If m and M denote the mean deviations about t | vedclass

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(A) The given data is $20, 5, 15, 2, 7, 3, 11$. The number of observations $n = 7$. First,calculate the mean $\bar{x} = \frac{20+5+15+2+7+3+11}{7} = \

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$\frac{1}{7}$

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(A) The given data is $20, 5, 15, 2, 7, 3, 11$. The number of observations $n = 7$. First,calculate the mean $\bar{x} = \frac{20+5+15+2+7+3+11}{7} = \frac{63}{7} = 9$. Mean deviation about mean $m = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|20-9|+|5-9|+|15-9|+|2-9|+|7-9|+|3-9|+|11-9|}{7} = \frac{11+4+6+7+2+4+2}{7} = \frac{36}{7}$. Wait,let us re-calculate: $11+4+6+7+2+4+2 = 36$. So $m = \frac{36}{7}$. Now,for median,arrange data in ascending order: $2, 3, 5, 7, 11, 15, 20$. Median $= \left(\frac{7+1}{2}ight)^{	ext{th}}$ term $= 4^{	ext{th}}$ term $= 7$. Mean deviation about median $M = \frac{\sum |x_i - 	ext{Median}|}{n} = \frac{|2-7|+|3-7|+|5-7|+|7-7|+|11-7|+|15-7|+|20-7|}{7} = \frac{5+4+2+0+4+8+13}{7} = \frac{36}{7}$. Wait,let us re-calculate: $5+4+2+0+4+8+13 = 36$. So $M = \frac{36}{7}$. Actually,$m = \frac{36}{7}$ and $M = \frac{36}{7}$. The mean of $m$ and $M$ is $\bar{x}^{\prime} = \frac{m+M}{2} = \frac{36/7 + 36/7}{2} = \frac{36}{7}$. The mean deviation about the mean of $m$ and $M$ is $\frac{|m-\bar{x}^{\prime}| + |M-\bar{x}^{\prime}|}{2} = \frac{|36/7 - 36/7| + |36/7 - 36/7|}{2} = 0$. Given the options,there might be a calculation error in the provided question's data or options. Re-evaluating the sum for $m$: $|20-9|=11, |5-9|=4, |15-9|=6, |2-9|=7, |7-9|=2, |3-9|=6, |11-9|=2$. Sum $= 11+4+6+7+2+6+2 = 38$. So $m = \frac{38}{7}$. Re-evaluating the sum for $M$: $|20-7|=13, |5-7|=2, |15-7|=8, |2-7|=5, |7-7|=0, |3-7|=4, |11-7|=4$. Sum $= 13+2+8+5+0+4+4 = 36$. So $M = \frac{36}{7}$. Mean of $m$ and $M$ is $\bar{x}^{\prime} = \frac{38/7 + 36/7}{2} = \frac{74/7}{2} = \frac{37}{7}$. Mean deviation about mean of $m$ and $M$ is $\frac{|38/7 - 37/7| + |36/7 - 37/7|}{2} = \frac{1/7 + 1/7}{2} = \frac{2/7}{2} = \frac{1}{7}$.

$x_i$	$3$	$6$	$9$	$12$	$13$	$15$	$21$	$22$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$

If $m$ and $M$ denote the mean deviations about the mean and about the median respectively of the data $20, 5, 15, 2, 7, 3, 11$,then the mean deviation about the mean of $m$ and $M$ is

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