For the following frequency distribution The variance is

Question

Aicharya · Accepted Answer

$\begin{array}{cccc}\boldsymbol{x}_{\mathbf{i}} & \boldsymbol{f}_{\mathbf{i}} & \boldsymbol{f}_{\mathrm{i}} \boldsymbol{x}_{\mathbf{i}} & \boldsymbol{f}_{\mathbf{i}} \mathbf{x}_{\mathbf{i}}{ }^2 \ 5 & 3 & 15 & 75 \ 6 & 7 & 42 & 252 \ 7 & 4 & 28 & 196 \ 8 & 2 & 16 & 128 \ 10 & \frac{4}{N=20} & \frac{40}{\sum f_i x_i=141} & \frac{400}{\sum f_i x_i^2=1051}\end{array}$ $\begin{aligned} & 	ext { Variance }=\frac{\sum f_i x_i^2}{N}-\left(\frac{\sum f_i x_i}{N}ight)^2 \ & =\frac{1051}{20}-\left(\frac{141}{20}ight)^2 \ & =52.55-(7.05)^2 \ & =52.55-49.7025 \ & =2.85\end{aligned}$

For the following frequency distribution The variance is

Solution