Let r be the range and {S^2} = frac{1}{{n - 1 | vedclass

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Question

(a) We have $\mathop {r = \max |{x_i} - {x_j}|}\limits_{\,\,\,\,\,\,\,\,\,\,\,\,\,i\, 
e j} $
and ${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{

Vedclass · Accepted Answer

$S \le r\sqrt {\frac{n}{{n - 1}}} $

Vedclass · Answer

(a) We have $\mathop {r = \max |{x_i} - {x_j}|}\limits_{\,\,\,\,\,\,\,\,\,\,\,\,\,i\, 
e j} $
and ${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} $
Now, ${({x_i} - \bar x)^2} = {\left( {{x_i} - \frac{{{x_1} + {x_2} + ..... + {x_n}}}{n}} ight)^2}$

$ = \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}$,

==> ${({x_i} - \bar x)^2} \le {r^2} \Rightarrow \sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2} \le n{r^2}} $

==> $\frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2} \le \frac{{n{r^2}}}{{(n - 1)}}} $==> ${S^2} \le \frac{{n{r^2}}}{{(n - 1)}}$

==> $S \le r\sqrt {\frac{n}{{n - 1}}} $.

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

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