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પ્રથમ $n$ પ્રાકૃતિક સંખ્યાઓનું પ્રમાણિત વિચલન = ………
$\sqrt {\frac{{{n^2}\, - \,\,1}}{2}} $
$\sqrt {\frac{{{n^2}\, - \,\,1}}{3}} $
$\sqrt {\frac{{{n^2}\, - \,\,1}}{4}} $
$\sqrt {\frac{{{n^2}\, - \,\,1}}{{12}}} $
Solution
$\because \,\,S.D.\,\, = \,\,\,\sqrt {\frac{1}{n}\,\Sigma x_i^2\,\, – \,\,{{\left( {\frac{1}{n}\Sigma {x_i}} \right)}^2}} $
તેથી પ્રથમ $ n $ પ્રાકૃતિક સંખ્યાના ${\text{S}}{\text{.D}}{\text{.}}\,\,\, = \,\,\,\,\sqrt {\frac{{\text{1}}}{{\text{n}}}\Sigma {n^2}\, – \,\,{{\left( {\frac{1}{n}\Sigma n} \right)}^2}} $
$ = \,\,\,\sqrt {\frac{1}{n}\,.\,\,\frac{{n(n\,\, + \,\,1)\,(2n\,\, + \,\,1)}}{6}\,\, – \,\,{{\left\{ {\frac{1}{n}\,\,.\,\frac{{n(n\,\, + \,\,1)}}{2}} \right\}}^2}} $
$ = \,\,\,\sqrt {\frac{{(n\,\, + \,\,1)\,(2n\,\, + \,\,1)}}{6}\,\, – \,\,\frac{{{{(n\,\, + \,\,1)}^2}}}{4}} \,\,\, = \,\,\sqrt {\frac{{(n\,\, + \,\,1)\,(n\,\, – \,\,1)}}{{12}}} \,\,\,\,\,\,\,\,\,\, = \,\,\sqrt {\frac{{{n^2}\, – \,\,1}}{{12}}} $
