Find the variance of the first n natural numb | vedclass

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(C) The variance $\sigma^2$ is given by the formula: $\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$. For the first $n$ natura

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

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(C) The variance $\sigma^2$ is given by the formula: $\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$. For the first $n$ natural numbers,$\sum x_i = \frac{n(n+1)}{2}$ and $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$. Substituting these values: $\sigma^2 = \frac{n(n+1)(2n+1)}{6n} - \left( \frac{n(n+1)}{2n} \right)^2$. $\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$. Taking the common denominator $12$: $\sigma^2 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12}$. $\sigma^2 = \frac{(n+1) [2(2n+1) - 3(n+1)]}{12} = \frac{(n+1)(4n+2 - 3n - 3)}{12} = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}$.

Marks	$1-3$	$3-5$	$5-7$	$7-9$
Number of students	$40$	$30$	$20$	$10$

Find the variance of the first $n$ natural numbers.

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