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Exercise


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May 08'23

Answer

Solution: D

With each load of coal having mean 1.5 and standard deviation 0.25, twenty loads have a mean of 20(1.5) = 30 and a variance of 20(0.0625) = 1.25. The total amount removed is normal with mean 4(7.25) = 29 and standard deviation 4(0.25) = 1. The difference is normal with mean 30 –29 = 1 and standard deviation sqrt(1.25 + 1) = 1.5. If D is that difference,

[[math]] \operatorname{P}( D \gt 0) = \operatorname{P}( Z \gt \frac{0-1}{1.5} = -0.67 ) = 0.7486. [[/math]]

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

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