Exercise


ABy Admin
May 14'23

Answer

Key: B

The number of repairs for each boat type has a binomial distribution. For power boats:

[[math]] \begin{aligned} &\operatorname{E}[ S ] = 100(0.3)(300) = 9, 000, \\ &\operatorname{E}[ S ] = 100(0.3)10, 000) + 100(0.3)(0.7)(300 2 ) = 2,190, 000 \end{aligned} [[/math]]

For sail boats:

[[math]] \begin{aligned} &\operatorname{E}[ S ] = 300(0.1)(1, 000) = 30, 000, \\ &\operatorname{E}[ S ] = 300(0.1)(400, 000) + 300(0.1)(0.9)(1, 0002 ) = 39, 000, 000 \end{aligned} [[/math]]

For luxury yachts:

[[math]] \begin{aligned} &\operatorname{E}[ S ] = 50(0.6)(5, 000) = 150, 000, \\ &\operatorname{E}[ S ] = 50(0.6)(0.4)(2, 000, 000) + 50(0.6)(0.4)(5, 0002 ) = 360, 000, 000 \end{aligned} [[/math]]

The sums are 189,000 expected and a variance of 401,190,000 for a standard deviation of 20,030. The mean plus standard deviation is 209,030.

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

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