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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Statistics
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The variance of $n$ natural number is

$\begin {array} {1 1} (A)\;\large\frac{n^2+1}{12} & \quad (B)\;\large\frac{n^2-1}{12} \\ (C)\;\large\frac{(n+1)(2n+1)}{6} & \quad (D)\;\bigg[ \large\frac{n(n+1)}{2} \bigg]^2 \end {array}$

 

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1 Answer

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$ \overline x = \large\frac{1+2+3+....+n}{n}$
$ = \large\frac{(n+1)}{2}$
$ \therefore \sigma^2 = \large\frac{ \Sigma x_i^2}{n} $ $ - ( \overline x)^2$
$ = \large\frac{ \Sigma n^2}{n}$ $ - \large\frac{(n+1)^2}{2}$
$ \large\frac{(n+1)(2n+1)}{6n} $ $ - \bigg( \large\frac{n+1}{2} \bigg)^2$
$ = \large\frac{n^2-1}{12}$
Ans : (B)
answered Jan 30, 2014 by thanvigandhi_1
 
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