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Math Help - the probability that this power supply will be inadequate on any given day

  1. #1
    Yan
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    the probability that this power supply will be inadequate on any given day

    In a certain city, the daily consumption of electric power in millions of kilowatt-hours can be treated as a random variable having a gamma distribution with α=3 and β=b. If the power plant of this city has a daily capacity of 12 million kilowatt-hours, what is the probability that this power supply will be inadequate on any given day?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Yan View Post
    In a certain city, the daily consumption of electric power in millions of kilowatt-hours can be treated as a random variable having a gamma distribution with α=3 and β=b. If the power plant of this city has a daily capacity of 12 million kilowatt-hours, what is the probability that this power supply will be inadequate on any given day?
    Let X be the rv representing the daily consumption (in millions of kw-h).

    So you are looking for \mathbb{P}(X>12)=1-\mathbb{P}(X\leqslant 12)

    The pdf of a gamma distribution with parameters (\alpha,\beta) is:
    g(x,\alpha,\beta)=x^{\alpha-1} \cdot \frac{\beta^\alpha e^{-\beta x}}{\Gamma(\alpha)}

    So here, it's g(x)=\frac{b^3}{\Gamma(3)} \cdot x^2 e^{-bx}

    \mathbb{P}(X\leqslant 12)=\int_0^{12} g(x) ~dx=\frac{b^3}{\Gamma(3)} \int_0^{12}x^2 e^{-bx} ~dx

    make twice an integration by parts and you'll be done.
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  3. #3
    Yan
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    sorry! i made a mistake on the question. the β is equal to 2 not b. β=2.
    and can you explain why i have to make twice an integration by parts? thanks!!!
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  4. #4
    Yan
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    is the equation should be

    The pdf of a gamma distribution with parameters (\alpha,\beta) is:
    g(x,\alpha,\beta)=x^{\alpha-1} \cdot \frac{\beta^{-\alpha} e^{-x/\beta }}{\Gamma(\alpha)}?
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  5. #5
    Moo
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    Quote Originally Posted by Yan View Post
    is the equation should be

    The pdf of a gamma distribution with parameters (\alpha,\beta) is:
    g(x,\alpha,\beta)=x^{\alpha-1} \cdot \frac{\beta^{-\alpha} e^{-x/\beta }}{\Gamma(\alpha)}?
    I don't know... the notations are made this way in the wikipedia.

    The pdf I gave you is if we're working on (\alpha,\beta). The pdf you gave is if we're working on (k,\theta)

    You can see that \alpha=k and \beta=\frac 1 \theta

    So here, it would depend on how you have been taught



    As for your mistake of b=2, it doesn't change anything to the result, since b was a constant, and that you can substitute it by any positive value you want

    And for the double integration by parts, it's because we have x^2 e^{-2x} in the integrand. We can't calculate an antiderivative of it. So make an integration by parts : the power x^2 will be transformed into x and the exponential remains. But we still cannot find an antiderivative. So integrate again by parts : the power x will be transformed to a constant and the exponential remains. But then you can compute an antiderivative, because the integrand will be an exponential.
    Just do it and you'll see (each time, take the u part as x^2 or x and the dv part as the exponential)
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  6. #6
    MHF Contributor matheagle's Avatar
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    LOL
    I was about to warn you.
    IN SOME books \beta is in the numerator
    and in other books it's in the denominator.
    I was going ask the person who submitted this thread to clarify.
    By just saying \beta is such and such it's not clear.
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