Revision as of 17:33, 26 May 2023 by Admin (Created page with "'''Key: C''' The average number of claims on a policy is 9720/6480 = 1.5. Using this as the parameter of a Poisson distribution yields the following table of actual (given) a...")
Exercise
May 26'23
Answer
Key: C
The average number of claims on a policy is 9720/6480 = 1.5. Using this as the parameter of a Poisson distribution yields the following table of actual (given) and expected (calculated) policies with given numbers of claims:
Number of Claims | ActualNumber of Policies | Poisson Probability | ExpectedNumber of Policies | Chi-square |
---|---|---|---|---|
0 | 1282 | 0.2231 | 1445.69 | 18.53 |
1 | 2218 | 0.3347 | 2168.86 | 1.11 |
2 | 1856 | 0.2510 | 1626.48 | 32.39 |
3 | 801 | 0.1255 | 813.24 | 0.18 |
4 | 235 | 0.0471 | 305.21 | 16.15 |
5 | 81 | 0.0141 | 91.37 | 1.18 |
6 or more | 7 | 0.0045 | 29.16 | 16.84 |
Total | 6480 | 1.0000 | 6480.01 | 86.38 |