For a discrete series (where all values are n | vedclass

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(C) The formula for Standard Deviation is $S.D. = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n}}$.The formula for Mean Deviation about mean is $M.D. = \frac{\

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$M.D. < S.D.$

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(C) The formula for Standard Deviation is $S.D. = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n}}$.The formula for Mean Deviation about mean is $M.D. = \frac{\sum|x_i - \bar{x}|}{n}$.Let $|x_i - \bar{x}| = y_i$. Then $(x_i - \bar{x})^2 = |x_i - \bar{x}|^2 = y_i^2$.Consider $(S.D.)^2 - (M.D.)^2 = \frac{\sum y_i^2}{n} - \left(\frac{\sum y_i}{n}\right)^2$.This expression represents the variance of the values $y_i$,which is always non-negative. Since not all values are equal,the variance is strictly greater than zero.Therefore,$(S.D.)^2 - (M.D.)^2 > 0$,which implies $(S.D.)^2 > (M.D.)^2$.Since both $S.D.$ and $M.D.$ are non-negative,we conclude $S.D. > M.D.$ or $M.D. < S.D.$.

Class	$0-5$	$5-10$	$10-15$	$15-20$	$20-25$	$25-30$
Frequency $(f_i)$	$3$	$4$	$7$	$11$	$2$	$5$

For a discrete series (where all values are not equal),what is the relationship between Mean Deviation $(M.D.)$,Mean $(Mean)$,and Standard Deviation $(S.D.)$?

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