The mean deviation from the median for the fo | vedclass

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(A) First,we calculate the cumulative frequency $(cf)$ and the median $(M)$:$x_i$$f_i$$cf$$|x_i - M|$$f_i|x_i - M|$$10$$6$$6$$|10-12|=2$$12$$11$$12$$1

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$0.75$

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(A) First,we calculate the cumulative frequency $(cf)$ and the median $(M)$:$x_i$$f_i$$cf$$|x_i - M|$$f_i|x_i - M|$$10$$6$$6$$|10-12|=2$$12$$11$$12$$18$$|11-12|=1$$12$$12$$18$$36$$|12-12|=0$$0$$13$$12$$48$$|13-12|=1$$12$Total frequency $N = \sum f_i = 6 + 12 + 18 + 12 = 48$.The median is the value corresponding to the $\frac{N}{2} = \frac{48}{2} = 24$th observation. Looking at the $cf$ column,the $24$th observation falls in the class where $x = 12$. Thus,$M = 12$.The mean deviation about the median is given by:$MD = \frac{\sum f_i|x_i - M|}{N} = \frac{12 + 12 + 0 + 12}{48} = \frac{36}{48} = 0.75$.

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

The mean deviation from the median for the following data is:
$x$ $10$ $11$ $12$ $13$
$f$ $6$ $12$ $18$ $12$

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The mean deviation from the median for the following data is:$x$$10$$11$$12$$13$$f$$6$$12$$18$$12$