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Thread: Uniform Distribution - probabilities

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    Uniform Distribution - probabilities




    Hi guys, I am trying to solve an exercise but it seems to bem somehow strange.
    It is an experiment where A believes that men are bigger than women but B doesn't believe that, since A wants to convince B thatn ran an experiment and called randomly a number and ask to the person who answered the phone to say his height and his wife's height (if he is married). So we have two number (m,w) and the difference d=m-w.
    So I have the difference groups: G1: 0 to 10cm, G2: 10 to 20cm...G10: 90 to 100cm and G0: less than 0cm.
    Assume that in the city of A and B the height distribution of married people is equally distributed between 1m and 2m, both for men and women.
    I wanna calculate the propabilities of each group above. I have the sample space X={G0, G1,...G10}
    and now I want to calculate p(G0), p(G1), ..., p(G10).

    Since they are uniform distributed the density should be f(x)=1/b-a = 1/(2-1) = 1
    P(G1)=P(0<=x<=0.1(10 cm))=0.1/1=0.1 which is not true, because the book gives these probabilities and I just don't know how to do it, it says:

    P(G0)=1/2
    P(G1)=9.5/100
    P(G2)=8.5/100
    ...
    P(G10)=0.5/100

    What am I missing?

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    Re: Uniform Distribution - probabilities

    Note G0 is essentially "The woman is taller than the man". Since you're treating the men and women heights as independent and having the same distribution this shouldn't be too unintuitive. Keep in mind that m and w might have uniform distributions but that doesn't mean that d does.
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    Re: Uniform Distribution - probabilities


    Yep, I got it already. d has triangular distribution!
    So I calculated the cumulative function for each point and then everything was right.

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