Question

The variance of first $n$ even natural numbers is

• A. $\frac{n^2-1}{12}$
• B. $\frac{n^2-1}{6}$
• C. $\frac{n^2-1}{3}$
• D. None of these

Answer 1

Answer: (C)

For the variates $x_1,x_2,\dots ,x_n$ variance is given by
${\sigma }^2=\frac{\Sigma x^2}{n}-{\left(\frac{\Sigma x}{n}\right)}^2$
$=\frac{2^2+4^2+\dots +{\left(2n\right)}^2}{n}-{\left(\frac{2+4+6+\dots +2n}{n}\right)}^2$
$\left(\because x_i=2,4,6,\dots ,2n\right)$
$=4\left[\frac{n\left(n+1\right)\left(2n+1\right)}{n\times 6}-{\left(\frac{n\left(n+1\right)}{n\times 2}\right)}^2\right]$
$=4\left[\frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{{\left(n+1\right)}^2}{4}\right]=4\left[\frac{n^2-1}{12}\right]=\frac{n^2-1}{3}$
Short Cut Method :
We know that variance of first n natural numbers is given by $\frac{n^2-1}{12}$
Now we need the variance of $2,4,6,\dots ,2n$ i.e., $1\times 2,2\times 2,3\times 2,4\times 2,\dots ,n\times 2$ natural numbers which indicate that each item is multiplied by 2 and we know that if each item is m ultiplied by a scalar $(k)$ then new variance will be $k^2$ of the former variance
New variance $=k^2$ of the variance of first n natural numbers
$=k^2\frac{\left(n^2-1\right)}{12}$ (putting k = 2)
$=\frac{4\left(n^2-1\right)}{12}=\frac{n^2-1}{3}$