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Calculate the mean deviation about median age for the age distribution of 100 persons given below: (12)

12.    Calculate the mean deviation about median age for the age distribution of \small 100 persons given below:

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

[Hint  Convert the given data into continuous frequency distribution by subtracting 0.5from the lower limit and adding 0.5 to the upper limit of each class interval]
 

 

 

Answers (1)
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Age

(in years)

Number

 f_i

Cumulative

Frequency c.f.

Mid

Points x_i

|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
 

 

 

 

 

\sum f_i|x_i - M|

=735

Now, N = 100, which is even.

The class interval containing \left (\frac{N}{2} \right)^{th} or 50^{th} item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median = l + \frac{\frac{N}{2}- C}{f}\times h

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

Therefore, Median = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38

Now, we calculate the absolute values of the deviations from median, |x_i - M|  and

\sum f_i|x_i - M| = 735

\therefore M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|

= \frac{735}{100} = 7.35

Hence, the mean deviation about the median is 7.35

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