Question

The variance of first 50 even natural numbers is?

Answer 1

$Variance\ = \ \frac{1}{n} SUMMATION_{1-n} [1^2+2^2+3^+ ..... +49^2 + 50^2] \\ \\. \ \ \ \ = \frac{1}{n} * n * \frac{(n+1)*(2n+1)}{6}\ \ = \ \ \frac{(n+1)(2n+1)}{6} \\ \\ .\ \ \ \ \ = \ 51 * 101 / 6 = 858.5$