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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Statistics

The S.D of first $n$ naturals is

$\begin {array} {1 1} (A)\;\large\frac{n+1}{2} & \quad (B)\;\sqrt{\large\frac{n / n+1}{2}} \\ (C)\;\sqrt {\large\frac{n^2-1}{12}} & \quad (D)\;None \: of \: these \end {array}$

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1 Answer

$ \sigma = \sqrt { \large\frac{1}{n} \Sigma x^2 - \bigg( \large\frac{ \Sigma x}{n} \bigg)^2}$
$ = \sqrt { \large\frac{n(n+1)(2n+1)}{6n}- \bigg( \large\frac{n(n+1)}{2n} \bigg)^2}$
$ = \sqrt{ \large\frac{n+1}{2} \bigg( \large\frac{2n+1}{3}-\large\frac{n+1}{2} \bigg)}$
$ = \sqrt{ \large\frac{n+1}{2} \bigg( \large\frac{n-1}{6} \bigg) }$
$ = \sqrt { \large\frac{n^2-1}{12}}$
Ans : (C)
answered Jan 25, 2014 by thanvigandhi_1
 

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