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

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Question

(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}}$.

Classes	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Frequency	$11$	$29$	$18$	$4$	$5$	$3$

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

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Classes $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $11$ $29$ $18$ $4$ $5$ $3$

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