Question

The age distribution of 100 life-insurance policy holders is as follows:

Age (on nearest birth day)	17-19.5	20-25.5	26-35.5	36-40.5	41-50.5	51-55.5	56-60.5	61-70.5
No. of persons	5	16	12	26	14	12	6	5

Calculate the mean deviation from the median age

Answer 1

To make this function continuous, we need to subtract 0.25 from the lower limit and add 0.25 to the upper limit of the class.

Age	Number of People \[f_i\]	Cumulative Frequency	Midpoints \[x_i\]	\[\left| d_i \right| = \left| x_i - 38 . 63 \right|\]	\[f_i \left| d_i \right|\]
16.75−19.75	5	5	18.25	20.38	101.9
19.75−25.75	16	21	22.75	15.88	254.08
25.75−35.75	12	33	30.75	7.88	94.56
35.75−40.75	26	59	38.25	0.38	9.88
40.75−50.75	14	73	45.75	7.12	99.68
50.75−55.75	12	85	53.25	14.62	175.44
55.75−60.75	6	91	58.25	19.62	117.72
60.75−70.75	5	96	65.75	27.12	135.6
	\[N = \sum ^8 _{i = 1} f_i = 96\]				\[\sum^8_{i = 1} f_i \left| d_i \right| = 988 . 86\]

\[N = 96\]
\[ \therefore \frac{N}{2} = 48\]
\[\text{ The cumulative frequency just greater than } \frac{N}{2} = 38\text{ is 59 and the corresponding class is }  35 . 75 - 40 . 75 . \]
\[\text{ Thus, the median class is 35 . 75 - 40 . 75 } \]
\[l = 35 . 75, f = 26, F = 33, h = 5\]
\[\text{ Median } = l + \frac{\frac{N}{2} - F}{f} \times h\]
\[ = 35 . 75 + \frac{\left( 48 - 33 \right)}{26} \times 5 \]
\[ = 35 . 75 + 2 . 88\]
\[ = 38 . 63\]
\[\text{ Mean deviation from the median }  = \frac{\mathit{\sum^8_{i = 1}} f_i \left| d_i \right|}{N}\]
\[ = \frac{988 . 86}{96}\]
\[ = 10 . 30\]