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 $0.5$from the lower limit and adding $0.5$ to the upper limit of each class interval]

 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 $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $50^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

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

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

Therefore, Median $\dpi{100} = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$

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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