A) \[S\le r\sqrt{\frac{n}{n-1}}\]
B) \[S=r\sqrt{\frac{n}{n-1}}\]
C) \[S\ge r\sqrt{\frac{n}{n-1}}\]
D) None of these
Correct Answer: A
Solution :
We have \[\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,i\,\ne j}{\mathop{r=\max |{{x}_{i}}-{{x}_{j}}|}}\,\] and \[{{S}^{2}}=\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}}\] Now, \[{{({{x}_{i}}-\bar{x})}^{2}}={{\left( {{x}_{i}}-\frac{{{x}_{1}}+{{x}_{2}}+.....+{{x}_{n}}}{n} \right)}^{2}}\] \[=\frac{1}{{{n}^{2}}}[({{x}_{i}}-{{x}_{1}})+({{x}_{i}}-{{x}_{2}})+....+({{x}_{i}}-{{x}_{i}}-1)\] \[+({{x}_{i}}-{{x}_{i}}+1)+.......\]\[+({{x}_{i}}-{{x}_{n}})]\le \frac{1}{{{n}^{2}}}{{[(n-1)r]}^{2}}\], \[[\because |{{x}_{i}}-{{x}_{j}}|\le r]\] Þ \[{{({{x}_{i}}-\bar{x})}^{2}}\le {{r}^{2}}\Rightarrow \sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}\le n{{r}^{2}}}\] Þ \[\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}\le \frac{n{{r}^{2}}}{(n-1)}}\]Þ \[{{S}^{2}}\le \frac{n{{r}^{2}}}{(n-1)}\] Þ \[S\le r\sqrt{\frac{n}{n-1}}\].
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