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Q.
The variance of first $n$ even natural numbers is
Statistics
Solution:
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}+\ldots+\left(2n\right)^{2}}{n}-\left(\frac{2+4+6+\ldots+2n}{n}\right)^{2} $
$\left(\because x_{i} =2,4, 6, \ldots, 2n\right)$
$=4\left[\frac{n\left(n+1\right)\left(2n+1\right)}{n\times6}-\left(\frac{n\left(n+1\right)}{n \times2}\right)^{2}\right]$
$=4\left[\frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{\left(n+1\right)^{2}}{4}\right]=4 \left[\frac{n^{2}-1}{12}\right]=\frac{n^{2}-1}{3}$ 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)
$=\frac{4 \left(n^{2}-1\right)}{12} = \frac{n^{2}-1}{3}$