The S.D. of the first n natural numbers is | vedclass

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(C) The $S.D.$ of the first $n$ natural numbers is given by the formula: $\sigma = \sqrt{\frac{1}{n}\sum x^2 - \left(\frac{\sum x}{n}\right)^2}$.Subst

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The $S.D.$ of the first $n$ natural numbers is given by the formula: $\sigma = \sqrt{\frac{1}{n}\sum x^2 - \left(\frac{\sum x}{n}\right)^2}$.Substituting the sums $\sum x = \frac{n(n + 1)}{2}$ and $\sum x^2 = \frac{n(n + 1)(2n + 1)}{6}$:$\sigma = \sqrt{\frac{n(n + 1)(2n + 1)}{6n} - \left[\frac{n(n + 1)}{2n}\right]^2}$$= \sqrt{\frac{(n + 1)(2n + 1)}{6} - \left(\frac{n + 1}{2}\right)^2}$$= \sqrt{\frac{n + 1}{2} \left(\frac{2n + 1}{3} - \frac{n + 1}{2}\right)}$$= \sqrt{\frac{n + 1}{2} \left(\frac{4n + 2 - 3n - 3}{6}\right)}$$= \sqrt{\frac{n + 1}{2} \cdot \frac{n - 1}{6}}$$= \sqrt{\frac{n^2 - 1}{12}}$.

The $S.D.$ of the first $n$ natural numbers is

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