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. STATISTICS 199 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
Step-by-Step Solution
Key Concept: For grouped data, the median is found using the formula \(\displaystyle \text{Median}=L+\frac{\frac{N}{2}-c_f}{f}\times h\), where \(L\) is the lower class boundary of the median class, \(c_f\) is the cumulative frequency before the median class, \(f\) is the frequency of the median class, \(h\) is the class width and \(N\) is the total number of observations.
1. Total number of policy holders \(N = 100\).
2. Position of the median in the ordered data: \(\frac{N}{2}=\frac{100}{2}=50\) (the 50th observation).
3. Identify the median class using the cumulative frequencies:
- Cumulative frequency just before 35 years = 45 ("Below 35").
- Cumulative frequency just after 35 years = 78 ("Below 40").
Since 45 < 50 ≤ 78, the median lies in the class 35–40 years.
4. Extract required values:
- Lower class boundary \(L = 35\).
- Class width \(h = 40-35 = 5\) years.
- Cumulative frequency before the median class \(c_f = 45\).
- Frequency of the median class \(f = 78-45 = 33\).
5. Apply the median formula:
\[\text{Median}= L + \frac{\frac{N}{2} - c_f}{f}\times h \]
\[\text{Median}= 35 + \frac{50 - 45}{33}\times 5 \]
\[\text{Median}= 35 + \frac{5}{33}\times 5 \]
\[\text{Median}= 35 + \frac{25}{33} \]
\[\text{Median}= 35 + 0.7576 \approx 35.76\text{ years}\]
6. Result: The median age of the policy holders (aged 18 – < 60) is approximately 35.8 years.
*Note*: The lower age limit of 18 years does not affect the calculation because the median class (35–40) lies well within the given range.
Correct Answer: ≈ 35.8 years