Question

Find the mean, standard deviation and variance of first n natural numbers.

Answer 1

First n natural numbers are $1,2,3,...,n.$

Mean, $\overline{x}=\frac{(1+2+3+\cdots +n)}{n}=\frac{1}{n}.\frac{1}{2}n(n+1)=\frac{1}{2}(n+1)$

Sum of first n natural number

$[\because (1+2+3+\cdots +n)=\frac{1}{2}n(n+1\left)\right]$

$\therefore$ variance, ${\sigma }^2=\frac{\sum x_i^2}{n}-{\overline{x}}^2$

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

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

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

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

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

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

$\therefore$ variance, ${\sigma }^2=\frac{(n^2-1)}{12}$

Standard deviation, $\sigma =\sqrt{\frac{n^2-1}{12}}=\frac{1}{2}.\sqrt{\frac{n^2-1}{3}}$