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# 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.

$\begin{array}{1 1}(A)\;8.94\\(B)\;5\\(C)\;7\\(D)\;15\end{array}$

Toolbox:
• The formula used is Combined $SD= \sqrt {\large\frac{n_1 \sigma_1^2+n_2 \sigma^2}{n_1+n_2} +\frac{n_1n_2 (\bar {x_1} -\bar {x_2})^2}{(n_1+n_2)^2}}$
Step 1:
Given $n_1=60 \qquad n_2=80$
$\bar{x_1}=650 \qquad \bar{x_2}=660$
$\sigma_1=8 \qquad \sigma_2=7$
The formula used is Combined $SD= \sqrt {\large\frac{n_1 \sigma_1^2+n_2 \sigma^2}{n_1+n_2} +\frac{n_1n_2 (\bar {x_1} -\bar {x_2})^2}{(n_1+n_2)^2}}$
$\qquad= \sqrt { \large\frac{60 \times 8^2 + 80 \times 7^2}{60+80}+ \frac{ 60 \times 80 (650-660)^2}{(60+80)^2}}$
$\qquad= \sqrt {\large\frac{ 3840+3920}{140}+ \frac{4800 \times 100}{(140)^2}}$
$\qquad= \sqrt {\large\frac{7760 \times 140+4800 \times 100}{(140^2)}}$
$\qquad= \large\frac{1}{140}$$\sqrt {1086400+480000} \qquad= \large\frac{1251.559}{140}$$=8.939$
$\qquad= 8.94$
Hence A is the correct answer.