May 05'23
Exercise
The table below shows the joint probability function of a sailor’s number of boating accidents and number of hospitalizations from these accidents this year.
Number of Hospitalizations from Accidents | |||||
Number of Accidents | 0 | 1 | 2 | 3 | |
0 | 0.700 | ||||
1 | 0.150 | 0.050 | |||
2 | 0.060 | 0.020 | 0.010 | ||
3 | 0.005 | 0.002 | 0.002 | 0.001 |
Calculate the sailor’s expected number of hospitalizations from boating accidents this year.
- 0.085
- 0.099
- 0.410
- 1.000
- 1.500
May 05'23
Solution: B
The marginal distribution for the probability of a given number of hospitalizations can be calculated by adding the columns. Then p(0) = 0.915, p(1) = 0.072, p(2) = 0.012, and p(3) = 0.001. The expected value is
0.915(0) + 0.072(1) + 0.012(2) + 0.001(3) = 0.099.