# If the variance of the following frequency distribution: $\begin{array}{|l|c|c|c|} \hline \mathrm{Class} & \mathrm{10-20} & \mathrm{20-30} & \mathrm{30-40} \\ \hline \mathrm{Frequency} & \mathrm{2} & \mathrm{x} & \mathrm{2} \\ \hline \end{array}$ is 50, then x is equal to _________   Option: 1 2 Option: 2 6 Option: 3 8 Option: 4 4

Variance is independent of shifting of origin

$\begin{array}{|l|c|c|c|} \hline \mathrm{Class} & \mathrm{10-20} & \mathrm{20-30} & \mathrm{30-40} \\ \hline \mathrm{Frequency} & \mathrm{2} & \mathrm{x} & \mathrm{2} \\ \hline \end{array}$

$\Rightarrow \begin{array}{|l|c|c|c|} \hline \mathrm{Class\;(x_i)} & \mathrm{15} & \mathrm{25} & \mathrm{35} \\ \hline \mathrm{Frequency\;(f_i)} & \mathrm{2} & \mathrm{x} & \mathrm{2} \\ \hline \end{array}\;\text{ or }\;\begin{array}{|l|c|c|c|} \hline \mathrm{Class\;(x_i)} & \mathrm{-10} & \mathrm{0} & \mathrm{10} \\ \hline \mathrm{Frequency\;(f_i)} & \mathrm{2} & \mathrm{x} & \mathrm{2} \\ \hline \end{array}$

\begin{aligned} &\Rightarrow \text { Variance }\left(\sigma^{2}\right)=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2} \mathrm{f}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}-(\bar{\mathrm{x}})^{2}\\ &\Rightarrow 50=\frac{200+0+200}{x+4}-0\\ &\Rightarrow 200+50 x=200+200\\ &\Rightarrow x=4 \end{aligned}

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