Question 3

We need to calculate the median age of the policy holders.

Step 1 — Convert to Class Intervals

The given data is in "less than" form. We need to convert it into class intervals with their frequencies. The policies are given from 18 years onwards.

| 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) | 100 | |

Step 2 — Identify Median Class

The total number of policy holders is $n = 100$. We calculate $n/2$.

$n/2 = 100/2$

$= 50$

We look for the cumulative frequency just greater than 50. This is 78. The class corresponding to this cumulative frequency is 35 - 40. So, the median class is 35 - 40.

From the median class: Lower limit ($l$) = 35 Frequency ($f$) = 33 Class size ($h$) = $40 - 35 = 5$ Cumulative frequency of the class preceding the median class ($cf$) = 45

Step 3 — Calculate Median

The formula for the median is:

$\text{Median} = l + \left[ \frac{n/2 - cf}{f} \right] \times h$

Now, we substitute the values into the formula.

$\text{Median} = 35 + \left[ \frac{50 - 45}{33} \right] \times 5$

$= 35 + \left[ \frac{5}{33} \right] \times 5$

$= 35 + \frac{25}{33}$

$= 35 + 0.7575...$

$= 35.76 \text{ (approximately)}$

$\boxed{35.76 \text{ years}}$

Answer

(i) The median age is 35.76 years.