Q3 of 25 Page 346

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.]


For median:


We have, total frequency, N = 534


N/2 = 534/2 = 267


Observe, cf = 336 is just greater than 267.


Thus, median class = 45-50


Median is given by



Where,


L = Lower class limit of median class = 45


N/2 = 267


cf = cumulative frequency of the class preceding median class = 244


f = frequency of the median class = 92


h = class interval of the median class = 5


Substituting these values in the formula of median, we get



⇒


⇒ Median = 45 + 115/92


⇒ Median = 45 + 1.25


⇒ Median = 46.25


Thus, the median is 46.25.


More from this chapter

All 25 →
1

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

 2

If the median of 60 observations, given below is 28.5, find the values of x and y.

 4

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table :


Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, ….., 171.5 – 180.5.)

 5

The following table gives the distribution of the life-time of 400 neon lamps


Find the median life time of a lamp.