- A$\sqrt {\frac{{{n^2}\, - \,\,1}}{2}} $
- B$\sqrt {\frac{{{n^2}\, - \,\,1}}{3}} $
- C$\sqrt {\frac{{{n^2}\, - \,\,1}}{4}} $
- ✓$\sqrt {\frac{{{n^2}\, - \,\,1}}{{12}}} $
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}}} $
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| વર્ગ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ |
| આવૃતિ | $\alpha$ | $110$ | $54$ | $30$ | $\beta$ |
જો બધીજ આવૃતિનો સરવાળો $584$ હોય અને મધ્યસ્થ $45$ હોય તો $|\alpha-\beta|$ મેળવો.