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Math Help - Bivariate Normal

  1. #1
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    Linear Function of Random Variables

    I've got this problem here, but I'm a little concerned about my answer, not sure where I'm going wrong.

    Making handcrafted pottery generally takes two major steps: wheel throwing and firing. The time of wheel throwing and the time of firing are normally distributed random variables with means of 40 min and 60 min and standard deviations of 2 min and 3 min, respectively
    Let X_1 be the time throwing the wheels.
    \mu_X_1 = 40 min
    \sigma_X_1 = 2 min

    Let X_2 be the time firing.
    \mu_X_2 = 60 min
    \sigma_X_2 = 3 min

    Y=X_1 + X_2

    E(Y)=E(X_1)+E(X_2)=40min + 60min = 100min
    V(Y)=\sigma_X_1 ^2 + \sigma_X_2 ^2 = 2^2 + 3^2 = 13min^2

    \therfore P(Y<=85)=P(z< \frac{85-100}{sqrt(13)})=0.0000159

    This answer seems extremely low, considering 85 is just outside the first standard deviation. Not sure where I messed up here.
    Last edited by Kasper; October 30th 2010 at 11:15 AM.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Kasper View Post
    I've got this problem here, but I'm a little concerned about my answer, not sure where I'm going wrong.



    Let X_1 be the time throwing the wheels.
    \mu_X_1 = 40 min
    \sigma_X_1 = 2 min

    Let X_2 be the time firing.
    \mu_X_2 = 60 min
    \sigma_X_2 = 3 min

    Y=X_1 + X_2

    E(Y)=E(X_1)+E(X_2)=40min + 60min = 100min
    V(Y)=\sigma_X_1 ^2 + \sigma_X_2 ^2 = 2^2 + 3^2 = 13min^2

    \therfore P(Y<=85)=P(z< \frac{85-100}{sqrt(13)})=0.0000159

    This answer seems extremely low, considering 85 is just outside the first standard deviation. Not sure where I messed up here.
    You haven't posted the whole question but, reading between the lines, your answer is correct.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You haven't posted the whole question but, reading between the lines, your answer is correct.
    Whoops had it posted, must have accidentally deleted it when I edited it. Thanks!
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