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Math Help - Normal distribution intervals.

  1. #1
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    Normal distribution intervals.

    An instructor has two classes, one size 30, the other size 64. From past experience, the students' mean exam score is 77 and standard deviation is 15.
    What is the approximate probability that the average test score in the size 30 class is more than 1 point higher than that of the size 64 class.

    This is what I have so far but don't know where to go after this:
    P(X_{30} > X_{64} + 1) = P(\frac{X_{30}-77}{15/\sqrt{30}} > \frac{X_{64}-77}{15/ \sqrt{64}} + ?)
    Last edited by BERRY; November 14th 2009 at 05:54 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by BERRY View Post
    An instructor has two classes, one size 30, the other size 64. From past experience, the students' mean exam score is 77 and standard deviation is 15.
    What is the approximate probability that the average test score in the size 30 class is more than 1 point higher than that of the size 64 class.

    This is what I have so far but don't know where to go after this:
    P(X_{30} > X_{64} + 1) = P(\frac{X_{30}-77}{15/\sqrt{30}} > \frac{X_{64}-77}{15/ \sqrt{64}} + ?)
    The mean class scores have means 77, and and variances 15^2/30 and 15^2/64 respectivly.

    Therefore the difference in means has mean 0 and variance 15^2/30+15^2/64.

    Now we want to know what is the probability (assume the difference is normall distributed) that a RV X \sim N(0, 15^2/30+15^2/64) is greater than 1.

    CB
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