A life insurance agent found the following data for distribution of ages of 100 policy holders:
| Age in years | Policy holders (frequency) | Cumulative frequency |
|---|---|---|
| 20-25 | 2 | 2 |
| 25-30 | 4 | 6 |
| 30-35 | 12 | 18 |
| 35-40 | 20 | 38 |
| 40-45 | 28 | 66 |
| 45-50 | 22 | 88 |
| 50-55 | 8 | 96 |
| 55-60 | 4 | 100 |
On a graph sheet draw an ogive using the given data. Take 2 cm = 5 years along one axis and 2 cm = 10 policy holders along the other axis. Use your graph to find:
(a) The median age.
(b) Number of policy holders whose age is above 52 years.
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Steps of construction :
Take 2 cm = 5 years on x-axis.
Take 1 cm = 10 policy holders on y-axis.
Plot the points (20, 0), (25, 2), (30, 6), (35, 18), (40, 38), (45, 66), (50, 88), (55, 96) and (60, 100).
Join the points by a free-hand curve.
Here, n = 100
Median = $\frac{n}{2} = \frac{100}{2}$ = 50th term
(a) Through J = 50 draw a horizontal line to meet the ogive at K. Through K, draw a vertical line to meet the x-axis at L. The abscissa of the point L represents 42.
Hence, the median age = 42 years.
(b) Through M = 52 draw a vertical line to meet the ogive at N. Through N, draw a horizontal line to meet the y-axis at O. The ordinate of the point O represents 91.
Hence, 91 people have their age less than or equal to 52.
∴ No. of people whose age is greater than 52 = 100 - 91 = 9.
Hence, number of policy holders whose age is above 52 years equal to 9.