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If $\mathrm{M}$ and $\sigma^2$ represent respectively the mean deviation from the mean and the variance for the data $1,3,5,7,11$, $13,17,19,23$ then $3\left(\sigma^2-M\right)=$
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136
$\begin{aligned} & \text {} \bar{X}=\frac{\Sigma x_i}{n}=\frac{99}{9}=11 \\ & \begin{aligned} M & =\frac{\Sigma\left|x_i^n-\bar{x}\right|^{\prime}}{n} \\ & =\frac{10+8+6+4+0+2+6+8+12}{9}=\frac{56}{9} \\ \sigma^2 & =\frac{\Sigma\left|x_i-\bar{x}\right|^2}{n}\end{aligned}\end{aligned}$
$\begin{aligned} & =\frac{(10)^2+8^2+6^2+4^2+2^2+6^2+8^2+12^2}{9}=\frac{464}{9} \\ & 3\left(\sigma^2-M\right)=3\left(\frac{464}{9}-\frac{56}{9}\right)=\frac{408 \times 3}{9}=136 .\end{aligned}$
$\begin{aligned} & =\frac{(10)^2+8^2+6^2+4^2+2^2+6^2+8^2+12^2}{9}=\frac{464}{9} \\ & 3\left(\sigma^2-M\right)=3\left(\frac{464}{9}-\frac{56}{9}\right)=\frac{408 \times 3}{9}=136 .\end{aligned}$
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