# Help with cdf

#### bigbob

##### New Member
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.

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#### asterisk

##### New Member
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)

#### bigbob

##### New Member
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?

#### asterisk

##### New Member
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

#### bigbob

##### New Member
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.

#### asterisk

##### New Member
Yes, so

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

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