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(Created page with "'''Solution: A''' Let N denote the number of warranty claims received. Then, <math display = "block"> 0.6 =\operatorname{P}( N =0) =e^{−c} ⇒ c =− \ln(0.6) =0.5108. </m...") |
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\begin{align*} | \begin{align*} | ||
&5000[ \operatorname{P}( N = 2) + 2 \operatorname{P}( N = 3) + 3\operatorname{P}( N = 4) + \cdots] \\ | &5000[ \operatorname{P}( N = 2) + 2 \operatorname{P}( N = 3) + 3\operatorname{P}( N = 4) + \cdots] \\ | ||
&= 5000[ \operatorname{P}( N = | &= 5000[ \operatorname{P}( N = 1)+ 2 \operatorname{P}( N = 2)+ 3\operatorname{P}( N = 3) + \cdots] − 5000[ \operatorname{P}( N = 1) + \operatorname{P}( N =+ 2) \operatorname{P}( N = 3) + \cdots] \\ | ||
&=5000 \operatorname{E}( N ) − 5000[1 − \operatorname{P}( N =0)] =5000(0.5108) − 5000(1 − 0.6) =554. | &=5000 \operatorname{E}( N ) − 5000[1 − \operatorname{P}( N =0)] =5000(0.5108) − 5000(1 − 0.6) =554. | ||
\end{align*} | \end{align*} | ||
Latest revision as of 19:54, 2 January 2024
Solution: A
Let N denote the number of warranty claims received. Then,
[[math]]
0.6 =\operatorname{P}( N =0) =e^{−c} ⇒ c =− \ln(0.6) =0.5108.
[[/math]]
The expected yearly insurance payments are:
[[math]]
\begin{align*}
&5000[ \operatorname{P}( N = 2) + 2 \operatorname{P}( N = 3) + 3\operatorname{P}( N = 4) + \cdots] \\
&= 5000[ \operatorname{P}( N = 1)+ 2 \operatorname{P}( N = 2)+ 3\operatorname{P}( N = 3) + \cdots] − 5000[ \operatorname{P}( N = 1) + \operatorname{P}( N =+ 2) \operatorname{P}( N = 3) + \cdots] \\
&=5000 \operatorname{E}( N ) − 5000[1 − \operatorname{P}( N =0)] =5000(0.5108) − 5000(1 − 0.6) =554.
\end{align*}
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.