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    Hey!!!


    I need some help!!! Let the distribution function F of a random variable X given in the following attachment. Calculate the following:
    P(X=-1), P(X<0), P(X<=0), P(X=1), P(X>5), P(X>=5), P(3<=X<=4).

    Thanks in advance!!!
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    (Original post by mathmari)
    Hey!!!


    I need some help!!! Let the distribution function F of a random variable X given in the following attachment. Calculate the following:
    P(X=-1), P(X<0), P(X<=0), P(X=1), P(X>5), P(X>=5), P(3<=X<=4).

    Thanks in advance!!!
    The graph gives you the cumulative probability: F(k) = P(X <= k). So, for instance, P(3 <= X <= 4) = P(X <= 4) - P(X < 3) - do you see why?
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    A ok... So P(X<0)=F(0-)=0.1, P(X<=0)=F(0)=0.2, P(3<=X<=4)=F(4)-F(3)=0.8-0.8=0, P(X>5)=P(X>=5)=0, P(X=-1)=F(-1+)-F(-1-)=0.1-0=0.1, P(X=1)=F(1+)-F(1-)=0.3-0.3=0...is this right???
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    (Original post by mathmari)
    A ok... So P(X<0)=F(0-)=0.1, P(X<=0)=F(0)=0.2, P(3<=X<=4)=F(4)-F(3)=0.8-0.8=0, P(X>5)=P(X>=5)=0, P(X=-1)=F(-1+)-F(-1-)=0.1-0=0.1, P(X=1)=F(1+)-F(1-)=0.3-0.3=0...is this right???
    ETA: The following line isn't correct, due to me writing this quickly and tiredly:
    Almost: P(3<=X<=4)=F(4)-F(3), but F(3) isn't 0.8.

    ETA: It should actually read:
    Almost: P(3<=X<=4) = P(X<=4) - P(X<3) = F(4)-F(3-).
    (Thanks to ghostwalker for pointing this out!)
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    How can I find this then???
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    (Original post by Smaug123)
    Almost: P(3<=X<=4)=F(4)-F(3), but F(3) isn't 0.8.
    I beg to differ. In fact you seem to be contradicting your first post.

    @OP

    P(3 <= X <= 4) = P(X <= 4) - P(X < 3)

    We know P(X <= 4) is F(4)

    The problem(?) lies with P(X < 3) as it does not include equality and the cdf has a discontinuity at X=3.

    \displaystyle P(X&lt;3) = \lim_{x\to 3^-}F(x)= F(3-)\text{ with your notation}

    I.e the limit of F(x) as x tends to 3 from below.

    Aside: This is the second mixed distribution I've seen on here in the last couple of weeks, and that's the only time I've seen them on here. Someone changed their syllabus?
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    (Original post by ghostwalker)
    I beg to differ.
    (Original post by mathmari)
    How can I find this then???
    Apologies, I wrote my reply sufficiently late at night that sleepiness got in the way ghostwalker is quite right, my notation was too sloppy for the problem at hand.
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    Thank you very much!!!!
 
 
 
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