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X & Y Normal Distribution

  1. Aug 13, 2013 #1
    1. The problem statement, all variables and given/known data

    X refers to score distribution in Math and Y refers to score distribution in Stat in a certain degree course exam. It is known that X~N(mean = 62, sigma=7) while Y~N(mean = 68, sigma=10). If X and Y are independent, find (i) P[X+Y>120]; (ii) P[X<Y]; (iii) P[X+Y>140]; (iv)P[100 < X+Y < 150].

    2. Relevant equations



    3. The attempt at a solution

    P(X+Y>120)...

    So I have to combine X & Y into one Norm Distribution by find there combined E(x) & V(x)

    E(X+Y)=E(X)+E(Y)=62+68=130

    V(X+Y)=V(X)+V(Y)=7^2 + 10^2 = 149, Standard Dev = sqrt(149)

    So where Z = X + Y > 120, Z~N(130,sqrt(149))

    P(X+Y>120) = P(Z>120)=1-PHI((130-120)/sqrt(149))

    Is this correct so far....


    P(X<Y)=P(X-Y<0)

    E(X-Y)=E(X)-E(Y) = 62-68=-6

    V(X-Y) = V(X)+V(Y)=sqrt(149)

    W~N(-6,sqrt(149))

    P(X-Y<0)=P(W<0)=PHI((-6-0)\sqrt(149))

    How does this look. The rest are basically the same I think if I got these two conceptualy correct. any issues?
     
  2. jcsd
  3. Aug 13, 2013 #2

    statdad

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    I haven't checked down to the arithmetic level but the approach looks good. One thing that helped me keep things straight notationally was this: When dealing with the first part, where you are working with the sum of two random variables, let [itex] S = X + Y [/itex]. Then the work you have shows that [itex] S [/itex] has a normal distribution with a particular mean and standard deviation, and I just found it easier to write
    [itex] P(S > 120) [/itex] instead of the way you had it. Similar for the difference.

    After a final glance I do see one item I believe you should reconsider. Do you REALLY want to write
    [tex]
    V(X-Y) = V(X) + V(Y) = sqrt(149)
    [/tex]

    (I am pointing my comment at what you have after the second equal sign) - remember in those steps you are finding a variance.
     
    Last edited by a moderator: Aug 14, 2013
  4. Aug 13, 2013 #3

    Ray Vickson

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    The equation ##V(X-Y) = V(X) + V(y)## is true if X and Y are independent, so what he wrote is correct. In fact, if ##a## and ##b## are constants, we have
    [tex] V(aX + bY) = a^2 V(X) + b^2 V(Y)[/tex] for independent (or uncorrelated) X and Y, whether normally-distributed or not. It is a general result.
     
  5. Aug 14, 2013 #4

    statdad

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    I am aware of all this. You missed my point, however: it is likely my comment wasn't clear. The OP is attempting to calculate the variance of X - Y: the written equation was (my comments added)
    [tex]
    V(X-Y) = \underbrace{V(X) + V(Y)}_{\text{Okay here}} = \overbrace{\text{sqrt}(149)}^{\text{My concern}}
    [/tex]

    For emphasis, compare the ending of the above work to the ending of the line in which the variance of [itex] V(X + Y) [/itex] was correctly calculated.

    Marginally related comment: apologies for missing the fact that I screwed up a tag in one of my earlier posts.
     
  6. Aug 14, 2013 #5

    Ray Vickson

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    Just to be clear: you (correctly) want him to write '149' instead of 'sqrt(149)'. To my shame, I missed that the first time around!
     
  7. Aug 14, 2013 #6

    statdad

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    "Just to be clear: you (correctly) want him to write '149' instead of 'sqrt(149)'. To my shame, I missed that the first time around!"

    No big deal. As I noted, I was not entirely clear in my original note. That comes, I am afraid, from reviewing student papers and wanting to point out issues for them without giving the whole thing away.
     
  8. Aug 14, 2013 #7

    Ray Vickson

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    I agree with the 'not giving away' part, but he did already have the correct answer---or at least, the correct numerical inputs---so I did not feel guilty about laying out some more detail.
     
  9. Aug 14, 2013 #8

    statdad

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    Probably true - actually, just plain "yup" on that. Finding a way to guide someone along giving support while pointing out subtle, tertiary level errors, is always a difficult task for me.
     
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