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Math Help - Discrete Probability Distribution II

  1. #1
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    Discrete Probability Distribution II

    Could anyone teach me how to solve these?

    Discrete Probability Distribution II-q9.jpg

    Discrete Probability Distribution II-q10.jpg

    Answer:
    Q9: (i)0.1108, (ii)0.6227, 2, 0.2681
    Q10: (i)0.0803, (ii)0.0498, (iii)0.0032
    Last edited by mr fantastic; October 13th 2010 at 06:38 PM.
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  2. #2
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    Quote Originally Posted by cloud5 View Post
    continued from the previous thread... Could anyone teach me how to solve these?

    Click image for larger version. 

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ID:	19313Click image for larger version. 

Name:	Q10.JPG 
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ID:	19314

    Answer:
    Q9: (i)0.1108, (ii)0.6227, 2, 0.2681
    Q10: (i)0.0803, (ii)0.0498, (iii)0.0032
    # 9

    You are asked to determine the most likely value of X or the mode of this poisson distribution.

    <br />
\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1} in which lambda is 2.2

    Consider for two cases, if P(X=x+1)>P(X=x), then 2.2>x+1, x<1.2 so x=1

    P(X=2)>P(X=1)>P(X=0)

    If P(X=x+1)<P(X=x), then 2.2<x+1 , x>1.2 so x=2,3,4,...

    P(x=2)>P(X=3)>P(X=4)>...

    Do you see that P(X=2) has the highest probability? That implies that the value of X most likely to occur is 2. Calculate its probability.

    (10) Adjust the mean to get 1/12.
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    # 9
    <br />
\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1} in which lambda is 2.2
    <br />
\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}

    How did you find this solution?
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  4. #4
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    Quote Originally Posted by cloud5 View Post
    <br />
\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}

    How did you find this solution?
    It can be proven.
    <br />
P(X=x+1)=\frac{e^{{-\lambda}}{(\lambda)^{x+1}}}{(x+1)!} -- 1

    P(X=x)=\frac{{e^{-\lambda}}{\lambda^x}}{x!} -- 2

    Divide equations 1 and 2, then simplify.
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