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Two sets each of 20 observations, have the same standard deviation 5 . The first set has a mean 17 and the second a mean 22. Determine the standatd deviation of the set obtained by combining the given two sets.
Solution:
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Verified Answer
We have, $n_1=20, \sigma_1=5, \bar{x}_1=17$
$$
\begin{aligned}
n_2 &=20, \sigma_2=5, \bar{x}_2=22 \\
\bar{x}_{12} &=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2} \\
&=\frac{(20)(17)+(20)(22)}{20+20}=\frac{340+440}{40}=19.5 \\
d_1 &=\bar{x}_1-\bar{x}_{12}=17-19.5=-2.5
\end{aligned}
$$
$$
\begin{aligned}
d_2 &=\bar{x}_2-\bar{x}_{12}=22-19.5=2.5 \\
\sigma_{12} &=\sqrt{\frac{n_1\left(\sigma_1^2+d_1^2\right)+n_2\left(\sigma_2^2+d_2^2\right)}{n_1+n_2}} \\
&=\sqrt{\frac{20(25+6.25)+20(25+6.25)}{20+20}} \\
&=\sqrt{\frac{1250}{40}}=\sqrt{31.25}=5.59
\end{aligned}
$$
$$
\begin{aligned}
n_2 &=20, \sigma_2=5, \bar{x}_2=22 \\
\bar{x}_{12} &=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2} \\
&=\frac{(20)(17)+(20)(22)}{20+20}=\frac{340+440}{40}=19.5 \\
d_1 &=\bar{x}_1-\bar{x}_{12}=17-19.5=-2.5
\end{aligned}
$$
$$
\begin{aligned}
d_2 &=\bar{x}_2-\bar{x}_{12}=22-19.5=2.5 \\
\sigma_{12} &=\sqrt{\frac{n_1\left(\sigma_1^2+d_1^2\right)+n_2\left(\sigma_2^2+d_2^2\right)}{n_1+n_2}} \\
&=\sqrt{\frac{20(25+6.25)+20(25+6.25)}{20+20}} \\
&=\sqrt{\frac{1250}{40}}=\sqrt{31.25}=5.59
\end{aligned}
$$
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