Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age(in years) 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55
Number 5 6 12 14 26 12 16 9
[Hint: Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

Solution:

The given data is not continuous.

Therefore,

it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows

Age (in years)	Number of person fi	c. f	Mid-point xi	x_i - M	f_i x_i - M
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	160
50.5 - 55.5	9	100	53	15	135
	100				735

The class interval containing the (N/2)^th or 50)^t^h item is 35.5 - 40.5

Therefore,

35.5 - 40.5 is the median class.

It is known that,

Median = (l + (N/2 - C))f' × h

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

Median,

M = 35.5 + (50 - 37)/26 × 5

= 35.5 + (13 × 5)/26

= 35.5 + 2.5

= 38

Thus, the mean deviation about the median is given by:

M.D. (M) = 1/N ⁸∑_{i = 1} f_i|x_i - M|

= 1/100 × 735

= 7.35

NCERT Solutions Class 11 Maths Chapter 15 Exercise 15.1 Question 12

Calculate the mean deviation about median age for the age distribution of 100 persons given below: Age(in years) 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 Number 5 6 12 14 26 12 16 9 [Hint: Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

Summary:

The mean deviation about the median for the given data is 7.35