Exercise


ABy Admin
May 09'23

Answer

Solution: E

Using the formulas for the variance and mean of the uniform distribution:

[[math]] \begin{align*} \operatorname{E}(X^2) = \operatorname{Var}((X) + \operatorname{E}(X)^2 &= \frac{(100-a)^2}{12} + \left(\frac{100+a}{2}\right)^2 \\ &= \frac{100^2 -200a + a^2 +3(100)^2 + 600a + 3a^2}{12} \\ &= \frac{40000 + 400a + 4a^2}{12} \\ &= \frac{19600}{3} \\ 0 &= 40000 -78400 + 400a + 4a^2 \\ 0 &= a^2 + 100a - 9600 \\ 0 &= (a-60)(a+160) \\ a &= 60. \end{align*} [[/math]]

Then, [math]Y[/math] is uniform on the interval 1.25(60) = 75 to 100. The 80th percentile is 75 + 0.8(25) = 95.

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

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