Question

The variance of first n natural numbers is ____________.

Answer 1

For first n natural numbers. 1, 2, 3, ...........n

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{1+2+3+........+n}{n}\\ & =& \frac{n\left(n+1\right)}{2n}\\ \mathrm{Mean}& =& \frac{n+1}{2}\end{array}$
$\begin{array}{rcl}\mathrm{Hence},\mathrm{Variance}\mathrm{is}{\sigma }^2& =& \frac{\sum _{}{\left(x_i\right)}^2}{n}-{\left(\mathrm{mean}\right)}^2\\ & =& \frac{1^2+2^2+3^2+....n^2}{n}-{\left(\frac{n+1}{2}\right)}^2\\ & =& \frac{n\left(n+1\right)\left(2n+1\right)}{6n}-{\left(\frac{n+1}{2}\right)}^2\\ & =& \frac{n+1}{2}\left[\frac{\left(2n+1\right)}{3}-\frac{n+1}{2}\right]\\ & =& \frac{n+1}{2}\left[\frac{4n+2-3n-3}{6}\right]\\ & & \end{array}$
$\mathrm{Variance}\mathrm{of}\mathrm{first}n\mathrm{natural}\mathrm{numbers}=\frac{n+1}{2}\left(\frac{n-1}{6}\right)=\frac{n^2-1}{12}.$