In any discrete series (when all values are n | vedclass

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(D) For any frequency distribution,the relationship between Mean Deviation $(M.D.)$ about the mean and Standard Deviation $(S.D.)$ is given by the ine

Vedclass · Accepted Answer

$M.D. \le S.D.$

Vedclass · Answer

(D) For any frequency distribution,the relationship between Mean Deviation $(M.D.)$ about the mean and Standard Deviation $(S.D.)$ is given by the inequality $M.D. \le S.D.$This is because the variance of the absolute deviations from the mean is always non-negative.Specifically,for a set of values,the $S.D.$ is defined as $\sqrt{\frac{1}{N}\sum f_i(x_i - \bar{x})^2}$ and the $M.D.$ about mean is $\frac{1}{N}\sum f_i|x_i - \bar{x}|$.By the properties of dispersion,the root mean square deviation is always greater than or equal to the mean absolute deviation,hence $M.D. \le S.D.$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
$f_i$	$2$	$18$	$30$	$45$	$35$	$20$	$6$	$3$

In any discrete series (when all values are not same),the relationship between $M.D.$ about mean and $S.D.$ is:

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Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

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