Question

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

Answer 1

STEP 1 : Assumption

Let the first $n$ natural numbers be $1,2,3,4,.......,n$.

Sum of first $n$ natural numbers $\sum _{}x=\frac{n(n+1)}{2}$

Sum of the square of the first $n$ natural numbers is $\sum _{}{x}^2=\frac{n(n+1)\left(2n+1\right)}{6}$

STEP 2 : Finding the standard deviation

The standard deviation, $\sigma =\sqrt{\frac{\sum _{}{x}^2}{n}-{\left(\frac{\sum _{}x}{n}\right)}^2}$

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

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

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

$\sigma =\sqrt{\left(\frac{n+1}{2}\right)\left[\frac{{\displaystyle (2n+1)}}{3}-\frac{{\displaystyle (n+1)}}{2}\right]}$

$\sigma =\sqrt{\left(\frac{n+1}{2}\right)\left[\frac{{\displaystyle 2(2n+1)-3(n+1)}}{6}\right]}$

$\sigma =\sqrt{\left(\frac{n+1}{2}\right)\left[\frac{{\displaystyle 4n+2-(3n+3)}}{6}\right]}$

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

$\sigma =\sqrt{\left(\frac{n+1}{2}\right)\left(\frac{{\displaystyle n-1}}{6}\right)}$

$\sigma =\sqrt{\frac{{\displaystyle n^2-1}}{12}}$

Hence, the standard deviation of the first $n$ natural numbers is $\sqrt{\frac{n^2-1}{12}}$.