A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
| Age (in years) | Number of policy holders |
| Below 20 | 2 |
| Below 25 | 6 |
| Below 30 | 24 |
| Below 35 | 45 |
| Below 40 | 78 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
Solution
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below
| Age (in years) | Number of policy holders (fi) | Cumulative frequency (cf) |
| 18 - 20 | 2 | 2 |
| 20 - 25 | 6 - 2 = 4 | 6 |
| 25 - 30 | 24 - 6 = 18 | 24 |
| 30 - 35 | 45 - 24 = 21 | 45 |
| 35 - 40 | 78 - 45 = 33 | 78 |
| 40 - 45 | 89 - 78 = 11 | 89 |
| 45 - 50 | 92 - 89 = 3 | 92 |
| 50 - 55 | 98 - 92 = 6 | 98 |
| 55 - 60 | 100 - 98 = 2 | 100 |
| Total (n) |
From the table, it can be observed that n = 100.
Cumulative frequency (cf) just greater than `n/2(100/2 = 50)` is 78, belonging to interval 35 - 40.
Therefore, median class = 35 - 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
`"Median" = l + ((n/2-cf)/f)xxh`
`= 35 + ((50-45)/33)xx5`
`= 35 + 25/33`
= 35.76
Therefore, median age is 35.76 years.
