Exercise


ABy Admin
May 07'23

Answer

Solution: E

For a single recruit, the probability of 0 pensions is 0.6, of 1 pension is 0.4(0.25) = 0.1, and of 2 pensions is 0.4(0.75) = 0.3. The expected number of pensions is 0(0.6) + 1(0.1) + 2(0.3) = 0.7. The second moment is 0(0.6) + 1(0.1) + 4(0.3) = 1.3. The variance is 1.3 – 0.49 = 0.81. For 100 recruits the mean is 70 and the variance is 81. The probability of providing at most 90 pensions is (with a continuity correction) the probability of being below 90.5. This is (90.5 – 70)/9 = 2.28 standard deviations above the mean. From the tables, this probability is 0.9887.

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

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