Question

An SBI health insurance agent found the following data for distribution of ages of 100 policy holders. Find the modal age and median age of the policy holders.

Hint:

Median class is the first class whose cumulative frequency is greater than or equal to \( N/2 \).

Answer 1

Given:
The distribution of ages of 100 policy holders:

Age (yrs)	Frequency
15–20	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

Part 1: Modal Age
Mode = Class with highest frequency.
Highest frequency = 33 (class 35–40).

Formula for mode in continuous data:
\[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)h \] Where:
l = 35
\(f_1 = 33\) (modal class frequency)
\(f_0 = 21\) (preceding class frequency)
\(f_2 = 11\) (succeeding class frequency)
h = 5 (class width)

Substitute:
\[ \text{Mode} = 35 + \left(\frac{33 - 21}{2(33) - 21 - 11}\right)5 \] \[ = 35 + \left(\frac{12}{66 - 32}\right)5 \] \[ = 35 + \left(\frac{12}{34}\right)5 \] \[ = 35 + 1.7647 \approx 36.76 \]
Modal Age ≈ 36.8 years

Part 2: Median Age
Median class = the class where cumulative frequency ≥ 50 (since N = 100).

Cumulative frequencies:
2, 6, 24, 45, 78, 89, 92, 98, 100

So, median class = 35–40.

Formula for median:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - c}{f}\right)h \] Where:
l = 35
N/2 = 50
c = 45 (cumulative frequency before median class)
f = 33 (median class frequency)
h = 5

Substitute:
\[ \text{Median} = 35 + \left(\frac{50 - 45}{33}\right)5 \] \[ = 35 + \left(\frac{5}{33}\right)5 \] \[ = 35 + 0.7575 \approx 35.76 \]
Median Age ≈ 35.8 years

Final Answers:
Modal age ≈ 36.8 years
Median age ≈ 35.8 years