Question

Prove that the standard deviation of the first n natural numbers is $\sigma =\sqrt{\frac{n^2-1}{12}}$.

Answer 1

The first n natural numbers are $1,2,3,.....,n$.
Their mean, $\overline{x}=\frac{\sum x}{n}=\frac{1+2+3+.....+n}{n}$
$=\frac{n(n+1)}{2n}=\frac{n+1}{2}$
Sum of the square of the first n natural numbers is $\sum x^2=\frac{n(n+1)(2n+1)}{6}$
Thus, the standard deviation $\sigma =\sqrt{\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2}$
$=\sqrt{\frac{n(n+1)(2n+1)}{6n}-{\left(\frac{n+1}{2}\right)}^2}$
$=\sqrt{\frac{(n+1)(2n+1)}{6}-{\left(\frac{n+1}{2}\right)}^2}$
$=\sqrt{\left(\frac{n+1}{2}\right)\left[\frac{(2n+1}{3}-\frac{(n+1)}{2}\right]}$
$=\sqrt{\left(\frac{n+1}{2}\right)\left[\frac{2(2n+1)-3(n+1)}{6}\right]}$
$=\sqrt{\left(\frac{n+1}{2}\right)\left(\frac{4n+2-3n-3}{6}\right)}$
$=\sqrt{\left(\frac{n+1}{2}\right)\left(\frac{n-1}{6}\right)}$
$=\sqrt{\frac{n^2-1}{12}}$.
Hence, the S.D. of the first n natural numbers is $\sigma =\sqrt{\frac{n^2-1}{12}}$.