}}}$
$=\sqrt{\frac{6 \times 64+8 \times 49}{14}+\frac{60 \times 80 \times 100}{140 \times 140}}$
$=\sqrt{\frac{192+196}{7}+\frac{1200}{49}=\sqrt{\frac{388}{7}+\frac{1200}{49}}}{\sqrt{\frac{2716+1200}{49}}}$
$=\sqrt{\frac{3915}{49}}=\sqrt{79.9}=8.9$
Variance of $^{10}C_0$ , $^{10}C_1$ , $^{10}C_2$ ,.... $^{10}C_{10}$ is
Let the mean and standard deviation of marks of class $A$ of $100$ students be respectively $40$ and $\alpha( > 0)$, and the mean and standard deviation of marks of class $B$ of $n$ students be respectively $55$ and $30-\alpha$. If the mean and variance of the marks of the combined class of $100+ n$ students are respectively $50$ and $350$,then the sum of variances of classes $A$ and $B$ is
The data is obtained in tabular form as follows.
${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |
The variance of $\alpha$, $\beta$ and $\gamma$ is $9$, then variance of $5$$\alpha$, $5$$\beta$ and $5$$\gamma$ is
The mean and standard deviation of $20$ observations are found to be $10$ and $2$, respectively. On respectively, it was found that an observation by mistake was taken $8$ instead of $12$ . The correct standard deviation is