Find the standard deviation of first n natural numbers is equal to $σ$

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

\sqrt{\frac{n^{2} + 1}{12}}

\sqrt{\frac{n^{2} - 1}{16}}

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

answer is A.

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Detailed Solution

Here,
1, 2, 3... n are the first n natural numbers.

∴

MEAN,

\overset{̅}{x} = \frac{\sum x}{n}

= \frac{1 + 2 + 3 + \dots \dots + n}{n}

= \frac{n (n + 1)}{2 n}

= \frac{n + 1}{2}

Thus, standard deviation

σ = \sqrt{\frac{\sum x^{2}}{n} - {(\frac{\sum x}{n})}^{2}}

= \sqrt{\frac{n (n + 1) (2 n + 1)}{6 n} - {(\frac{n + 1}{2})}^{2}}

= \sqrt{\frac{(n + 1) (2 n + 1)}{6} - {(\frac{n + 1}{2})}^{2}}

= \sqrt{(\frac{n + 1}{2}) [\frac{2 (2 n + 1)}{3} - \frac{(n + 1)}{2}]}

= \sqrt{(\frac{n + 1}{2}) (\frac{4 n + 2 - 3 n - 3}{6})}

= \sqrt{(\frac{n + 1}{2}) (\frac{n - 1}{6})}

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

Hence, the S.D. of the first n natural numbers is

σ = \sqrt{\frac{n^{2} - 1}{12}}

Correct option is 1.

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