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Math Help - Question on cumulative distribution function.

  1. #1
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    Unhappy 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 ?
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  2. #2
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    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.
    Your question shows a real confusion about this material.
    You need to sit down with a live tutor.

    P(a<Y<b)=F_Y(b-)-F_Y(a)
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  3. #3
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    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
    Last edited by Sneaky; October 8th 2010 at 04:08 PM.
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  4. #4
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    Quote Originally Posted by Sneaky View Post
    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.
    PLEASE, seek real live help.
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  5. #5
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    hmm what am i doing wrong?
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  6. #6
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    Quote Originally Posted by Sneaky View Post
    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.
    Last edited by Plato; October 8th 2010 at 04:39 PM.
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  7. #7
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    Quote Originally Posted by Sneaky View Post
    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<a)
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  8. #8
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    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?
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  9. #9
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    Quote Originally Posted by Sneaky View Post
    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.

    Quote Originally Posted by Sneaky View Post
    [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 ....
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  10. #10
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    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
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  11. #11
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    Quote Originally Posted by Sneaky View Post
    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.
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