Question

Question 3
A life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

$\begin{array}{cc}Age\left(inyears\right)& Numberofpolicyholders\\ Below20& 2\\ Below25& 6\\ Below30& 24\\ Below35& 45\\ Below40& 78\\ Below45& 89\\ Below50& 92\\ Below55& 98\\ Below60& 100\end{array}$

Answer 1

Here the class width is not same. There is no need to adjust the frequencies according to class intervals. Now, the given frequency table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years, we can define class intervals with their respective cumulative frequency as below:

$\begin{array}{ccc}Age\left(inyears\right)& Numberofpolicyholders\left(f_i\right)& Cumulativefrequency\left(c_f\right)\\ 18-20& 2& 2\\ 20-25& 6-2=4& 6\\ 25-30& 24-6=18& 24\\ 30-35& 45-24=21& 45\\ 35-40& 78-45=33& 78\\ 40-45& 89-78=11& 89\\ 45-50& 92-89=3& 92\\ 50-55& 98-92=6& 98\\ 55-60& 100-98=2& 100\\ Total\left(n\right)& & \end{array}$

Now from table we observe that n = 100.

Cumulative frequency $c_f$ just greater than $\frac{n}{2}(i.e,\frac{100}{2}=50)$ is 78 belonging to interval 35 - 40.

So, median class = 35 - 40

Lower limit $l$ of median class = 35

Class size $h$ = 5

Frequency $f$ of median class = 33

Cumulative frequency $c_f$ of class preceding median class = 45

$Median=l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h=35+\left(\frac{50-45}{33}\right)\times 5=35+\frac{25}{33}$

= 35.76

So, median age is 35.76 years.