A certain brand of refrigerator has a useful life that is normally distributed with mean 10 years and standard deviation 3 years. The useful lives of these refrigerators are independent.

Calculate the probability that the total useful life of two randomly selected refrigerators will exceed 1.9 times the useful life of a third randomly selected refrigerator.

• 0.407
• 0.444
• 0.556
• 0.593
• 0.604

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

Solution: C

Let [math]X, Y,[/math] and [math]Z[/math] be the three lifetimes. We want

[[math]]\operatorname{P} ( X + Y \gt 1.9 Z ) = \operatorname{P} (W = X + Y − 1.9 Z \gt 0) . [[/math]]

A linear combination of independent normal variables is also normal. In this case W has a mean of 10 + 10 – 1.9(10) = 1 and a variance of 9 + 9 + 1.9(1.9)(9) = 50.49 for a standard deviation of 7.106. Then the desired probability is that a standard normal variable exceeds (0 – 1)/7.106 = –0.141. Interpolating in the normal tables gives 0.5557 + (0.5596 – 0.5557)(0.1) = 0.5561, which rounds to 0.556.

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