So this question comes from SOA/CAS Exam P of November 2009, I'm not sure why the solution is the way it is. The question says this:
An insurance company sells an auto insurance policy that covers losses incurred by a policyholder, subject to a deductible of 100. Loses incurred follow an exponential distribution with mean 300. What is the $95^{th}$ percentile of actual losses that exceed the deductible?
I know that $\lambda=\frac1\mu$ where $\mu$ is the mean. So $\lambda=\frac{1}{300}$. But here is where it got crazy (to me at least). This is what the solution says:
The probability that a claim exceeds the $100$ deductible is:$1-F(100)=1-\left(1-e^{-\frac{100}{300}}\right)=e^{-\frac13}=0.7165$ The $95^{th}$ percentile of losses that exceed the deductible is the $96.42^{th}$ percentile of all losses, since:$1-0.05\times0.7165=0.9642$ This is found as: $\begin{align}0.9642=F(x')=1-e^{\frac{-x'}{300}}\\ \Rightarrow x'=-300ln(.0358)=999\end{align}$
So what I'm confused by is 1) Why the cumulative distribution function was used. And 2) How the $2$nd line was come up with in the solution. What's with this 96.42th percentile? Otherwise I know why they solved for $x'$. If you read this whole thing, I give you much thanks, I know it's long.