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) \[S<r\sqrt{\frac{n}{n-1}}\]
Correct Answer: D
Solution :
[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\[S<r\sqrt{\frac{n}{n-1}}\]
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