Long Answer Questions 2 — Question 10
Back to all questionsA 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 |
(a) If the mean age of the policy holders is 35.65 years, find the unknown frequency f.
(b) Find the median class of the distribution.
(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 | 18 | 74 |
| 40-45 | 14 | 88 |
| 45-50 | 8 | 96 |
| 50-55 | 4 | 100 |
Since, n = 100, which is even.
Median = = 50th term.
Steps of construction :
Plot class interval on x-axis and cumulative frequency on y-axis.
Mark points (20, 7), (25, 19), (30, 34), (35, 56), (40, 74), (45, 88), (50, 96) and (55, 100).
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.
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.