Find the mean deviation and its coefficient about the median for the following data. | Statistics | SciPiPupil

Question: Find the mean deviation and its coefficient about the median for the following data.

Class	0-6	6-12	12-18	18-24	24-30
Frequency	8	10	12	9	5

Solution:

Arranging given data in cumulative frequency (c.f.) table,

Class	Frequency	c.f.
0-6	8	8
6-12	10	18
12-18	12	30
18-24	9	39
24-30	5	44
	N=44

Median (M) class = $\dfrac{N}{2}^{th} class$

$= \dfrac{44}{2}^{th} class$

$= 22^{nd} class$

In c.f. table, just greater value than 22 is 30 whose corresponding class is (12-18). So, median class = (12-18).

So we know,

l = 12, i = 6, f = 12, c.f. = 18, $\frac{N}{2}$ = 22

M = $l + \dfrac{i}{f} × \left ( \dfrac{N}{2} - c.f. \right ) $

$= 12 + \dfrac{6}{12} × (22-18)$

$= 12 + 2$

$= 14$

Now,

Arranging the data in a table:

Class	Frequency (f)	Mid Value(x)	|x-M|	f|x-M|
0-6	6	3	11	88
6-12	7	9	5	50
12-18	15	15	4	12
18-24	16	21	7	63
24-30	4	27	13	65
				$\sum$f|x-M| = 278

Now,

Mean Deviation about the median (M.D.) = $\dfrac{\sum f|x-M|}{N}$

$= \dfrac{278}{44}$

$= 6.318$

And,

Coefficient of M.D. = $\dfrac{M.D.}{M}$

$= \dfrac{6.318}{14}$

$= 0.451$

Hence, the required mean deviation about the median of the above data is 6.318 and the coefficient of mean deviation is 0.451.

Related Notes and Solutions:

Here is the website link to the notes of Statistics of Class 10.

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