Another Actuary Question Here is another question I have while practicing for the P/! exam:
An automobile policy owner has a 10% chance of having an accident in a $3$ 
year period. Given that an accident occurs, the payout for the accident follows the following distribution
$$f(x) = 0.5e^{-0.5x}$$
with $x$ representing thousands of dollars. If no accident takes place, the driver receives a rebate of $\$100$. What is the standard deviation of the payout?
Since $x$ appears to be following an exponential distribution, I thought that $\mu$ would be $2$ (or since $x$ is in thousands of dollars, the average payout to be $\$2000$ assuming there is an accident)
I know form the answer key that the answer is $\$851$, but I'm stuck as to why. 
 A: You have calculated the mean assuming there is an accident as $\$2000$.  You could also calculate the standard deviation as $\$2000$ assuming there is an accident, making $E[X\mid A]=2000$ and $E[X^2\mid A]=2000^2+2000^2=8000000$ 
Similarly you could calculate $E[X\mid A^c]=100$ and $E[X^2\mid A^c]=0^2+100^2=10000$
That then makes 


*

*$E[X]=0.1\times 2000 + 0.9\times 100=\$290$ 

*$E[X^2]=0.1\times 8000000 + 0.9\times 10000=809000$ 

*$\text{Var}(X)=809000-290^2=724900$ 

*the overall standard deviation $\sqrt{724900}\approx \$851$ 

A: In this case, if accident then the payout has mean $2000$ with variance $2000^2$. If no accident then the payout has mean $100$ with variance $0$. By the law of total variance,
\begin{align*}
\text{total var} &= \text{mean}(\text{var}) + \text{var}(\text{mean})\\
&= 0.1 \times 2000^2 + (0.9 \times 100^2 + 0.1 \times 2000^2 - 290^2)\\
&= 724900,
\end{align*}
and $\sqrt{724900} = 851.41$.
I just learned this in a probability class. I only vaguely understand it conceptually.
