The mean deviation from the median for the fo | vedclass

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(C) First,arrange the data in ascending order of $x_i$ and calculate the cumulative frequency:$x_i$$f_i$Cumulative Frequency$|x_i - M|$$f_i |x_i - M|$

Vedclass · Accepted Answer

$4.97$

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(C) First,arrange the data in ascending order of $x_i$ and calculate the cumulative frequency:$x_i$$f_i$Cumulative Frequency$|x_i - M|$$f_i |x_i - M|$$3$$3$$3$$10$$30$$6$$4$$7$$7$$28$$9$$5$$12$$4$$20$$12$$2$$14$$1$$2$$13$$4$$18$$0$$0$$15$$5$$23$$2$$10$$21$$4$$27$$8$$32$$22$$3$$30$$9$$27$Here,$N = \Sigma f_i = 30$.The median is the value corresponding to the cumulative frequency just greater than $\frac{N}{2} = 15$.The cumulative frequency $18$ corresponds to $x_i = 13$. Thus,Median $(M) = 13$.The sum $\Sigma f_i |x_i - 13| = 30 + 28 + 20 + 2 + 0 + 10 + 32 + 27 = 149$.Mean deviation from median $= \frac{\Sigma f_i |x_i - M|}{N} = \frac{149}{30} \approx 4.97$.

Income per day	Number of persons
$0-100$	$4$
$100-200$	$8$
$200-300$	$9$
$300-400$	$10$
$400-500$	$7$
$500-600$	$5$
$600-700$	$4$
$700-800$	$3$

The mean deviation from the median for the following distribution (corrected to $2$ decimals) is
$x_i$ $3$ $6$ $9$ $12$ $13$ $15$ $21$ $22$
$f_i$ $3$ $4$ $5$ $2$ $4$ $5$ $4$ $3$

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Let the mean of $6$ observations $1, 2, 4, 5, x,$ and $y$ be $5$ and their variance be $10$. Then their mean deviation about the mean is equal to $........$.

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

The mean deviation from the median for the following distribution (corrected to $2$ decimals) is$x_i$$3$$6$$9$$12$$13$$15$$21$$22$$f_i$$3$$4$$5$$2$$4$$5$$4$$3$