Question

Write the variance of first n natural numbers.

Answer 1

Variance = ${\sum }_{x^2/n}-({\sum }_{x/n}{)}^2$

But $\sum x=1+2+3+...+n=n(n+\frac{1}{2})$

$\sum x^2=n(n+1)\frac{(2n+1)}{6}$

Substituting these values we get

variance = $\frac{n^2-1}{12}$