Question

Find out the coefficient of mean deviation about median in the following series.

Age in years	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
No. of persons	8	12	16	20	37	25	19	13

Answer 1

X	f	cf
0 - 10	8	8
10 - 20	12	20
20 - 30	16	36
30 - 40	20	56
40 - 50	37	93
50 - 60	25	118
60 - 70	19	137
70 - 80	13	150
	N = 150

`"N"/2 = 150/2` = 75

The class interval corresponding to cumulative frequency 75 is (40 - 50).

So, the corresponding values from the median class are L = 40, pcf = 56, f = 37, C = 10, `"N"/2` = 75.

Median = `"L" + (("N"/2 - "pcf"))/"f" xx "C"`

= `40 + ((75 - 56)/37) xx 10`

= `40 + 190/37`

= 40 + 5.1351

= 45.1351

= 45.14

Now we calculate the mean deviation about the median of 45.11

X	f	M	|D| = |M − 45.14|	f|D|
0 - 10	8	5	40.14	321.12
10 - 20	12	15	30.14	361.68
20 - 30	16	25	20.14	322.24
30 - 40	20	35	10.14	202.8
40 - 50	37	45	0.14	5.18
50 - 60	25	55	9.86	246.50
60 - 70	19	65	19.86	377.34
70 - 80	13	75	29.86	388.18
	N = 150			∑f|D| = 2225.04

Mean deviation about median = `(sum"f"|"D"|)/"N" = 2225.04/150` = 14.83