If the mean deviation about the mean is $m$ and variance is $\sigma^2$ for the following data, then $m+\sigma^2=$ \begin{array}{|c|c|c|c|c|c|}\hlinex & 1 & 3 & 5 & 7 & 9 \\hlinef & 4 & 24 & 28 & 16 & 8 \\hline\end{array}

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Aicharya · Accepted Answer

$\begin{array}{|c|c|c|c|c|c|}\hline \boldsymbol{x}_{\mathbf{i}} & \boldsymbol{f}_{\mathbf{i}} & \boldsymbol{x}_{\mathrm{i}} f_{\mathbf{i}} & \left|\boldsymbol{x}_{\boldsymbol{i}}-\overline{\boldsymbol{x}}\right| & \boldsymbol{f}_{\boldsymbol{i}}\left|\boldsymbol{x}_{\boldsymbol{i}}-\overline{\boldsymbol{x}}\right| & \boldsymbol{x}_{\boldsymbol{i}}^{\mathbf{2}} \boldsymbol{f}_{\boldsymbol{i}} \\\hline 1 & 4 & 4 & 4 & 16 & 4 \\\hline 3 & 24 & 72 & 2 & 48 & 216 \\\hline 5 & 28 & 140 & 0 & 0 & 700 \\\hline 7 & 16 & 112 & 2 & 32 & 784 \\\hline 9 & 8 & 72 & 4 & 32 & 648 \\\hline Total & 80 & 400 & & 128 & 2352 \\\hline\end{array}$ $\text {Mean }=\frac{\Sigma x_i f_i}{\Sigma f_i}=\frac{400}{80}=5$ $\text { Mean deviation }=m=\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}=\frac{128}{80}=\frac{8}{5}$ Variance $\left(\sigma^2\right)=\frac{\Sigma x_i^2 f_i}{\Sigma f_i}-(\bar{x})^2$ $=\frac{2352}{80}-25=\frac{352}{80}=\frac{22}{5}$
Now, $m+\sigma^2=\frac{8}{5}+\frac{22}{5}=\frac{30}{5}=6$

If the mean deviation about the mean is $m$ and variance is $\sigma^2$ for the following data, then…

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