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Math Help - Exponential Distribution

  1. #1
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    Exponential Distribution

    Weights of elephants approximately follow an exponential distribution with the mean of 2.5 tons. One hundred elephants are being transported on a ship that has a cargo limit of 300 tons. What is the chance this ship will sink?

    I think i kind of understand this problem but unsure.

    lambda = 1/2.5 = 0.40
    exp. dist. x < 300

    so 1- e^-(0.40)(100) = 1 so it will sink?
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  2. #2
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    Quote Originally Posted by aaronrj View Post
    Weights of elephants approximately follow an exponential distribution with the mean of 2.5 tons. One hundred elephants are being transported on a ship that has a cargo limit of 300 tons. What is the chance this ship will sink?

    I think i kind of understand this problem but unsure.

    lambda = 1/2.5 = 0.40
    exp. dist. x < 300

    so 1- e^-(0.40)(100) = 1 so it will sink?
    You're told the pdf of X, the random variable 'weight of a single elephant'. I suggest you think about what the pdf of Y = 100X is and then calculate Pr(Y > 300).
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  3. #3
    MHF Contributor matheagle's Avatar
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    I would use the normal approximation, with n=100.

    By the Central Limit Theorem

     P\left(\sum_{i=1}^{100}X_i>300\right)=P\left(\bar X>3\right)

    \approx P\left(Z>{3-\mu\over \sigma/\sqrt{100}}\right)

    NOW use your distribution..
    You have a mean of 2.5 and a variance of 2.5 via the exponential distribution...

    \approx P\left(Z>3.16227766\right)

    ----------------------------------------------------

    If you don't want to approximate this, then use that fact that a sum of iid exponentials is a gamma.

    Either way you need to calculate  P\left(\sum_{i=1}^{100}X_i>300\right)
    Last edited by matheagle; March 2nd 2010 at 11:36 PM.
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    Quote Originally Posted by matheagle View Post
    I would use the normal approximation, with n=100.

    By the Central Limit Theorem

     P\left(\sum_{i=1}^{100}X_i>300\right)=P\left(\bar X>3\right)

    \approx P\left(Z>{3-\mu\over \sigma/\sqrt{100}}\right)

    NOW use your distribution..
    You have a mean of 2.5 and a variance of 2.5 via the exponential distribution...

    \approx P\left(Z>3.16227766\right)

    ----------------------------------------------------

    If you don't want to approximate this, then use that fact that a sum of iid exponentials is a gamma.

    Either way you need to calculate  P\left(\sum_{i=1}^{100}X_i>300\right)
    You're too kind (I know some would beg to differ!!). I came back to fix my basic error (and I blame my physio for having made it) but was glad to see your reply.
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  5. #5
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    so if (Z > 3.16) do find P(Z > 3.16) = 1-.9992 according to the normal table?

    Using the gamma: (2.5 x 1< 3) givs approx. 0.7576 ? Im confused.
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  6. #6
    MHF Contributor matheagle's Avatar
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    no problem.....
    The only reason I write any post is to help others
    wait, who says I'm mean?
    Last edited by matheagle; March 3rd 2010 at 03:47 PM.
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  7. #7
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    Come on please just point me in the right direction. I am studying for a midterm and I would not ask if I understood.
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  8. #8
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    Quote Originally Posted by aaronrj View Post
    so if (Z > 3.16) do find P(Z > 3.16) = 1-.9992 according to the normal table?

    Using the gamma: (2.5 x 1< 3) givs approx. 0.7576 ? Im confused.
    To confirm the answer using the normal approximation, note that if X1, X2, .... Xn ~ Exp(lambda) then Y = X1 + X2 + ... + Xn ~ Gamma(n, 1/lambda).
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