The variance of the first 50 even natural numbers is

Question

Aicharya · Accepted Answer

Given, 50 even natural numbers are $2,4,6, \ldots .100$ Mean, $\bar{x}=\frac{\sum x i}{n}=\frac{2+4+6+\ldots+98+100}{50}$ $\begin{aligned} & \bar{x}=\frac{2(1+2+3+\ldots+49+50)}{50} \ & \bar{x}=\frac{2 	imes 50 	imes 51}{2 	imes 50}=51 \end{aligned}$ Variance, $\sigma^2=\frac{\sum x_1^2}{n}-(\bar{x})^2$ $\begin{aligned} & =\frac{\left(2^2+4^2+6^2+\ldots .+98^2+100^2ight)}{50}-(51)^2 \ & =\frac{2^2\left(1^2+2^2+3^2+\ldots .+49^2+50^2ight)-(51)^2 	imes 50}{50} \ & =\frac{4(50)(51)(101)-(51) 	imes(51) 	imes 50 	imes 6}{6 	imes 50}=833 \end{aligned}$

The variance of the first 50 even natural numbers is

Solution