The variance of first 50 even natural numbers is

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Variance }=\frac{\sum x_{\mathrm{i}}^2}{\mathrm{n}}-(\bar{x})^2 \ & =\frac{\left(2^2+4^2+\ldots+100^2ight)}{50}-\left(\frac{2+4+\ldots+100}{50}ight)^2 \ & =\frac{4\left(1^2+2^2+\ldots .+50^2ight)}{50}-(51)^2 \ & =4\left(\frac{50 	imes 51 	imes 101}{50 	imes 6}ight)-(51)^2 \ & =3434-2601 \ & =833\end{aligned}$

The variance of first 50 even natural numbers is

Solution