Question

find the standard deviation of first n natural numbers

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{first}\mathrm{n}\mathrm{natural}\mathrm{numbers}\mathrm{are}: \\ 1,2,3,4,..........\mathrm{n}. \\ \mathrm{Mean}\mathrm{of}\mathrm{first}\mathrm{n}\mathrm{natural}\mathrm{numbers},\overline{\mathrm{x}}=\frac{1+2+3+......+\mathrm{n}}{\mathrm{n}}=\frac{1}{\mathrm{n}}\times \left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]=\frac{\mathrm{n}+1}{2} \\ \mathrm{Now},\mathrm{variance},{\mathrm{\sigma }}^2=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{{\mathrm{x}}_{\mathrm{i}}}^2}}{\mathrm{n}}-\left(\overline{\mathrm{x}}\right){}^2 \\ =\frac{1^2+2^2+....{\mathrm{n}}^2}{\mathrm{n}}-{\left(\frac{\mathrm{n}+1}{2}\right)}^2 \\ =\frac{1}{\mathrm{n}}\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}-\frac{\left(\mathrm{n}+1\right)\left(\mathrm{n}+1\right)}{4} \\ =\frac{\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}-\frac{\left(\mathrm{n}+1\right)\left(\mathrm{n}+1\right)}{4} \\ =\left(\mathrm{n}+1\right)\left[\frac{\left(2\mathrm{n}+1\right)}{6}-\frac{\left(\mathrm{n}+1\right)}{4}\right] \\ =\left(\mathrm{n}+1\right)\left[\frac{4\mathrm{n}+2-3\mathrm{n}-3}{12}\right] \\ =\frac{\left(\mathrm{n}+1\right)\left(\mathrm{n}-1\right)}{12} \\ =\frac{{\mathrm{n}}^2-1}{12} \\ \mathrm{So},\mathrm{variance},{\mathrm{\sigma }}^2=\frac{{\mathrm{n}}^2-1}{12} \\ \mathbf{Now}\mathbf{,}\mathbf{}\mathbf{standard}\mathbf{}\mathbf{deviation}\mathbf{,}\mathbf{}\mathbf{\sigma }\mathbf{}\mathbf{=}\mathbf{}\sqrt{\frac{{\mathbf{n}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}{\mathbf{12}}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\sqrt{\frac{{\mathbf{n}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}{\mathbf{3}}} \end{gathered}$$