The S.D. of the first n natural numbers is | vedclass

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Question

(c) $S. D.$ of first $n$ natural numbers $ = \sqrt {\frac{1}{n}\Sigma {x^2} - {{\left( {\frac{{\Sigma x}}{n}} \right)}^2}} $,

$ = \sqrt {\frac{{n(n

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) $S. D.$ of first $n$ natural numbers $ = \sqrt {\frac{1}{n}\Sigma {x^2} - {{\left( {\frac{{\Sigma x}}{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} - {{\left( {\frac{{n + 1}}{2}} \right)}^2}} = \sqrt {\frac{{n + 1}}{2}\left( {\frac{{2n + 1}}{3} - \frac{{n + 1}}{2}} \right)} $

$ = \sqrt {\frac{{n + 1}}{2}\left( {\frac{{4n + 2 - 3n - 3}}{6}} \right)} $

$ = \sqrt {\frac{{{n^2} - 1}}{{12}}} $.

$X$	$c$	$2c$	$3c$	$4c$	$5c$	$6c$
$f$	$2$	$1$	$1$	$1$	$1$	$1$

The $S.D$. of the first $n$ natural numbers is

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