Calculatethe mean deviation about median age for the age distribution of 100persons given below:

Age	Number
16-20	5
21-25	6
26-30	12
31-35	14
36-40	26
41-45	12
46-50	16
51-55	9

The givendata is not continuous. Therefore, it has to be converted intocontinuous frequency distribution by subtracting 0.5 from the lowerlimit and adding 0.5 to the upper limit of each class interval.

The tableis formed as follows.

Age	Number	Cumulative frequency (c.f.)	Mid-point
15.5-20.5	5	5	18	20	100
20.5-25.5	6	11	23	15	90
25.5-30.5	12	23	28	10	120
30.5-35.5	14	37	33	5	70
35.5-40.5	26	63	38	0	0
40.5-45.5	12	75	43	5	60
45.5-50.5	16	91	48	10	100
50.5-55.5	9	100	53	15	135
	100				735

The classinterval containing theor50^th item is 35.5 – 40.5.

Therefore,35.5 – 40.5 is the median class.

It isknown that,

Here, l= 35.5, C = 37, f = 26, h = 5, and N = 100

Thus, meandeviation about the median is given by,