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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.
| Age (in yrs) | Number of policy holders |
|---|---|
| 15 - 20 | 2 |
| 20 - 25 | 4 |
| 25 - 30 | 18 |
| 30 - 35 | 21 |
| 35 - 40 | 33 |
| 40 - 45 | 11 |
| 45 - 50 | 3 |
| 50 - 55 | 6 |
| 55 - 60 | 2 |
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Age (in years) | f | cf
15 - 20 | 2 | 2
20 - 25 | 4 | 6
25 - 30 | 18 | 24
30 - 35 | 21 | 45
35 - 40 | 33 | 78
40 - 45 | 11 | 89
45 - 50 | 3 | 92
50 - 55 | 6 | 98
55 - 60 | 2 | 100
Total | 100
For correct table (1 Mark)
Modal class = 35-40 (1/2 Mark)
Mode = $35 + \frac{33-21}{2(33)-21-11} \times 5$ (1 Mark)
$= \frac{625}{17} = 36.7(\text{approx.})$ (1/2 Mark)
Modal age = 36.7 years (approx.)
$\frac{n}{2} = 50$, Median class = 35 – 40 (1/2 Mark)
Median = $35 + \frac{50-45}{33} \times 5$ (1 Mark)
$= \frac{1180}{33} = 35.7(\text{approx.})$ (1/2 Mark)
Median age = 35.7 years (approx.)
15 - 20 | 2 | 2
20 - 25 | 4 | 6
25 - 30 | 18 | 24
30 - 35 | 21 | 45
35 - 40 | 33 | 78
40 - 45 | 11 | 89
45 - 50 | 3 | 92
50 - 55 | 6 | 98
55 - 60 | 2 | 100
Total | 100
For correct table (1 Mark)
Modal class = 35-40 (1/2 Mark)
Mode = $35 + \frac{33-21}{2(33)-21-11} \times 5$ (1 Mark)
$= \frac{625}{17} = 36.7(\text{approx.})$ (1/2 Mark)
Modal age = 36.7 years (approx.)
$\frac{n}{2} = 50$, Median class = 35 – 40 (1/2 Mark)
Median = $35 + \frac{50-45}{33} \times 5$ (1 Mark)
$= \frac{1180}{33} = 35.7(\text{approx.})$ (1/2 Mark)
Median age = 35.7 years (approx.)