Exercise
May 01'23
Answer
Solution: B
Denote the insurance payment by the random variable [math]Y[/math]. Then
[[math]]
Y = \begin{cases}
0, \quad 0 \lt X \leq C \\
X-C, \quad C \lt X \lt 1
\end{cases}
[[/math]]
Now we are given that
[[math]]
\begin{align*}
0.64 = \operatorname{P}[Y \lt 0.5 ) = \operatorname{P}[ 0 \lt X \lt 0.5 + C ) &= \int_0^{0.5 + C} 2x dx \\
&= x^2 \Big |_0^{0.5 + C} \\
&= (0.5 + C)^2.
\end{align*}
[[/math]]
Therefore, solving for [math]C[/math], we find [math]C = ±0.8 − 0.5[/math]. Finally, since [math]0 \lt C \lt 1 [/math] , we conclude that [math]C = 0.3[/math].