# 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 $\small 6$ $\small 8$ $\small 14$ $\small 16$ $\small 4$ $\small 2$

 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 $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $\dpi{100} 25^{th}$ item is 20-30. Therefore, 20-30 is the median class.

We know,

Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$

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

Therefore, Median $\dpi{100} = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85$

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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