The standard deviation of the first n natural | vedclass

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(D) The standard deviation ($S$.$D$.) is given by the formula: $S.D. = \sqrt{\frac{1}{n} \sum x_i^2 - \left( \frac{1}{n} \sum x_i \right)^2}$.For the

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

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(D) The standard deviation ($S$.$D$.) is given by the formula: $S.D. = \sqrt{\frac{1}{n} \sum x_i^2 - \left( \frac{1}{n} \sum x_i \right)^2}$.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:$S.D. = \sqrt{\frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} - \left( \frac{1}{n} \cdot \frac{n(n+1)}{2} \right)^2}$$S.D. = \sqrt{\frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}}$$S.D. = \sqrt{\frac{2(n+1)(2n+1) - 3(n+1)^2}{12}}$$S.D. = \sqrt{\frac{(n+1) [2(2n+1) - 3(n+1)]}{12}}$$S.D. = \sqrt{\frac{(n+1) [4n + 2 - 3n - 3]}{12}}$$S.D. = \sqrt{\frac{(n+1)(n-1)}{12}} = \sqrt{\frac{n^2 - 1}{12}}$.

$C$.$I$.	$0$ - $6$	$6$ - $12$	$12$ - $18$
f_i	$2$	$4$	$6$

The standard deviation of the first $n$ natural numbers is . . . . . . .

Similar Questions

The standard deviation of the following distribution is:
$C$.$I$. $0$ - $6$ $6$ - $12$ $12$ - $18$
f_i $2$ $4$ $6$

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Find the standard deviation of the first $n$ natural numbers.

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