નીચે આપેલ આવૃતિ વિતરણ માટે મધ્યક અને પ્રમાણિત વિચલન મેળવો 

$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Marks } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text { Frequency } & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$

 

$\begin{array}{|c|r|r|r|r|r|} \hline \text { Marks } & f_{i} & f_{i} x_{i} & d_{i}=x_{i}-\bar{x} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 2 & 1 & 2 & -4 & -4 & 16 \\ \hline 3 & 6 & 18 & -3 & -18 & 54 \\ \hline 4 & 6 & 24 & -2 & -12 & 24 \\ \hline 5 & 8 & 40 & -1 & -8 & 8 \\ \hline 6 & 8 & 48 & 0 & 0 & 0 \\ \hline 7 & 2 & 14 & 1 & 2 & 2 \\ \hline 8 & 2 & 16 & 2 & 4 & 8 \\ \hline 9 & 3 & 27 & 3 & 9 & 27 \\ \hline 10 & 0 & 0 & 4 & 0 & 0 \\ \hline 11 & 2 & 22 & 5 & 10 & 50 \\ \hline 12 & 1 & 12 & 6 & 6 & 36 \\ \hline 13 & 0 & 0 & 7 & 0 & 0 \\ \hline 14 & 0 & 0 & 8 & 0 & 0 \\ \hline 15 & 0 & 0 & 9 & 0 & 0 \\ \hline 16 & 1 & 16 & 10 & 10 & 100 \\ \hline \text { Total } & \Sigma f_{i}=40 & \Sigma f_{i} x_{i}=239 & & \Sigma f_{i} d_{i}=-1 & \Sigma f_{i} x_{i}^{2}=325 \\ \hline \end{array}$

$\therefore \quad$ Mean $\bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{239}{40}=5.975 \approx 6$

and $\sigma=\sqrt{\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}-\left(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\right)^{2}}=\sqrt{\frac{325}{40}-\left(\frac{-1}{40}\right)^{2}}$

$=\sqrt{8.125-0.000625}=\sqrt{8.124375}=2.85$