Question

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

Age:	16-20	21-25	26-30	31-35	36-40	41-45	46-50	51-55
Number of persons	5	6	12	14	26	12	16	9

Answer 1

Since the function is not continuous, we subtract 0.5 from the lower limit of the class and add 0.5 to the upper limit of the class so that the class interval remains same, while the function becomes continuous.

Thus, the mean distribution table will be as follows:

Age	Number of Persons $f_i$	Midpoint $x_i$	Cumulative Frequency	$\left|d_i\right|=\left|x_i-38\right|$	$f_i{d}_i$
15.5−20.5	5	18	5	20	100
20.5−25.5	6	23	11	15	90
25.5−30.5	12	28	23	10	120
30.5−35.5	14	33	37	5	70
35.5−40.5	26	38	63	0	0
40.5−45.5	12	43	75	5	60
45.5−50.5	16	48	91	10	160
50.5−55.5	9	53	100	15	135
	$N=\sum _{i=1}^8{f}_i=100$				$\sum _{i=1}^8{f}_i{d}_i=735$

$$\begin{gathered} N=100 \\ \Rightarrow \frac{N}{2}=50 \end{gathered}$$
Thus, the cumulative frequency slightly greater than 50 is 63 and falls in the median class 35.5−40.5.

$$\begin{gathered} \therefore l=35.5,F=37,f=26,h=5 \\ \mathrm{Median}=l+\frac{{\displaystyle \frac{N}{2}}-F}{f}\times h \\ =35.5+\frac{\left(50-37\right)}{26}\times 5 \\ =35.5+2.5 \\ =38 \\ \mathrm{Mean}\mathrm{deviation}\mathrm{about}\mathrm{the}\mathrm{median}\mathrm{age}=\frac{{\displaystyle \sum _{i=1}^8}{f}_i\left|d_i\right|}{\mathrm{N}} \\ =\frac{735}{100} \\ =7.35 \end{gathered}$$


Thus, the mean deviation from the median age is 7.35 years.