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If the variance of the data $2,3,5,8,12$ is $\sigma^2$ and the mean deviation from the median for this data is $\mathrm{M}$, then $\sigma^2-\mathrm{M}=$
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The correct answer is:
$10.2$
(a) Observations : 2, 3, 5, 8, 12
$$
\begin{aligned}
& \text { Mean }=\frac{2+3+5+8+12}{5}=6 \\
& \therefore \quad \sigma^2=\frac{\sum\left(x_i-\bar{x}\right)^2}{n}=\frac{16+9+1+4+36}{5}=\frac{66}{5} \\
& \therefore \quad \sigma^2=13.2 \\
& \text { Median }=5=m
\end{aligned}
$$
$\therefore \quad$ Mean deviation about median
$$
\begin{aligned}
= & \frac{\Sigma\left|x_i-m\right|}{n}=\frac{3+2+0+3+7}{5}=3 \\
M=3 \Rightarrow \sigma^2-M & =13.2-3=10.2 .
\end{aligned}
$$
$$
\begin{aligned}
& \text { Mean }=\frac{2+3+5+8+12}{5}=6 \\
& \therefore \quad \sigma^2=\frac{\sum\left(x_i-\bar{x}\right)^2}{n}=\frac{16+9+1+4+36}{5}=\frac{66}{5} \\
& \therefore \quad \sigma^2=13.2 \\
& \text { Median }=5=m
\end{aligned}
$$
$\therefore \quad$ Mean deviation about median
$$
\begin{aligned}
= & \frac{\Sigma\left|x_i-m\right|}{n}=\frac{3+2+0+3+7}{5}=3 \\
M=3 \Rightarrow \sigma^2-M & =13.2-3=10.2 .
\end{aligned}
$$
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