The mean deviation about the median of the fo | vedclass

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(A) To find the mean deviation about the median,we first calculate the cumulative frequency $(c.f.)$:$x$$f$$c.f.$$|x_i - 	ext{Median}|$$f_i |x_i -

Vedclass · Accepted Answer

$7.5$

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(A) To find the mean deviation about the median,we first calculate the cumulative frequency $(c.f.)$:$x$$f$$c.f.$$|x_i - 	ext{Median}|$$f_i |x_i - 	ext{Median}|$$6$$4$$4$$18$$72$$12$$7$$11$$12$$84$$18$$9$$20$$6$$54$$24$$18$$38$$0$$0$$30$$15$$53$$6$$90$$36$$10$$63$$12$$120$$42$$5$$68$$18$$90$Total frequency $N = \Sigma f_i = 68$.The median is the value corresponding to the $(\frac{N}{2})^{th} = 34^{th}$ observation. From the $c.f.$ column,the $34^{th}$ observation lies in the class where $x = 24$.So,$	ext{Median} = 24$.The mean deviation about the median is given by:$	ext{M.D.}(	ext{Median}) = \frac{\Sigma f_i |x_i - 	ext{Median}|}{\Sigma f_i} = \frac{72 + 84 + 54 + 0 + 90 + 120 + 90}{68} = \frac{510}{68} = 7.5$.

Class Interval	$0$–$2$	$2$–$4$	$4$–$6$	$6$–$8$	$8$–$10$
Frequency	$1$	$3$	$4$	$1$	$2$

The mean deviation about the median of the following distribution is:
$x$ $6$ $12$ $18$ $24$ $30$ $36$ $42$
$f$ $4$ $7$ $9$ $18$ $15$ $10$ $5$

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The mean deviation about the mean for the following data is
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The mean deviation about the median of the following distribution is:$x$$6$$12$$18$$24$$30$$36$$42$$f$$4$$7$$9$$18$$15$$10$$5$