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Exercise
May 05'23
Answer
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.