Question

A life insurance agent found the following data of age distribution of 100 policy holders, where f is an unknown frequency.

Age in years	No. of Policy Holders
15 – 20	7
20 – 25	12
25 – 30	15
30 – 35	22
35 – 40	f
40 – 45	14
45 – 50	8
50 – 55	4

1. If the mean age of the policy holders is 35.65 years, find the unknown frequency f. 
2. Find the median class of the distribution.

Answer 1

a. Given,

Total no. of policy holders = 100

∴ 7 + 12 + 15 + 22 + f + 14 + 8 + 4 = 100

⇒ 82 + f = 100

⇒ f = 100 – 82 = 18.

Hence, f = 18.

b. Cumulative frequency distribution table:

Age in years	No. of Policy Holders	Cumulative frequency
15 – 20	7	7
20 – 25	12	19
25 – 30	15	34
30 – 35	22	56
35 – 40	f	74
40 – 45	14	88
45 – 50	8	96
50 – 55	4	100

Since, n = 100, which is even.

Median = `n/2`th term

= `100/2`

= 50th term

Steps of construction:

1. Plot class interval on x-axis and cumulative frequency on y-axis.
2. Mark points (20, 7), (25, 19), (30, 34), (35, 56), (40, 74), (45, 88), (50, 96) and (55, 100).
3. Draw a free hand curve passing through the points marked and starting from the lower limit of first class and terminating at upper limit of the last class.
4. From point A = 50 draw a line parallel to x-axis touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at C.

From graph,

C = 33.75, which lies between 30 – 35.

Hence, median class = 30 – 35.