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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
$S \le r\sqrt {\frac{n}{{n - 1}}} $
$S = r\sqrt {\frac{n}{{n - 1}}} $
$S \ge r\sqrt {\frac{n}{{n - 1}}} $
None of these
Solution
(a) We have $\mathop {r = \max |{x_i} – {x_j}|}\limits_{\,\,\,\,\,\,\,\,\,\,\,\,\,i\, \ne 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}$,
==> ${({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}}} $.
Similar Questions
What is the standard deviation of the following series
| class | $0-10$ | $10-20$ | $20-30$ | $30-40$ |
| Freq | $1$ | $3$ | $4$ | $2$ |
