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Math Help - If the random variable X follows Poisson distribution such that

  1. #1
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    If the random variable X follows Poisson distribution such that

    If a random variable X follows Poisson distribution such that

    P(X=1)=P(X=2)

    Compute
    i) the mean of distibution
    ii) P(X=0)
    iii) standard deviation of the distribution

    Solution

    Poisson distribution p(x, \lambda) = \frac{\lambda^x}{x!} . e^{- \lambda}

    since P(X=1)=P(X=2) \Rightarrow P(X=1, \lambda)=P(X=2, \lambda) \Rightarrow  \frac{\lambda^1}{1!} . e^{- \lambda} = \frac{\lambda^2}{2!} . e^{- \lambda}

    \lambda = 2 .................Am i correct ??????
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  2. #2
    MHF Contributor matheagle's Avatar
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    looks good, the mean is lambda and the standard deviation is the square root of lambda
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  3. #3
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    Quote Originally Posted by Athena View Post
    If a random variable X follows Poisson distribution such that

    P(X=1)=P(X=2)

    Compute
    i) the mean of distibution
    ii) P(X=0)
    iii) standard deviation of the distribution

    Solution

    Poisson distribution p(x, \lambda) = \frac{\lambda^x}{x!} . e^{- \lambda}

    since P(X=1)=P(X=2) \Rightarrow P(X=1, \lambda)=P(X=2, \lambda) \Rightarrow \frac{\lambda^1}{1!} . e^{- \lambda} = \frac{\lambda^2}{2!} . e^{- \lambda}

    \lambda = 2 .................Am i correct ??????
    Asked here: http://www.mathhelpforum.com/math-he...n-problem.html

    Thread closed.
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