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Probability, Correlation, Variance Statistics Homework Help.

  1. Apr 11, 2014 #1
    1. The problem statement, all variables and given/known data
    1. Assume each birth in a hospital on a given day is independent of one another, and each birth, P(boy)=0.48. What is the probability that the 8th baby born is the 5th girl.

    2. Two random variables X and Y have joint distribution given by. What is their correlation.
    J7cHcIB.png

    3. Z is drawn randomly from {1,2,3,4,5}. Independently of Z, 4 other RVs Y1,...,Y4 are drawn randomly, with replacement, from {1,...,6}. Define X as the number of Yjs less than or equal to Z. What is the variance of X?
    2. Relevant equations
    2. Corr(X,Y)=Cov(X,Y)/((Var(X),Var(Y)^0.5)
    where Cov(X,Y)=E(X,Y)-E(X)E(Y)


    3. The attempt at a solution
    I think that most of my problem comes from not being able to apply the formulas and theory since my teacher doesn't do very many problems, and mainly does proofs which I don't find too helpful.

    Anyways my theory.
    1. So I think that I need to find the probability that out of the first 7 babies, 4 are girls, and I'm almost certain that this is a binomial. But from there, I don't know what to do. Then multiply by the chance that the 8th born, is a girl (0.52).
    Edit: I got this one! I found the forumla for probability and applied it.

    2. This is an issue what I don't know how to intepret the data. To find Var, I need to Expectation, but how do I find it? Which numbers do I multiple to get it.
    I also need to find Cov(X,Y) which require E(X,Y), which I am also clueless to find when the data is in this form.
    Is it E(X)=-1*(0.1+0.1+0.1)+0(0.1+0.15+0.1)+1(0.15+0.1+0.1)
    and E(Y)=-1(0.35)+1(0.3)
    When I tried this, the answer was almost certainly incorrect, since I got a positive answer, and the real answer I know is -ve.
    I thought that maybe the above might be correct, and I just don't know how to compute E(X,Y) correctly.

    3. This last one, I was given a hint. Var(X)=var[4z/5]+E((4z/5)(1-(z/5))
    I still don't know how to do this question even with this forumla. I assume that there are only a couple more steps (2?), but I don't know how to continue with the question.

    Thanks for any help
     
    Last edited: Apr 11, 2014
  2. jcsd
  3. Apr 12, 2014 #2

    haruspex

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    Yes
    So post your working.
    You mean E(XY). It's the same way you calculate the expected value of any r.v. List the values of XY, multiply each by its probability, and add them up.
    That's rather a specific hint. I mean, it's unclear how to adapt that to a different question. Anyway, not sure what's stopping you from applying it.
    What's the relationship between var(aZ) and var(Z) for a constant a?
    How can you expand E(aZ(b-cZ))?
    (Btw, you should be using capital Z inside Var() and E(), as I have, because these refer to random variables. The lower case letters are used for individual values of the random variables.)
     
  4. Apr 12, 2014 #3
    Sorry you were correct for E(XY)
    With E(XY), how would you do the two together?
    Would it be something like:
    (-1*-1*0.1)+(-1*0*0.1)+(-1*1*0.1)+...+(1*1*0.1)?

    With that hint, it's meant specifically for this questions. Anyhow.
    Do you mean for a constant a, it can be brought to the find of it?
    E(aZ(b-cZ))=aE(Z(b-cZ))?

    Thanks for replying, I'm really struggling with the subject.
     
  5. Apr 12, 2014 #4

    haruspex

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    Yes
    Yes. E is a linear function. E(aX+bY) = aE(X)+bE(Y).
    But be careful with Var - that is not linear. To see what happens to Var(aX), use the formula Var(X) = E(X2)-E(X)2.
     
  6. Apr 12, 2014 #5
    Ok so i know I'm meant to use the forumla Corr(X,Y)=E(XY)-E(X)E(Y) / sqrt[Var(X)Var(Y)]

    What I've got so far
    E(X)=0.05
    E(X^2)=0.65
    E(Y)=-0.05
    E(Y^2)=-0.65
    Var(X)=Var(Y)=0.6475
    E(XY)=-0.05
    When I put it into the forumla, the answer the I get isn't correct.
    Is there something that I'm doing wrong?

    For the last question, I'm still not sure what I'm meant to do. What am I looking for?
    E(Z)=(1*0.2)+(2*0.2)+3*(0.2)+(4*0.2)+(5*0.2)??
    What am I meant to apply to this Z/5?

    Thanks
     
  7. Apr 12, 2014 #6

    haruspex

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    I trust you mean (E(XY)-E(X)E(Y)) / sqrt[Var(X)Var(Y)]
    Looks correct to there, so you'll have to post the rest of your working. I get -0.073.
    Yes
    Not yet. Expand E((4Z/5)(1-(Z/5)) first, using the linearity formula I gave you.
     
  8. Apr 12, 2014 #7
    OH MY LORD. My friend gave me an answer to the wrong question! -0.073 was correct! I thought that I waas wrong the whole time. Thanks a bunch :D

    Var(X)=Var(4Z/5)+4E[z(1-(z/5))]
    For Var(4Z/5), you said that it doesn't scale linearly, and to use E(X) and E(X^2), but how do I apply either of them in this situation?

    Thanks
     
  9. Apr 12, 2014 #8

    haruspex

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    I need to some attempt from you here.
    Var(X) = E(X2)-E(X)2.
    Suppose Y=aX, some constant a.
    So what do you get for Var(Y)?
     
  10. Apr 12, 2014 #9
    Ok so I'm not sure if this is correct, but I'll try and explain my logic behind this.
    So X is the number of Ys Less than or equal to Z.
    For Z=1, there is only one thing thess than or =to 1 (1/6) from Y, which is one. For 2, there are 2 things, 1 and also 2.
    so I get
    E(X)=(1*1/6)+(2*2/6)+(3*3/6)+(4*4/6)+(5*4/6)+(6*4/6)
    E(X^2)=(1*1/6)+(4*2/6)...+(25*4/6)+(36*4/6)
     
    Last edited: Apr 12, 2014
  11. Apr 12, 2014 #10
    For 4E(Z/5 - Z^2/25)
    E(Z/5-Z^2/25)=(1/5 * 1/5)+(2/5 * 1/5)+...+(5/5 * 1/5)+ [(1^2/25) *1/5]+[(2^2/25)*1/5] +[(3^2/25)*1/5]+...+[(6^2/25)*1/5]
    Then multiply that above answer by 4.
    And this ans my post above, that would be the answer?
     
  12. Apr 12, 2014 #11

    haruspex

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    No, it's rather more complex than that.
    E.g. suppose Z=3. What is the probability that none of the Y are <= 3? That is the probability P(X=0|Z=3). It would be a lot of work to write out all 25 combinations of P(X=x|Z=z), so see if you can work out the general formula.
     
  13. Apr 12, 2014 #12

    haruspex

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    Yes, except that you turned a minus into a plus. Don't forget the Var[4z/5] term of the hint.
    Btw, I don't know how that hint formula was derived, and it gives me a different answer than the one I get if I don't use it. But I may have made a mistake, of course.
     
  14. Apr 12, 2014 #13
    Is the formula that I'm meant to be using nCr*p^r*(1-p)^(n-r)?
     
  15. Apr 12, 2014 #14
    I have the answer to the question. Could you tell me the anser that you get. If you get different answers for the hint method and the method that you have chosen to do, I don't want to be doing something incorrectly.
     
  16. Apr 12, 2014 #15
    Hint is derived from
    X|Z=z - B(4,(Z/5))
    Var(X)=Var[E(X|Z)]+E[Var(X|Z)]
    so Var(X)=[4Z/5]+E[4(Z/5)(1-(Z/5))]
     
  17. Apr 12, 2014 #16

    haruspex

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    If not using the hint, yes.
     
  18. Apr 12, 2014 #17
    Could you tell me if the answer that you got without/without the hint is 1.67?
     
  19. Apr 12, 2014 #18

    haruspex

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    You mean X|(Z=z) = z - B(4,(z/5))?
    That doesn't look right to me. E.g. P(X=4|Z=1) = 6-4. The RHS is not going to give a power of six in the denominator.
    Using the hint I get 48/25. What do you get? I'm fairly certain the right answer is a bit less.
     
  20. Apr 12, 2014 #19
    I wasn't sure how to work out a general formula for use with the hint, but if you got 48/25, I doubt that you're wrong, and suspect that the hint is incorrect-which is a bummer.
    How would you calculate the answer without the hint?

    The reason that I used this "Is the formula that I'm meant to be using nCr*p^r*(1-p)^(n-r)?", was because I thought that it was a binomial since it's with replacement.
     
  21. Apr 12, 2014 #20
    Oh if you missed the answer on the previous page, it was 1.67 so you were right in thinking that it was less.
     
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