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A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. [Policies are given only to persons having age 18 years onwards but less than 60 years.

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The given distribution being of the less than type, 25, 30, 35, give the upper limits of corresponding class intervals. So the classes should be 20 – 25, 25 – 30, 30 – 35, ………. 55 – 60. 

Observe that from the given distribution 2 persons with age less than 20. i.e., frequency of the class below 20 is 2. 

Now there are 6 persons with age less than 25 and 2 persons with age less than 20.

∴ The number of persons with age in the interval 20 – 25 is 6 – 2 = 4. 

Similarly, the frequencies can be calculated as shown in table. 

Number of observations = 100 n = 100

\(\frac{n}{2}=\frac{100}{2}\) = = 50, which lies in the class 35-40 

∴ 35 – 40 is the median class and lower boundary l = 35 

cf = 45; 

h = 5; 

f = 33

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