Exponential dist expected value help

Quote:

Originally Posted by Danneedshelp

Q: The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula $C=100+40Y+3Y^{2}$ relates the cost C of completing this operation to the square of the time to completion. Find the mean and variance of C.

My work:

Since $V(C)=E(C^{2})-[E(C)]^{2}$ I need to solve for $E(C^{2})$ and $[E(C)]^{2}$ first.

$E(C^{2})=9E(Y^{4})+240E(Y^{3})+2200E(Y^{2})+8000E( Y)+10000$

So, I started with finding $E(Y^{4})$ ,

$E(Y^{4})=\int_{0}^{\infty}y^{4}e^{\frac{-y}{10}}dy$ , by u-sub we have $u=\frac{y}{10}\Rightarrow\\10du=dy$ , so $10\int_{0}^{-\infty}u^{4}e^{-u}du=<br /> 10\int_{0}^{-\infty}u^{3-1}e^{-u}du=10\Gamma(3)$

and so, if that is correct, shouldn't $E(Y^{3})=10\Gamma(2)$ and so, down to $E(Y)=\beta=10$ ?

Nevertheless, I keep ending up with a negative variance when I use this method, since my $[E(C)]^{2}$ gets so large. Do I just have to crank out every integrand by way of integration by parts instead of using the gamma function?

Some direction would be great,

Thanks

Your integrals are wrong:

$E(Y^n) = \int_0^{+\infty} y^n e^{-y/10} \, dy = 10^{n+1} \Gamma(n+1)$ .