Question

Calculate the mean and the variance of first n natural numbers.

Answer 1

We have, first n natural numbers,

i.e. 1, 2, 3, 4, ..., n.

$\therefore$ Mean = $\frac{1+2+3+4+...+n}{n}$

$\Rightarrow$ Mean = $\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}[\because \sum n=\frac{n(n+1)}{2}]$

and $Var\left(x\right)=\frac{\sum x_i^2}{n}-\left({\overline{x}}^2\right)$

= $\frac{1^2+2^2+3^2+...+n^2}{n}-{\left(\frac{n+1}{2}\right)}^2$

= $\frac{n(n+1)(2n+1)}{6n}-\frac{(n+1{)}^2}{4}$

$[\because \sum n^2=\frac{n(n+1)(2n+1)}{6}]$

= $\frac{(n+1)(2n+1)}{6}-\frac{(n+1{)}^2}{4}$

= $\frac{n+1}{2}(\frac{2n+1}{3}-\frac{n+1}{2})$

= $\frac{n+1}{2}\left(\frac{4n+2-3n-3}{6}\right)$

= $\frac{n+1}{2}\left(\frac{n-1}{6}\right)=\frac{n^2-1}{12}$

Hence, the mean of first n natural numbers is $\frac{n+1}{2}$ and the variance of first n natural numbers is $\frac{n^2-1}{12}$