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    Help with cdf




    I'm having trouble with grasping cdf's. If I have a cumulative distribution function
    F(x) =
    .0 ----- x<-2
    .2 ----- -2 ≤ x< 0
    .5 ----- 0 ≤ x < 1
    .7 ----- 1 ≤ x < 3
    .9 ----- 3 ≤ x < 4
    1.0 ----- x ≥ 4
    Im trying to evaluate P(-1 < X < 3) and P(X=4). For P(-1 < X < 3) im doing P(3≤ X<4)-P(-2≤ X<0) which gives me .7. For P(X=4) I subtract P(3<=X<4) from 1 giving me 1-.9=.10.
    Im not sure I am doing this right though.
    Last edited by bigbob; 10-06-2013 at 05:40 PM.

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    Re: Help with cdf

    P(X=4) = .10 is correct

    P(3≤ X<4)-P(-2≤ X<0) will give you P(-1 < X < 4) not P(-1 < X < 3)

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    Re: Help with cdf

    Quote Originally Posted by asterisk View Post
    P(X=4) = .10 is correct

    P(3≤ X<4)-P(-2≤ X<0) will give you P(-1 < X < 4) not P(-1 < X < 3)
    I should do P(1<=X<3)-P(-2<=X<0) does this include the positive 1 if I do?

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    Re: Help with cdf

    Yes, positive 1 is included in P(1<=X<3)

    Just so we're on e same page, this is a cumulative distribution function of a discrete probability distribution.

    The graph of the cdf looks like this



    From the graph you should be able to calculate the pdf

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    Re: Help with cdf

    Quote Originally Posted by asterisk View Post
    Yes, positive 1 is included in P(1<=X<3)

    Just so we're on e same page, this is a cumulative distribution function of a discrete probability distribution.

    The graph of the cdf looks like this



    From the graph you should be able to calculate the pdf
    Right so the pmf values would be the difference between the two points (vertically) for that number. Unless it is undefined.

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    Re: Help with cdf


    Yes, so

    P(-2) = .2
    P(-1) = 0
    P(0) = .3

    Note while the cdf is not continuous, it is defined for all points.

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