Question

The variance of first $n$ natural numbers, is

• A. $\frac{n+1}{2}$
• B. $\frac{n^2+1}{12}$
• C. $\frac{n^2-1}{6}$
• D. $\frac{n^2-1}{12}$

Answer 1

The correct option is D $\frac{n^2-1}{12}$

First $n$ natural number are, $x=1,2,3,.............n$
So mean of first $n$ natural number is $\overline{x}=\frac{\sum x_i}{n}=\frac{n+1}{2}$
$\therefore$ variance, ${\sigma }^2=\frac{\sum x_i^2}{n}-(\overline{x}{)}^2$
$=\frac{\frac{1}{6}n(n+1)(2n+1)}{n}-{\left(\frac{n+1}{2}\right)}^2=\frac{1}{12}(n+1)(n-1)=\frac{n^2-1}{12}$