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Probability - continuous random variables

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Ok, I have 2 questions:

    1. Nicotine levels in smokers can be modelled by a normal random variable with mean 315 and variance 1312. What is the probability, if 20 smokers are tested, that at most one has a nicotine level higher than 500?

    2. fX,Y (x,y) = xe-x-y 0<x<y<[tex]\infty[/tex]
    Find c.
    Find the marginal probability density functions.

    2. Relevant equations



    3. The attempt at a solution

    1. I have worked out that each smoker individually has a 0.079 probability of having a nicotine level higher than 500, but I'm not sure about the at most one section. Would I need to work out 1-P(at least 2 have a nicotine level higher than 500) where P(at least 2 have a nicotine level higher than 500) = P(2 have a nicotine level higher than 500) + P(3 do) + P(4 do) + ... + P(20 do). In this case would it be a geometric sum with a = 0.0792, r = 0.079 and n = 19?

    2. It is the limits of integration that I am finding confusing here. I know to find c I need to do the double integral equal to one, but is it [tex]\int[/tex][tex]^{infinity}_{0}[/tex][tex]\int[/tex][tex]^{y}_{0}[/tex] fX,Y (x,y) dx dy (i.e. integrating x between 0 and y and y between 0 and infinity)?

    Similarly for the marginal distributions would the limits when integrating with respect to y be 0 and infinity and x be 0 and y?
     
  2. jcsd
  3. Feb 9, 2010 #2

    LCKurtz

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    P(at most 2) = P(0) + P(1). And if your p = .079 is correct and X is the number with nicotine over 500, then isn't X binomial(20,p)?

    Your limits are correct. Click on the expression below to see how to render it in tex:

    [tex]\int_0^\infty \int_0^y f(x,y)dxdy[/tex]
     
  4. Feb 9, 2010 #3

    LCKurtz

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    I didn't notice this final question. For the marginals your limits are determined by the non-zero region. So when integrating in the x direction you will go from 0 to y and when integrating in the y direction, it will go from x to infinity. Your nonzero domain is in the first quadrant above the line y = x.
     
  5. Feb 9, 2010 #4
    Thanks a lot, really helpful :)
     
  6. Feb 9, 2010 #5
    Ah ok, so fX(x) = [tex]\int_x^\infty f(x,y) dxdy[/tex] not fX(x) = [tex]\int_0^\infty f(x,y) dxdy[/tex]?
     
  7. Feb 9, 2010 #6

    vela

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    Probably just a typo, but you shouldn't have dx in the integrals because you're only integrating over y. And yes, the limits on your first integral is what LCKurtz meant.
     
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