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Question

Mathematics Question on Statistics

The standard deviation of the first n natural numbers is

A

n2+112\frac{\sqrt{{{n}^{2}}+1}}{12}

B

n2−112\frac{{{n}^{2}}-1}{12}

C

n2−112\sqrt{\frac{{{n}^{2}}-1}{12}}

D

n2+112\frac{{{n}^{2}}+1}{12}

Answer

n2−112\sqrt{\frac{{{n}^{2}}-1}{12}}

Explanation

Solution

Since, Σn=n(n+1)2\Sigma n=\frac{n(n+1)}{2} and Σn2=n(n+1) (2n+1)6\Sigma {{n}^{2}}=\frac{n(n+1)\,(2n+1)}{6}
∴\therefore SD=Σx2n−(Σxn)2SD=\sqrt{\frac{\Sigma {{x}^{2}}}{n}-{{\left( \Sigma \frac{x}{n} \right)}^{2}}}
=Σn2n−(Σnn)2=\sqrt{\frac{\Sigma {{n}^{2}}}{n}-{{\left( \frac{\Sigma n}{n} \right)}^{2}}}
=n(n+1)(2n+1)6n−(n+(n+1)2n)2=\sqrt{\frac{n(n+1)(2n+1)}{6n}-{{\left( \frac{n+(n+1)}{2n} \right)}^{2}}}
=(n+1)(2n+1)6−(n+1)24=\sqrt{\frac{(n+1)(2n+1)}{6}-\frac{{{(n+1)}^{2}}}{4}}
=n2−112=\sqrt{\frac{{{n}^{2}}-1}{12}}