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Probability related to Normal Distribution

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Variable A has mean 55 and variance 9, variable B has mean 65 and variance 25. If A and B are normally distributed, find P (B > A)


    2. Relevant equations
    z = (x - μ) / σ



    3. The attempt at a solution
    Can this be solved? What is the meaning of P (B > A)? The probability of random variable B is bigger than random variable A?

    Thanks
     
  2. jcsd
  3. Jan 22, 2013 #2
    I suppose the problem assumes A and B are statistically independent. Then the formal way to do this is:
    [itex]P(B>A) = \int_{b=-\infty}^{+\infty} \int_{a=-\infty}^{b}f_A(a)f_B(b)dadb[/itex]
    This is a very nasty integral.
     
  4. Jan 22, 2013 #3

    Ray Vickson

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    Yes, P(B > A) means exactly what you said. It can be solved quite easily if A and B are independent, but I cannot give more details now: first you have to make a serious attempt to solve it yourself.
     
  5. Jan 22, 2013 #4
    Sorry I don't get the idea how to do the integration and why can it be like that

    I don't have idea how to start. The question I saw until now always contained number such as P (B > 23). Please give me clue how to start solving this problem. What is the first step I need to take? I can't find the value of z because I don't know the value of the random variable.

    Thanks
     
  6. Jan 22, 2013 #5

    Ray Vickson

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    Look at the random variable Y = B - A.
     
  7. Jan 22, 2013 #6
    OK maybe I get it

    Let: Y = B - A

    E(Y) = E(B) - E(A) = 10
    σ(Y) = σ(B) + σ(A) = 34

    P(B > A) = P(B - A > 0) = P(Y > 0) = P(Z > -1.715) = 0.9568

    Do I get it right?

    Thanks
     
  8. Jan 22, 2013 #7

    Ray Vickson

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    It looks right.
     
  9. Jan 23, 2013 #8
    ok thanks
     
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