# A life insurance

In this case, we are given less than (or below) cumulative frequency distribution,we need to convert it into normal frequency distribution.
So, we need to find class intervals and corresponding frequency.

Since, the difference between ages in each class is 5, we can take the first class interval as 15 - 20 and its frequency will be same as frequency of below 20 class.
Also, for other class, class interval will can be found as following and corresponding frequency can be find by subtracting the previous frequency from the cumulative frequency. As per the question,

N=100 Hence,

Median class = 35-45

Cumulative frequency = 100

Lower limit, l = 35

cf = 45

f = 33

h = 5

Now,

Median can be calculated as: where,
l = lower limit of median class
n = total frequency of the data
cf = cumulative frequency of the class before median class
f = frequency of the median class Median = 35 + Median = 35.75 years

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A data has 25 obsRS Aggarwal - Mathematics