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(D) First,we calculate the mean $(\bar{x})$:$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(2 	imes 1) + (4 	imes 2) + (6 	imes 3) + (8 	imes 2)

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$\frac{32}{9}$

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(D) First,we calculate the mean $(\bar{x})$:$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(2 	imes 1) + (4 	imes 2) + (6 	imes 3) + (8 	imes 2) + (10 	imes 1)}{1 + 2 + 3 + 2 + 1} = \frac{2 + 8 + 18 + 16 + 10}{9} = \frac{54}{9} = 6$.Now,calculate the mean deviation about the mean $(MD_{\bar{x}})$:$MD_{\bar{x}} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{1|2-6| + 2|4-6| + 3|6-6| + 2|8-6| + 1|10-6|}{9} = \frac{4 + 4 + 0 + 4 + 4}{9} = \frac{16}{9}$.Next,calculate the median $(M)$:Total frequency $N = 9$. The median is the value corresponding to the $\frac{N+1}{2}$-th observation,which is the $5$-th observation. Looking at the cumulative frequencies $(1, 3, 6, 8, 9)$,the $5$-th observation falls in the group where $x_i = 6$. Thus,$M = 6$.Calculate the mean deviation about the median $(MD_M)$:$MD_M = \frac{\sum f_i |x_i - M|}{\sum f_i} = \frac{1|2-6| + 2|4-6| + 3|6-6| + 2|8-6| + 1|10-6|}{9} = \frac{16}{9}$.The sum of the mean deviation from the mean and the mean deviation from the median is:$\frac{16}{9} + \frac{16}{9} = \frac{32}{9}$.

$x_i$	$2$	$4$	$6$	$8$	$10$
$f_i$	$1$	$2$	$3$	$2$	$1$

${x_i}$	$5$	$7$	$9$	$10$	$12$	$15$
${f_i}$	$8$	$6$	$2$	$2$	$2$	$6$

Consider the following distribution:
$x_i$ $2$ $4$ $6$ $8$ $10$
$f_i$ $1$ $2$ $3$ $2$ $1$

The sum of the mean deviation from the mean and the mean deviation from the median of this distribution is:

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Consider the following distribution:$x_i$$2$$4$$6$$8$$10$$f_i$$1$$2$$3$$2$$1$The sum of the mean deviation from the mean and the mean deviation from the median of this distribution is: