# Thread: Question on cumulative distribution function.

1. ## Question on cumulative distribution function.

suppose Fy(y)=y^3 for 0<=y<1/2 and Fy(y)=1-y^3 for 1/2<=y<=1. Compute these.

1.
P(1/3<Y<3/4)
2.
P(Y=1/3)
3.
P(Y=1/2)

I am not sure i did the first one right, i did integrals and got the final answer to be
approx. 0.12

Is that what i am supposed to do for this and is 1 right?

for 2. is it
1/27 ?
and 3. is it
1- 1/27 ?

2. Before you take offence at this remark, stop to consider that I have taught this for years. Therefore, I can easily recognize a student in real trouble.
You need to sit down with a live tutor.

$P(a

3. P(1/3<Y<3/4)
P(1/3<Y<1/2) + P(1/2<=Y<3/4)
( Fy(1/2-) - Fy(1/3) ) + ( Fy(3/4-) - Fy(1/2-) )
Fy(1/2-) - Fy(1/3) + Fy(3/4-) - Fy(1/2-)

Now I have to use
Fy(y)=y^3 for 0<=y<1/2 and Fy(y)=1-y^3 for 1/2<=y<=1

(1/8-) - 1/27 + (37/64-) - (1/8-)
= ~935/1728
Is this right for 1. ?
for 2 i am getting 1/27
for 3 i am getting 7/8

4. Originally Posted by Sneaky
P(1/3<Y<3/4)
P(1/3<Y<1/2) + P(1/2<=Y<3/4)
( Fy(1/2-) - Fy(1/3) ) + ( Fy(3/4-) - Fy(1/2-) )
Fy(1/2-) - Fy(1/3) + Fy(3/4-) - Fy(1/2-) Now I have to use
Fy(y)=y^3 for 0<=y<1/2 and Fy(y)=1-y^3 for 1/2<=y<=1

1/8 - 1/27 + 37/64 - 1/8
=935/1728
Is this right for 1. ?
for 2 i am getting 1/27
for 3 i am getting 7/8
You have proven my point.
You are profoundly confused.
This sort of website is not the place to end that kind of confusion.

5. hmm what am i doing wrong?

6. Originally Posted by Sneaky
hmm what am i doing wrong?
What you are doing wrong is: not understanding that there is no way for a site such as this it can correct your profound lack of knowledge of all of this.
You have no idea of the basic principles are.
In another posting, you did not even know that the total sum of probability is 1.

Get yourself a live tutor. Or just fail.

7. Originally Posted by Sneaky
suppose Fy(y)=y^3 for 0<=y<1/2 and Fy(y)=1-y^3 for 1/2<=y<=1. Compute these.

1.
P(1/3<Y<3/4)
2.
P(Y=1/3)
3.
P(Y=1/2)

I am not sure i did the first one right, i did integrals and got the final answer to be
approx. 0.12

Is that what i am supposed to do for this and is 1 right?

for 2. is it
1/27 ?
and 3. is it
1- 1/27 ?

You really need to understand cdfs. Plato has already told you how to go for (1).

For (2) and (3), use $P(X=a) = P(X \leq a) + P(X

8. the only thing i see is
1.
Fy(3/4-) - Fy(1/3)
=37/64- - 1/27
=935/1728-
2.
Fy(1/3) - Fy(1/3-)
=0+
3.
Fy(1/2) - Fy(1/2-)
=0+
If this is not right, how do I evaluate it to get a numerical value?

9. Originally Posted by Sneaky
suppose Fy(y)=y^3 for 0<=y<1/2 and Fy(y)=1-y^3 for 1/2<=y<=1.
[snip]
This does not define a cdf. The least of it's troubles is that it's a decreasing function for $\frac{1}{2} \leq y \leq 1$.

And if it's meant to define a pdf, then there ought to be a rule for y > 1 because what's been posted does not integrate to 1.

Go back and re-check the question please. And be clear to us on whether it's meant to be a cdf or a pdf.

Originally Posted by Sneaky
[snip]
2.
P(Y=1/3)
3.
P(Y=1/2)

[snip]
for 2. is it
1/27 ?
and 3. is it
1- 1/27 ?
You should have been taught that if Y is a continuous random variable then Pr(Y = a) = 0 ....

10. Suppose Fy(y)=y^3 for 0<=y<1/2, and Fy(y)=1-y^3 for 1/2<=y<=1. Compute each of the following:
a) P(1/3<Y<3/4)
b) P(Y=1/3)
c) P(Y=1/2)

the section name is cumulative distribution function
chapter name is random variables and distributions

11. Originally Posted by Sneaky
Suppose Fy(y)=y^3 for 0<=y<1/2, and Fy(y)=1-y^3 for 1/2<=y<=1. Compute each of the following:
a) P(1/3<Y<3/4)
b) P(Y=1/3)
c) P(Y=1/2)

the section name is cumulative distribution function
chapter name is random variables and distributions
It is NOT a cdf. I suggest you ask your instructor to clarify this question because as far as I can see it is flawed.