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Homework Help: Normal distribution head-scratcher

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data

    A study investigates the performance of students in an exam, which is known to follow a normal distribution, with a mean of 63 and a standard deviation of 21.

    What is the probability that the difference between two randomly selected
    students is more than 5 marks?

    2. Relevant equations


    Any ideas? Many thanks :)
  2. jcsd
  3. Feb 17, 2010 #2
    Try using

    Y=X1-X2 as a new random variable with mean 0 and standard deviation of sqrt(2)*21


    P(Z>.052) = 1-P(Z<.052)

    so now you can find the values in a z-table.

    I haven't thought of the other case in which X2>X1, but this is at least how you should think about the question.

    With such a high deviation intuitively you should be getting a high probability.
    Last edited: Feb 17, 2010
  4. Feb 21, 2010 #3
    How did you get 0.052? Mean=0, SD= sqrt(42)?
  5. Feb 21, 2010 #4
    For normal distributions you can use the properties if X1 and X2 are two independent random variables then Y=aX1+bX2 is a random variable with normal distribution and mean a*u1+b*u2 and variance a^2*sigma1^2+b^2*sigma2^2

    So in our case Y=X1-X2, mean = 1*63+-1*63=0
    variance = 1^2*(21^2)+(-1)^2*(21^2)=21^2+21^2=2*21^2

    You can't look this up. so you look up Z which is (Y-mean)/std.
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