Question

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

Age(in years)	Number of policyholders
Below $20$	$2$
Below $25$	$6$
Below$30$	$24$
Below $35$	$45$
Below $40$	$78$
Below $45$	$89$
Below $50$	$92$
Below $55$	$98$
Below $60$	$100$

Answer 1

Hint: Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limit. The formula used to find the median of a given data is as follows: $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
Where $l$ is the lower limit of median class, $n$ is the sum of all frequencies, $cf$ is the cumulative frequency before the median class, $f$ is the frequency of median class and $h$ is the size of median class.

Complete step-by-step answer:
The policies were given only to persons with age $18$ years onward but less than $60$ years. Therefore, class intervals with their respective cumulative frequency can be defined as below:

Age(in years)	Number of policy holders $\left( f \right)$	Cumulative frequency $\left( {cf} \right)$
$18 - 20$	$2$	$2$
$20 - 25$	$6 - 2 = 4$	$6$
$25 - 30$	$24 - 6 = 18$	$24$
$30 - 35$	$45 - 24 = 21$	$45$
$35 - 40$	$78 - 45 = 33$	$78$
$40 - 45$	$89 - 78 = 11$	$89$
$45 - 50$	$92 - 89 = 3$	$92$
$50 - 55$	$98 - 92 = 6$	$98$
$55 - 60$	$100 - 98 = 2$	$100$
	$n = \sum {f = 100} $

From the table, we obtain $n = 100$$ \Rightarrow \dfrac{n}{2} = 50$
Cumulative frequency $\left( {cf} \right)$just greater than $\dfrac{n}{2}$$\left( {i.e.,50} \right)$ is $78$, which lies in the interval $35 - 40$.
Therefore, median class=$35 - 40$
Lower limit of the median class, $l = 35$
Frequency of the median class, $f = 33$
Cumulative frequency of the class preceding the median class, $cf = 45$
Class size, $h = 5$
Therefore, $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
$Median = 35 + \left( {\dfrac{{50 - 45}}{{33}}} \right) \times 5$
$ \Rightarrow Median = 35 + \dfrac{{25}}{{33}}$
$ \Rightarrow Median = 35.76$
Therefore, median age is $35.76$ years.

Note: The cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the sum of all frequencies, since all frequencies will already have been added to the previous total.