Variance of first n natural numbers is qquad | vedclass

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(D) The variance $\sigma^2$ of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - (\bar{x})^2$.Here,

Vedclass · Accepted Answer

$\frac{n^2-1}{12}$

Vedclass · Answer

(D) The variance $\sigma^2$ of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - (\bar{x})^2$.Here,$\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\bar{x} = \frac{n+1}{2}$.Substituting these values:$\sigma^2 = \frac{1}{n} \left( \frac{n(n+1)(2n+1)}{6} \right) - \left( \frac{n+1}{2} \right)^2$$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$Taking $\frac{n+1}{2}$ as a common factor:$\sigma^2 = \frac{n+1}{2} \left( \frac{2n+1}{3} - \frac{n+1}{2} \right)$$\sigma^2 = \frac{n+1}{2} \left( \frac{4n+2 - 3n-3}{6} \right)$$\sigma^2 = \frac{n+1}{2} \left( \frac{n-1}{6} \right)$$\sigma^2 = \frac{n^2-1}{12}$

$x_i$	$5$	$7$	$9$	$11$
$f_i$	$3$	$2$	$1$	$2$

Variance of first $n$ natural numbers is $\qquad$ .

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