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 $x_i$	$\left|d_i\right|=\left|x_i-38.63\right|$	$f_{\mathrm{i}}\left|d_{\mathrm{i}}\right|$
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	9.38
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
	$\mathit{}N=\underset{i=1}{\overset{8}{\sum f_i}}=96$				$\sum _{i=1}^8{f}_i\left|d_i\right|=988.36$

$$\begin{gathered} N=96 \\ \therefore \frac{N}{2}=48 \\ \mathrm{The}\mathrm{cumulative}\mathrm{frequency}\mathrm{just}\mathrm{greater}\mathrm{than}\frac{N}{2}=38\mathrm{is}59\mathrm{and}\mathrm{the}\mathrm{corresponding}\mathrm{class}\mathrm{is}35.75-40.75. \\ \mathrm{Thus},\mathrm{the}\mathrm{median}\mathrm{class}\mathrm{is}35.75-40.75 \\ l=35.75,f=26,F=33,h=5 \\ \mathrm{Median}=l+\frac{{\displaystyle \frac{N}{2}}-F}{f}\times h \\ =35.75+\frac{\left(48-33\right)}{26}\times 5 \\ =35.75+2.88 \\ =38.63 \\ \mathrm{Mean}\mathrm{deviation}\mathrm{from}\mathrm{the}\mathrm{median}=\frac{{\displaystyle \mathrm{\sum }_{i=1}^8}{f}_i\left|d_i\right|}{N} \\ =\frac{988.36}{96} \\ =10.29 \end{gathered}$$