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Math Help - Exponential dist expected value help

  1. #1
    Senior Member Danneedshelp's Avatar
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    Exponential dist expected value help

    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
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  2. #2
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    Quote Originally Posted by Danneedshelp View Post
    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).
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