• # question_answer Let r be the range of $n(\forall n\ge 1)$ observations ${{x}_{1}},{{x}_{2}},....,{{x}_{n}}$,if$S=\sqrt{\frac{\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\overline{x})}^{2}}}}{n-1}}$then A) $S<r\sqrt{\frac{{{n}^{2}}+1}{n-1}}$ B) $S\ge r\sqrt{\frac{n}{n-1}}$ C) $S=r\sqrt{\frac{n}{n-1}}$         D) $S<r\sqrt{\frac{n}{n-1}}$

[d]: Here range = r = largest value - smallest value $=Max\left| {{x}_{i}}-{{x}_{j}} \right|(i\ne j)$and${{S}^{2}}=\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\overline{x})}^{2}}}$ Now,${{({{x}_{i}}-\overline{x})}^{2}}={{\left[ {{x}_{i}}-\frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} \right]}^{2}}$ $=\frac{1}{{{n}^{2}}}{{\left[ ({{x}_{i}}-{{x}_{1}})+({{x}_{i}}-{{x}_{2}})+...+({{x}_{i}}-{{x}_{n}}) \right]}^{2}}$ $=\frac{1}{{{n}^{2}}}[({{x}_{i}}-{{x}_{1}})+({{x}_{i}}-{{x}_{2}})+...+({{x}_{i}}-{{x}_{i-1}})$ $+({{x}_{i}}-{{x}_{i+1}})+...+{{({{x}_{i}}-{{x}_{n}})}^{2}}]$ $\Rightarrow$${{({{x}_{i}}-\overline{x})}^{2}}\le \frac{1}{{{n}^{2}}}{{[(n-1)r]}^{2}}$$(\because |{{x}_{i}}-{{x}_{j}}|\le r)$ $\Rightarrow$$\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{({{x}_{i}}\overline{x})}^{2}}\le \frac{1}{{{n}^{2}}(n-1)}}\sum\limits_{{}}^{{}}{{{[(n-1)r]}^{2}}}$ (summing up and dividing by (n - 1) both sides) $\frac{1}{{{n}^{2}}}\frac{1}{n-}n{{(n-1)}^{2}}{{r}^{2}}=\frac{n-1}{n}{{r}^{2}}<\frac{n}{n-1}{{r}^{2}}$ $\left( \because \forall n>1,n>\frac{1}{n} \right)$ Therefore${{S}^{2}}<\frac{n}{n-1}.{{r}^{2}}$or$S<r\sqrt{\frac{n}{n-1}}$