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Probability-distribution function F Watch

1. 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).

Attached Images

2. (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).

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?
3. 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???
4. (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.

Almost: P(3<=X<=4) = P(X<=4) - P(X<3) = F(4)-F(3-).
(Thanks to ghostwalker for pointing this out!)
5. How can I find this then???
6. (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.

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?
7. (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.
8. Thank you very much!!!!

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Updated: April 15, 2013
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