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    Poisson Distribution Help




    Hey,

    I need some help with the following exercise (big test on Sunday).

    Q: A lab researching cancer cells found that transformed epithelial cells divide at a rate of 5 divisions/hour.
    1) What is the probability that in four hours, the cells will divide 25 times?
    Lamda = 5/hour. P(X=25) = e^(-20) x (20^25)/(25!) = 0.0445. No problems so far.
    2) Researchers saw that over four hours, the cells divided 22 times. What is the probability that 7 divisions occurred in the first hour?
    This is what I have no idea how to calculate.
    The answer is 0.139.

    Thanks!!!

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    Re: Poisson Distribution Help

    Quote Originally Posted by HadasL View Post
    2) Researchers saw that over four hours, the cells divided 22 times. What is the probability that 7 divisions occurred in the first hour?
    This is what I have no idea how to calculate.
    The answer is 0.139.
    This isn't actually a poisson probability calculation. Think about what distribution would be more appropriate here.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Poisson Distribution Help

    Hey Dason,

    Thank you for your fast reply! 22 divisions in four hours - we'd expect 5.5 an hour... Don't know if it's the stress or lack of knowledge but I just can't figure it out

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    Re: Poisson Distribution Help

    By the way, this isn't homework I need to hand in or anything... this is just me trying to learn for a big test and not quite succeeding.

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    Re: Poisson Distribution Help

    Note that there are 22 divisions in total in the four hours. So what are the possible number of divisions in the first hour?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Poisson Distribution Help

    Well, it could be anything between 0 and 22. So there are 23 possibilities.

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    Re: Poisson Distribution Help

    What distribution(s) do you know that take support only a finite number of possibilities (between 0 and N - which is 22 in this case).
    I don't have emotions and sometimes that makes me very sad.

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    Re: Poisson Distribution Help

    Well, there's binomial and uniform - but what will the probability be? At first I thought it was 1/5.5 or 1/5 but that doesn't make much sense. Doing the binomial for x=7 when X~B(22, 0.2) didn't work out. I know I'm missing something elementary here and feel quite stupid to be honest.

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    Re: Poisson Distribution Help

    You only looking at 1 hour out of the 4 total hours. So in that time you would expect 1/4 of the divisions to take place in that hour.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Poisson Distribution Help

    Ok, I expect 5.5 and there's 7. P(X=7), E(X)=5.5, N=22. But I still don't understand which probability to use.

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    Re: Poisson Distribution Help

    Quote Originally Posted by HadasL View Post
    But I still don't understand which probability to use.
    Quote Originally Posted by Dason View Post
    You only looking at 1 hour out of the 4 total hours. So in that time you would expect 1/4 of the divisions to take place in that hour.
    \hspace{1cm}
    I don't have emotions and sometimes that makes me very sad.

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    HadasL (10-01-2013)

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    Re: Poisson Distribution Help


    You can actually use conditional probability and the independent increment property of Poisson process to calculate it.

    Let X be the number of divisions in the first hour, and N be the number of divisions in the first 4 hours.

    By the independent increment property, X and N - X are independent, as N - X is the number of divisions in the last 3 hours.

    For x \in \{0, 1, \ldots, 22\},

    \Pr\{X = x|N = 22\} = \frac {\Pr\{X = x, N = 22\}} {\Pr\{N = 22\}}

    = \frac {\Pr\{X = x, N - X = 22 - x \}} {\Pr\{N = 22\}}

    = \frac {\Pr\{X = x\}\Pr\{N - X = 22 - x \}} {\Pr\{N = 22\}}

    and let you fill the remaining parts.

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