Find the variance of the first n natural numb | vedclass

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Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $12$: $\sigma^2 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12}$. $\sigma^2 = \frac{(n+1) [2(2n+1) - 3(n+1)]}{12} = \frac{(n+1)(4n+2 - 3n - 3)}{12} = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}$.

Student	Marks
Ravi	$25, 50, 45, 30, 70, 42, 36, 48, 35, 60$
Hashina	$10, 70, 50, 20, 95, 55, 42, 60, 48, 80$

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

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