excans:237dbe9dad: Difference between revisions
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1-e^{-50/100} - (1-e^{-40/100}) &= 1-e^{-r/100} - (1-e^{-60/100}) \\ | 1-e^{-50/100} - (1-e^{-40/100}) &= 1-e^{-r/100} - (1-e^{-60/100}) \\ | ||
e^{-40/100} - e^{-50/100} &= e^{-60/100} - e^{-r/100} \\ | e^{-40/100} - e^{-50/100} &= e^{-60/100} - e^{-r/100} \\ | ||
e^{-r/100} &= e^{-60/100} - e^{-40/100} + e^{-50/100} | e^{-r/100} &= e^{-60/100} - e^{-40/100} + e^{-50/100} = 0.4850 \\ | ||
-r/100 &= \ln(0.4850) = -0.7236 \\ | -r/100 &= \ln(0.4850) = -0.7236 \\ | ||
r &= 72.36. | r &= 72.36. | ||
Latest revision as of 22:39, 2 May 2023
Solution: D
The cumulative distribution function for the exponential distribution is
[[math]]
F(x) = 1-e^{-\lambda x} = 1-e^{-x/\mu} = 1-e^{-x/100}, x \gt 0.
[[/math]]
From the given probability data,
[[math]]
\begin{align*}
F(50) - F(40) &= F(r) - F(60) \\
1-e^{-50/100} - (1-e^{-40/100}) &= 1-e^{-r/100} - (1-e^{-60/100}) \\
e^{-40/100} - e^{-50/100} &= e^{-60/100} - e^{-r/100} \\
e^{-r/100} &= e^{-60/100} - e^{-40/100} + e^{-50/100} = 0.4850 \\
-r/100 &= \ln(0.4850) = -0.7236 \\
r &= 72.36.
\end{align*}
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.