Statistics-And-Probability Question 631
Question: Let r be the range and $ S^{2}=\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{(x_{i}-\bar{x})}^{2}}} $ be the S.D. of a set of observations $ x_1,x_2,…x_{n}, $ then
Options:
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
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We have $ r=\max | x_{i}-x_{j} | $ and $ i\ne j $ $ S^{2}=\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{(x_{i}-\bar{x})}^{2}}} $ Now, consider $ {{(x_{i}-\bar{x})}^{2}}={{( x_{i}-\frac{x_{i}+x_2+…+x_{n}}{n} )}^{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}} $
$ \Rightarrow {{(x_{i}-\bar{x})}^{2}}\le r^{2}\Rightarrow \sum\limits_{i=1}^{n}{{{(x_{i}-\bar{x})}^{2}}\le nr^{2}} $
$ \Rightarrow \frac{1}{n-1}\sum\limits_{i=1}^{n}{{{(x_{i}-\bar{x})}^{2}}\le \frac{nr^{2}}{(n-1)}\Rightarrow S\le r\sqrt{\frac{n}{n-1}}} $ .
BETA
