किसी समूह के प्रेक्षणों {x_1},,{x_2},,.....{x | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$X$	$1$	$3$	$5$	$7$	$9$
$(f)$	$4$	$24$	$28$	$\alpha$	$8$

किसी समूह के प्रेक्षणों ${x_1},\,{x_2},\,.....{x_n}$ के लिये परिसर $r$ तथा मानक विचलन ${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} $ हैं, तब

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