1. ## 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.

2. ## 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)

3. ## Re: Help with cdf

Originally Posted by asterisk
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?

4. ## 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

5. ## Re: Help with cdf

Originally Posted by asterisk
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.

6. ## 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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