Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook