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Math Help - Exponential dsn - expected cost of a project

  1. #1
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    Exponential dsn - expected cost of a project

    Hi all,

    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 completeting this operation to the square of the time to completion. Find the mean and variance of C

    So mean of C;
    E(C)=E(100+40Y+3Y^2)=100+40\cdot E(Y)+3\cdot [E(Y)]^2=100+40\cdot 10+3\cdot 100=800

    Just not sure I have calculated variance correctly; E(C^2)=E[(100+40Y+3Y^2)^2]= E(100^2+1600Y^2+9Y^4+8000Y+600Y^2+240Y^3)=640000

    So var(C)=640000-800^2=0
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  2. #2
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    Quote Originally Posted by Robb View Post
    Hi all,

    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 completeting this operation to the square of the time to completion. Find the mean and variance of C

    So mean of C;
    E(C)=E(100+40Y+3Y^2)=100+40\cdot E(Y)+3\cdot [E(Y)]^2=100+40\cdot 10+3\cdot 100=800

    Just not sure I have calculated variance correctly; E(C^2)=E[(100+40Y+3Y^2)^2]= E(100^2+1600Y^2+9Y^4+8000Y+600Y^2+240Y^3)=640000

    So var(C)=640000-800^2=0
    You have made the very bad mistake of thinking that E(Y^n) = [E(Y)]^n.

    You're given E(Y) = 10. To get E(Y^n) you need to use the pdf: E(Y^n) = \int_0^{+\infty} y^n f(y) \, dy.

    E(C) = E(100 + 40Y + 3Y^2) = 100 + 40 E(Y) + 3 E(Y^2).

    Var(C) = E(C^2) - [E(C)]^2

    and note that C^2 = (100 + 40Y + 3Y^2)^2.
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