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Math Help - Probability question

  1. #1
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    Oct 2009
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    Probability question

    I've been stuck on this question for a while...

    Joe has a midterm in his 1:00pm geography class. Unfortunately, his computer class, which is scheduled to end at 12:50pm, usually runs overtime. The amount of time it takes that it ends late is normally distributed with a mean of 3 minutes and a standard deviation of 2 minutes. It then takes him an amount of time to get to his geography classroom which is normally distributed with mean 4 minutes and standard deviation 1 minute. What is the probability that Joe is late for his midterm? (Use the fact that the sum of two independent normal random variables is also normally distributed.)

    Thanks
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  2. #2
    Member
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    Dec 2008
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    Auckland, New Zealand
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    For 2 random variables A and B:
    E(A+B) = E(A) + E(B)
    Var(A+B) = Var(A)+Var(B)

    Which means in this case the mean time from the scheduled end of class until he can get to his test is 3+4 = 7 minutes.
    The variance of this time is 2^2 + 1^2 = 5 minutes, so the standard deviation is \sqrt{5} = 2.236 minutes

    You can use a calculating utility to solve the problem now, or convert to Z values and look up the probability in the standard normal distribution table.
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  3. #3
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    Thank you very much for your help!
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