Exercise


May 01'23

Answer

Solution: D

Note that

[[math]] \operatorname{E}[X] = \int_{-2}^0 \frac{-x^2}{10} dx + \int_0^4 \frac{x^2}{10} dx = \frac{-x^3}{30} \Big |_{-2}^0 + \frac{x^3}{30} \Big |_0^4 = \frac{28}{15}. [[/math]]

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

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