Revision as of 19:18, 5 May 2023 by Admin (Created page with "'''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...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

00