Exercise
ABy Admin
May 02'23
Answer
Solution: E
This is a conditional probability. The solution is
[[math]]
\begin{align*}
0.95 = \operatorname{P}[ X \leq p | X \gt 100 ] &= \frac{P[100 \leq X \leq p ]}{\operatorname{P}[X \gt 100]} = \frac{F(p) - F(100)}{1-F(100)} \\
&= \frac{1 − e^{-p /300} − 1 + e^{−100/300}}{1-1 + e^{-100/300}} \\ &= {e ^{-100/300} - e^{-p/300}}{e^{-100/300}} \\ &= 1 - e^{-(p-100)/300} \\
0.05 &= e^{-(p-100)/300 } \\
-2.9957 &= -(p-100)/300 \\
p &= 999
\end{align*}
[[/math]]