The variance of the first n natural numbers i | vedclass

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Question

(a) Variance $ = {({\rm{S}}{\rm{.D}}{\rm{.}})^2}$$ = \frac{1}{n}\Sigma {x^2} - {\left( {\frac{{\Sigma x}}{n}} \right)^2}$,$\left( {\because \;\;\bar x

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) Variance $ = {({\rm{S}}{\rm{.D}}{\rm{.}})^2}$$ = \frac{1}{n}\Sigma {x^2} - {\left( {\frac{{\Sigma x}}{n}} \right)^2}$,$\left( {\because \;\;\bar x = \frac{{\Sigma x}}{n}} \right)$

$ = \frac{{n(n + 1)\;(2n + 1)}}{{6n}} - {\left( {\frac{{n(n + 1)}}{{2n}}} \right)^2} $

$= \frac{{{n^2} - 1}}{{12}}$.

$\mathrm{x}$	$-2$	$-1$	$3$	$4$	$6$
$\mathrm{P}(\mathrm{X}=\mathrm{x})$	$\frac{1}{5}$	$\mathrm{a}$	$\frac{1}{3}$	$\frac{1}{5}$	$\mathrm{~b}$

The variance of the first $n$ natural numbers is

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