Exercise


ABy Admin
May 03'23

Answer

Solution: B

Let [math]X[/math] be normal with mean 10 and variance 4. Let [math]Z[/math] have the standard normal distribution. Let [math]p[/math] = 12th percentile. Then

[[math]] 0.12 = \operatorname{P}(X \leq p ) = \operatorname{P}\left( \frac{X-10}{2} \leq \frac{p-10}{2} \right) = \operatorname{P}( Z \leq \frac{p-10}{2}). [[/math]]

From the tables, [math]\operatorname{P}( Z ≤ −1.175) =0.12.[/math] Therefore,

[[math]] \frac{p-10}{2} = -1.175; p-10 = -2.35; p = 7.65. [[/math]]

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

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