Question

The Life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to the

persons whose age is 18 years onwards but less than 60 years.

Age in years	No. of Policy holders
$\mathrm{Below}20$	$2$
$\mathrm{Below}25$	$6$
$\mathrm{Below}30$	$24$
$\mathrm{Below}35$	$45$
$\mathrm{Below}40$	$78$
$\mathrm{Below}45$	$89$
$\mathrm{Below}50$	$92$
$\mathrm{Below}55$	$98$
$\mathrm{Below}60$	$100$

Answer 1

Step 1: Construct the table class intervals with their respective cumulative frequency

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals.

The given frequency table is less than the type represented by upper-class limits.

Median Class is the class having Cumulative frequency($\mathrm{cf}$) just greater than $\frac{\mathrm{n}}{2}$

$\mathrm{Median}=\mathrm{l}+\left[\frac{(\frac{\mathrm{n}}{2}-\mathrm{cf})}{\mathrm{f}}\right]\times \mathrm{h}$

Class size, $\mathrm{h}$

Number of observations, $\mathrm{n}$

The lower limit of median class,$\mathrm{l}$

Frequency of median class,$\mathrm{f}$

Cumulative frequency of class preceding median class, $\mathrm{cf}$

Class intervals with their respective cumulative frequency can be defined below

Age in years	No. of Policy holders	Class Interval	Frequency	Cumulative frequency
$\mathrm{Below}20$	$2$	$18-20$	$2$	$2$
$\mathrm{Below}25$	$6$	$20-25$	$4$	$2+4=6$
$\mathrm{Below}30$	$24$	$25-30$	$18$	$6+18=24$
$\mathrm{Below}35$	$45$	$35-40$	$21$	$24+21=45\left(f\right)$
$\mathrm{Below}40$	$78$	$45-50$	$33\left(\mathrm{f}\right)$	$45+33=78$
$\mathrm{Below}45$	$89$	$55-60$	$11$	$78+11=89$
$\mathrm{Below}50$	$92$	$65-70$	$3$	$89+3=92$
$\mathrm{Below}55$	$98$	$75-80$	$6$	$92+6=98$
$\mathrm{Below}60$	$100$	$85-90$	$2$	$98+2=100$

Step 2: Calculate the median

From the table, it can be observed that

$\begin{array}{rcl}\mathrm{n}& =& 100\\ & \Rightarrow & \frac{\mathrm{n}}{2}=50\end{array}$

Cumulative frequency $\mathrm{cf}$ just greater than $50\mathrm{is}78$, belonging to class-interval $35-40$

Therefore, the median class $=35-40$

Class size $\mathrm{h}=5$

Lower limit of median class,$\mathrm{l}=35$

Frequency of median class, $\mathrm{f}=33$

Cumulative frequency of class preceding median class, $\mathrm{cf}=45$

Median$=\mathrm{l}+\left[\frac{(\frac{\mathrm{n}}{2}-\mathrm{cf})}{\mathrm{f}}\right]\times \mathrm{h}$

$=35+\left[\frac{(50-45)}{33}\right]\times 5$

$=35+\frac{25}{33}$

$=35+0.76$

$=35.76$

A life insurance agent found the following data for the distribution of ages of $100$ policyholders.

If policies are given only to persons aged 18 years onwards but less than 60 years, the median age is $35.76$years.

Therefore, the median age is $35.76$ years.