ABy Admin
May 07'23
Exercise
A probability distribution of the claim sizes for an auto insurance policy is given in the table below:
Claim Size | Probability |
---|---|
20 | 0.15 |
30 | 0.10 |
40 | 0.05 |
50 | 0.20 |
60 | 0.10 |
70 | 0.10 |
80 | 0.30 |
Calculate the percentage of claims that are within one standard deviation of the mean claim size.
- 45%
- 55%
- 68%
- 85%
- 100%
ABy Admin
May 07'23
Solution: A
Let X denote claim size. Then
E[X] = [20(0.15) + 30(0.10) + 40(0.05) + 50(0.20) + 60(0.10) + 70(0.10) + 80(0.30)] = (3 + 3 + 2 + 10 + 6 + 7 + 24) = 55
E[X2] = 400(0.15) + 900(0.10) + 1600(0.05) + 2500(0.20) + 3600(0.10) + 4900(0.10) + 6400(0.30) = 60 + 90 + 80 + 500 + 360 + 490 + 1920 = 3500
Var[X] = E[X2] – (E[X])2 = 3500 – 3025 = 475 and Var[ X ] = 21.79 .
Now the range of claims within one standard deviation of the mean is given by
[55.00 – 21.79, 55.00 + 21.79] = [33.21, 76.79]
Therefore, the proportion of claims within one standard deviation is
0.05 + 0.20 + 0.10 + 0.10 = 0.45 .