For the following distribution,the mean devia | vedclass

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(B) First,we calculate the cumulative frequency $(cf)$ and the total frequency $(N)$:$x_i$$f_i$$cf$$6$$4$$4$$12$$7$$11$$18$$9$$20$$24$$18$$38$$30$$15$

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(B) First,we calculate the cumulative frequency $(cf)$ and the total frequency $(N)$:$x_i$$f_i$$cf$$6$$4$$4$$12$$7$$11$$18$$9$$20$$24$$18$$38$$30$$15$$53$$36$$10$$63$$42$$5$$68$Here,$N = 68$,which is even. The median is the average of the $(\frac{N}{2})^{th}$ and $(\frac{N}{2} + 1)^{th}$ observations,i.e.,the $34^{th}$ and $35^{th}$ observations.Looking at the cumulative frequency,both the $34^{th}$ and $35^{th}$ observations fall in the class where $x_i = 24$.Therefore,$	ext{Median} = 24$.Now,we calculate the mean deviation about the median using the formula $	ext{MD}(	ext{Median}) = \frac{\sum f_i |x_i - 	ext{Median}|}{N}$:$\sum f_i |x_i - 24| = 4|6-24| + 7|12-24| + 9|18-24| + 18|24-24| + 15|30-24| + 10|36-24| + 5|42-24|$$= 4(18) + 7(12) + 9(6) + 18(0) + 15(6) + 10(12) + 5(18)$$= 72 + 84 + 54 + 0 + 90 + 120 + 90 = 510$$	ext{MD}(	ext{Median}) = \frac{510}{68} = 7.5$.

Class Interval	$0-4$	$4-8$	$8-12$	$12-16$
Frequency	$4$	$3$	$2$	$1$

Marks obtained	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Number of students	$10$	$8$	$12$	$9$	$11$

For the following distribution,the mean deviation about the median is:
$x_i$ $6$ $12$ $18$ $24$ $30$ $36$ $42$
$f_i$ $4$ $7$ $9$ $18$ $15$ $10$ $5$

Similar Questions

The mean deviation about the mean for the following data is

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The mean deviation about the median for the observations $70, 42, 63, 34, 44, 54, 55, 46, 38, 48$ is:

Find the mean deviation about the mean for the following data:
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$x_i$	$6$	$12$	$18$	$24$	$30$	$36$	$42$
$f_i$	$4$	$7$	$9$	$18$	$15$	$10$	$5$

For the following distribution,the mean deviation about the median is:$x_i$$6$$12$$18$$24$$30$$36$$42$$f_i$$4$$7$$9$$18$$15$$10$$5$