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

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Question

(A) The range $r$ is defined as $r = \max |x_i - x_j|$ for $i 
eq j$.Given the variance $S^2 = \frac{1}{n - 1} \sum_{i = 1}^n (x_i - \bar{x})^2$.It i

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The range $r$ is defined as $r = \max |x_i - x_j|$ for $i 
eq j$.Given the variance $S^2 = \frac{1}{n - 1} \sum_{i = 1}^n (x_i - \bar{x})^2$.It is a known statistical inequality that for any set of $n$ observations,the standard deviation $S$ and the range $r$ satisfy the relation $S \le r \sqrt{\frac{n}{n - 1}}$.This is derived from the fact that $\sum (x_i - \bar{x})^2 \le n \frac{(n-1)}{n} r^2$ is not the correct bound,but rather $\sum (x_i - \bar{x})^2 \le \frac{n}{4} r^2$ for specific cases,however,the standard inequality for the sample standard deviation $S$ in terms of range $r$ is $S \le r \sqrt{\frac{n}{n-1}}$.

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

Let $r$ be the range and $S^2 = \frac{1}{n - 1} \sum_{i = 1}^n (x_i - \bar{x})^2$ be the variance of a set of observations $x_1, x_2, \dots, x_n$. Then:

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