Question

The variance of first $n$ natural numbers is

• A. $\frac{n(n+1)}{2}$
• B. $\frac{(n+1)(n+5)}{12}$
• C. $\frac{(n+1)(n-5)}{12}$
• D. $\frac{\left(n^2-1\right)}{12}$

Answer 1

Answer: (D)

Variance of first $n$ natural numbers
$=\frac{\Sigma n^2}{n}-{\left(\frac{\Sigma n}{n}\right)}^2$
$=\frac{n(n+1)(2n+1)}{6n}-{\left(\frac{n(n+1)}{2n}\right)}^2$
$=(n+1)\left[\frac{2n+1}{6}-\frac{(n+1)}{4}\right]$
$=\frac{(n+1)}{12}[4n+2-3n-3]$
$=\frac{(n+1)}{12}\times (n-1)=\frac{n^2-1}{12}$