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.
Given: $n_{1}=60, \bar{x}_{1}=650, s_{1}=8$ and $n_{2}=80, \bar{x}_{2}=660, s_{2}=7$.
The formula for the combined standard deviation $\sigma$ is:
$\sigma = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2}(\bar{x}_{1} - \bar{x}_{2})^{2}}{(n_{1} + n_{2})^{2}}}$
Substituting the values:
$\sigma = \sqrt{\frac{60(8)^{2} + 80(7)^{2}}{60 + 80} + \frac{60 \times 80(650 - 660)^{2}}{(60 + 80)^{2}}}$
$\sigma = \sqrt{\frac{60(64) + 80(49)}{140} + \frac{4800(-10)^{2}}{(140)^{2}}}$
$\sigma = \sqrt{\frac{3840 + 3920}{140} + \frac{480000}{19600}}$
$\sigma = \sqrt{\frac{7760}{140} + \frac{4800}{196}} = \sqrt{55.428 + 24.489} = \sqrt{79.917} \approx 8.94$ hours.
The formula for the combined standard deviation $\sigma$ is:
$\sigma = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2}(\bar{x}_{1} - \bar{x}_{2})^{2}}{(n_{1} + n_{2})^{2}}}$
Substituting the values:
$\sigma = \sqrt{\frac{60(8)^{2} + 80(7)^{2}}{60 + 80} + \frac{60 \times 80(650 - 660)^{2}}{(60 + 80)^{2}}}$
$\sigma = \sqrt{\frac{60(64) + 80(49)}{140} + \frac{4800(-10)^{2}}{(140)^{2}}}$
$\sigma = \sqrt{\frac{3840 + 3920}{140} + \frac{480000}{19600}}$
$\sigma = \sqrt{\frac{7760}{140} + \frac{4800}{196}} = \sqrt{55.428 + 24.489} = \sqrt{79.917} \approx 8.94$ hours.
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View SolutionConsider the following frequency distribution:
Value $4$ $5$ $8$ $9$ $6$ $12$ $11$ Frequency $5$ $f_1$ $f_2$ $2$ $1$ $1$ $3$
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List-$I$ List-$II$ $(P) \ 7f_1+9f_2$ is equal to $(1) \ 146$ $(Q) \ 19\alpha$ is equal to $(2) \ 47$ $(R) \ 19\beta$ is equal to $(3) \ 48$ $(S) \ 19\sigma^2$ is equal to $(4) \ 145$ $(5) \ 55$
| Value | $4$ | $5$ | $8$ | $9$ | $6$ | $12$ | $11$ |
| Frequency | $5$ | $f_1$ | $f_2$ | $2$ | $1$ | $1$ | $3$ |
Suppose that the sum of the frequencies is $19$ and the median of this frequency distribution is $6$. For the given frequency distribution,let $\alpha$ denote the mean deviation about the mean,$\beta$ denote the mean deviation about the median,and $\sigma^2$ denote the variance. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.
| List-$I$ | List-$II$ |
| $(P) \ 7f_1+9f_2$ is equal to | $(1) \ 146$ |
| $(Q) \ 19\alpha$ is equal to | $(2) \ 47$ |
| $(R) \ 19\beta$ is equal to | $(3) \ 48$ |
| $(S) \ 19\sigma^2$ is equal to | $(4) \ 145$ |
| $(5) \ 55$ |
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