11.     Find the mean deviation about median for the following data : 

Marks \small 0-10 \small 10-20 \small 20-30 \small 30-40 \small 40-50 \small 50-60
Number of girls \small 6 \small 8 \small 14 \small 16 \small 4 \small 2

 

Answers (1)

Marks

Number of 

Girls f_i

Cumulative

Frequency c.f.

Mid

Points x_i

|x_i - M| f_i|x_i - M|
0-10 6 6 5 22.85 137.1
10-20 8 14 15 12.85 102.8
20-30 14 28 25 2.85 39.9
30-40 16 44 35 7.15 114.4
40-50 4 48 45 17.15 68.6
50-60 2 50 55 27.15 54.3
 

 

 

 

 

\sum f_i|x_i - M|

=517.1

Now, N = 50, which is even.

The class interval containing \left (\frac{N}{2} \right)^{th} or 25^{th} item is 20-30. Therefore, 20-30 is the median class.

We know,

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

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85

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

\sum f_i|x_i - M| = 517.1

\therefore M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|

= \frac{517.1}{50} = 10.34

Hence, the mean deviation about the median is 10.34

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