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Math Help - Poisson Problem

  1. #1
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    Poisson Problem

    If the number of accidents occuring in an industrial plant during aa day is given by a Poisson random variable with parameter 3.
    Find
    i) probability that no accident occurs on a day
    ii)the expected number of accidents per day.
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  2. #2
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    Quote Originally Posted by zorro View Post
    If the number of accidents occuring in an industrial plant during aa day is given by a Poisson random variable with parameter 3.
    Find
    i) probability that no accident occurs on a day
    ii)the expected number of accidents per day.
    These questions are totally trivial and suggest a complete lack of effort on your part in trying to understand this material. Go to your class notes and/or textbook and read the section on the Poisson distribution.
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  3. #3
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    Then is this right

    probability of an accident happening in a day is
    <br />
p(x;\lambda) \ = \ \frac{\lambda^x \ e^{- \lambda}}{x!} \ = \ \frac{3^x \ e^{-3}}{x!} = \frac{3^x \ 0.050}{x!}<br />


    probability of accident happening is
    <br />
\left ( 1 - \frac{3^x \ 0.050}{x!} \right)<br />

    Expected no. of accidents per day is
    <br />
E(X) = \sum x . p(x;\lambda) = \sum x \ . \ \frac{3^x \ . \ 0.050}{x!}<br />
    Last edited by mr fantastic; January 7th 2009 at 11:49 PM.
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  4. #4
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    Quote Originally Posted by zorro View Post
    probability of an accident happening in a day is
    <br />
p(x;\lambda) \ = \ \frac{\lambda^x \ e^{- \lambda}}{x!} \ = \ \frac{3^x \ e^{-3}}{x!} = \frac{3^x \ 0.050}{x!}<br />


    probability of accident happening is
    <br />
\left ( 1 - \frac{3^x \ 0.050}{x!} \right)<br />

    Expected no. of accidents per day is
    <br />
E(X) = \sum x . p(x;\lambda) = \sum x \ . \ \frac{3^x \ . \ 0.050}{x!}<br />
    \Pr(\text{accident}) = 1 - \Pr(\text{no accident}) = 1 - \Pr(X = 0) = 1 - e^{-3} = \, ....

    E(X) = \lambda = 3.

    You need to thoroughly review this topic - the questions, especially the second question, are extremely straightforward.
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  5. #5
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    hw did u get E[x]

    Quote Originally Posted by mr fantastic View Post
    \Pr(\text{accident}) = 1 - \Pr(\text{no accident}) = 1 - \Pr(X = 0) = 1 - e^{-3} = \, ....

    E(X) = \lambda = 3.

    You need to thoroughly review this topic - the questions, especially the second question, are extremely straightforward.


    Isnt

    E[x] \ =\ \sum x .\ [1-p(x;\lambda)]= \ 0 \ . \ [1-p(x;\lambda) ]=0
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  6. #6
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    Quote Originally Posted by zorro View Post
    Isnt

    E[x] \ =\ \sum x .\ [1-p(x;\lambda)]= \ 0 \ . \ [1-p(x;\lambda) ]=0
    No.

    E[X] = \sum x \cdot p(x; \, \lambda).

    And the result of this calculation (which can be used without proof I would have thought unless the calculation is specifically asked for: http://www.mathhelpforum.com/math-he...tributiom.html) is E[X] = \lambda.
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