An SBI health insurance agent found the following data for

Q) An SBI health insurance agent found the following data for distribution of ages of 100 policy holders. The health insurance policies are given to persons of age 15 years and onwards, but less than 60 years.

Find the modal age and median age of the policy holders.

(Q 35 – 30/2/1 – CBSE 2026 Question Paper)

Ans:

(i) Modal Age:

Step 1: Since the modal class is the class with the highest frequency.

Here, in th egiven table, highest frequency is 33,

∵ highest frequency belongs to class “35 – 40”

∴ the modal class is “35 – 40”.

Step 2: Now mode of the grouped data is calculated by:

Mode = $L + \left [\frac{(f_1 - f_0)}{(2 f_1 - f_0 - f_2)}) \times h$

Here,

L = lower class limit of modal class = 35

f₁= frequency of modal class = 33

f₀= frequency of class proceeding to modal class = 21

f₂= frequency of class succeeding to modal class = 11

h = class size = 40 – 35 = 5

Let’s put values in the formula and solve:

Mode = $L + \left [\frac{(f_1 - f_0)}{(2 f_1 - f_0 - f_2)}\right] \times h$

= $35 + \left [\frac{(33 - 21)}{(2 (33) - 21 - 11)}\right] \times 5$

= $35 + \left [\frac{12}{(66 - 32)}\right] \times 5 = 35 + \frac{12 \times 5}{34}$

= 35 + $\frac{60}{34}$ = 35 + 1.77 = 36.77

Therefore, the modal age of the policyholders is 36.77 years.

(ii) Median Age:

Step 3: To calculate the median value, let’s re-organize the data:

To find the median, we need to first identify median class of the data.

We know that, Median class is the class where the cumulative frequency crosses 50% of total of frequencies.
Here, in the given data, total of frequencies is 100, hence 50% of it is 50.
At “35-40” class, cumulative frequency is crossing 50
Hence, our Median class = 35-40

Step 4: Next, the median value of a grouped data is given by:

Median = $L+ \left [\frac{\frac{n}{2}-c_f}{f}\right] \times h$