Question

The standard deviation of the first n natural numbers is

• A. $ \frac{\sqrt{{{n}^{2}}+1}}{12} $
• B. $ \frac{{{n}^{2}}-1}{12} $
• C. $ \sqrt{\frac{{{n}^{2}}-1}{12}} $
• D. $ \frac{{{n}^{2}}+1}{12} $

Answer 1

Answer: (C)

Since, $ \Sigma n=\frac{n(n+1)}{2} $ and $ \Sigma {{n}^{2}}=\frac{n(n+1)\,(2n+1)}{6} $
$ \therefore $ $ SD=\sqrt{\frac{\Sigma {{x}^{2}}}{n}-{{\left( \Sigma \frac{x}{n} \right)}^{2}}} $
$ =\sqrt{\frac{\Sigma {{n}^{2}}}{n}-{{\left( \frac{\Sigma n}{n} \right)}^{2}}} $
$ =\sqrt{\frac{n(n+1)(2n+1)}{6n}-{{\left( \frac{n+(n+1)}{2n} \right)}^{2}}} $
$ =\sqrt{\frac{(n+1)(2n+1)}{6}-\frac{{{(n+1)}^{2}}}{4}} $
$ =\sqrt{\frac{{{n}^{2}}-1}{12}} $