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प्रश्न
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 years.
| Age (in years) | Number of policy holders |
| Below 20 | 2 |
| 20 - 25 | 4 |
| 25 - 30 | 18 |
| 30 - 35 | 21 |
| 35 - 40 | 33 |
| 40 - 45 | 11 |
| 45 - 50 | 3 |
| 50 - 55 | 6 |
| 55 - 60 | 2 |
उत्तर
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
| Class Interval | Number of policy holders (f) | Cumulative frequency (cf) |
| Below 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 |
It is given 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 of median class (l) = 35
Class size (h) = 5
Frequency of median class (f) = 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 + 0.76
= 35.76
Therefore, the median age is 35.76 years.
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