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Exercise
ABy Admin
May 02'23
Answer
Solution: D
The distribution function of an exponential random variable [math]T[/math] with parameter [math]\theta [/math] is given by [math]F(t) = 1 − e^{− t \theta} , t \gt 0 [/math]. Since we are told that [math]T[/math] has a median of four hours, we may determine [math]\theta[/math] as follows:
[[math]]
\frac{1}{2} = F(4) = 1-e^{-4/\theta} \Rightarrow \theta = \frac{4}{\ln(2)}.
[[/math]]
Therefore,
[[math]]
\operatorname{P}(T ≥ 5 ) = 1 − F ( 5 ) = e^{−5 \theta} = e^{-\frac{5\ln(2)}{4}} = 2^{-5/4} = 0.42.
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.