A life-insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.
Age in years below 20 below 25 below 30 below 35 below 40 below 45 below 50 below 55 below 60
number of policyholders 2 6 24 45 78 89 92 98 100

Question

A life-insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.
Age in years below 20 below 25 below 30 below 35 below 40 below 45 below 50 below 55 below 60
number of policyholders 2 6 24 45 78 89 92 98 100

Vedantu Content Team · Accepted Answer

Hint: We find the values of class-limits and the class marks. We put all of them on a table. We have been the cumulative frequency in a less-than type form. From that value we find the frequencies, median class. We use the formula of median to find the solution of the problem.Complete step by step answer:We assume the frequencies as ${{f}_{i}}$ and the class marks as ${{x}_{i}}$.We need to find the class-limits and the class marks.We have been given the less-than type cumulative frequencies.We use simple subtraction form to find the frequencies.Total frequency is $n=100$. As in case of less-than type cumulative frequencies we have the last one as the total frequency.Also, the value of $\dfrac{n}{2}=\dfrac{100}{2}=50$.From the cumulative frequency we can find the median class will be 34.5-39.5.class intervalsclass limitsclass marks (${{x}_{i}}$)cumulative frequency (${{F}_{i}}$)frequency (${{f}_{i}}$)18-1914.5-19.5172220-2419.5-24.5226425-2924.5-29.527241830-3429.5-34.532452135-3934.5-39.537783340-4439.5-44.542891145-4944.5-49.54792350-5449.5-54.55298655-5954.5-59.5571002Total $n=100$We also the formula of median as $median\left( {{x}_{i}} \right)=l+\dfrac{\dfrac{n}{2}-{{F}_{l}}}{{{f}_{me}}}\times c$.Here l is the lower limit of the median class. ${{F}_{l}}$ denotes the cumulative frequency of the previous class of that median class. ${{f}_{me}}$ denotes the frequency of the median class. Also, c is the class width of the frequency table. In our problem the value of c is 5.So, we put the values in the equation and get $median\left( {{x}_{i}} \right)=l+\dfrac{\dfrac{n}{2}-{{F}_{l}}}{{{f}_{me}}}\times c=34.5+\dfrac{\dfrac{100}{2}-45}{37}\times 5$We solve this equation to get the value of median.So, $$median\left( {{x}_{i}} \right)=34.5+\dfrac{\dfrac{100}{2}-45}{37}\times 5=34.5+\dfrac{25}{37}=34.5+\text{0}\text{.675}=\text{35}\text{.176}$$The median value of the given frequency distribution table is 35.176.Note:We need to remember the median is the value of the frequency being at the most middle point. In most of the cases we have frequency given but, in this case, we have cumulative frequency and we needed to find the actual frequency. So, instead of finding mean we use the formula of median as it considers the density of a cumulative grouped data. We need to be careful finding the median class.

Age in years	below 20	below 25	below 30	below 35	below 40	below 45	below 50	below 55	below 60
number of policyholders	2	6	24	45	78	89	92	98	100

class intervals	class limits	class marks (${{x}_{i}}$)	cumulative frequency (${{F}_{i}}$)	frequency (${{f}_{i}}$)
18-19	14.5-19.5	17	2	2
20-24	19.5-24.5	22	6	4
25-29	24.5-29.5	27	24	18
30-34	29.5-34.5	32	45	21
35-39	34.5-39.5	37	78	33
40-44	39.5-44.5	42	89	11
45-49	44.5-49.5	47	92	3
50-54	49.5-54.5	52	98	6
55-59	54.5-59.5	57	100	2
Total				$n=100$