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Q.
Let $r$ be the range and $S^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}$ be the S.D. of a set of observations $x_{1}, x_{2}, \ldots . x _{ n }$, then
Statistics
Solution:
We have $r=\displaystyle\max _{i \neq j}\left|x_{i}-x_{j}\right|$ and
$S ^{2}=\frac{1}{n-1} \displaystyle\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}$
Now, consider $\left(x_{i}-\bar{x}\right)^{2}=\left(x_{i}-\frac{x_{1}+x_{2}+\ldots .+x_{n}}{n}\right)^{2}$
$=\frac{1}{n^{2}}\left[\left(x_{i}-x_{1}\right)+\left(x_{i}-x_{2}\right)+\ldots .+\left(x_{i}-x_{i}-1\right)\right] $
$\left.+\left(x_{i}-x_{i}+1\right)+\ldots+\left(x_{i}-x_{n}\right)\right] \leq \frac{1}{n^{2}}[(n-1) r]^{2}$
$\Rightarrow \left(x_{i}-\bar{x}\right)^{2} \leq r^{2} $
$\Rightarrow \displaystyle\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \leq n r^{2} $
$\Rightarrow \frac{1}{n-1} \displaystyle\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \leq \frac{n r^{2}}{(n-1)} $
$\Rightarrow S \leq r \sqrt{\frac{n}{n-1}}$