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 :
| \[\because \,\,r=\underset{i\,\ne \,j}{\mathop{\max }}\,\,\,|{{x}_{i}}-{{x}_{f}}|\] and \[{{S}^{2}}=\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}}\] | |
| Now, consider \[{{({{x}_{i}}-\bar{x})}^{2}}={{\left( {{x}_{i}}-\frac{{{x}_{1}}+{{x}_{2}}+....+{{x}_{n}}}{n} \right)}^{2}}\] | |
| \[=\frac{1}{{{n}^{2}}}[({{x}_{i}}-{{x}_{1}})+({{x}_{i}}-{{x}_{2}})+....+({{x}_{i}}-{{x}_{i}}-1)\]\[+({{x}_{i}}-{{x}_{i}}+1)+....+({{x}_{i}}-{{x}_{n}}){{]}^{2}}\le \frac{1}{{{n}^{2}}}{{[(n-1)r]}^{2}}\] | \[[\because \,\,|{{x}_{i}}-{{x}_{j}}|\le r]\] |
| \[\Rightarrow \,\,{{({{x}_{i}}-\bar{x})}^{2}}\le {{r}^{2}}\Rightarrow \sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}}\le n{{r}^{2}}\] | |
| \[\Rightarrow \,\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}\le \frac{n{{r}^{2}}}{(n-1)}}\Rightarrow {{S}^{2}}\le \frac{n{{r}^{2}}}{(n-1)}\] | |
| \[\Rightarrow S\le r\sqrt{\frac{n}{n-1}}\] | |
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