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Math Help - normal distribution question

  1. #1
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    normal distribution question

    I'm looking at an example. There's one part that I don't understand.

    The random variable Y has a normal distribution with mean 10 and standard deviation 2. Find the value of the constant a such that P(10-a<Y<10+a)=0.95.

    Y~N(10,4)
    Given P(10-a<Y<10+a)=0.95,
    Therefore, P(Y<10-a)=0.025
    10-a=6.080072028
    a=3.92

    What I don't get it is how does P(10-a<Y<10+a)=0.95
    gives me P(Y<10-a)=0.025 ???

    Any help is much appreciated. Thanks.
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  2. #2
    MHF Contributor
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    A little arithmetic.

    P(10-a < Y < 10+a) = 0.95

    P(10-a < Y AND Y < 10+a) = 0.95

    1 - P(10-a < Y AND Y < 10+a) = 1 - 0.95

    P(10-a > Y OR Y > 10+a) = 0.05

    P(10-a > Y) + P(Y > 10+a) = 0.05

    2*P(10-a > Y) = 0.05

    P(Y<10-a)=0.025
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  3. #3
    MHF Contributor
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    see the attached picture.



    The area under the distribution is 100%, and you are told that 95% is in the range (10-a,10+a). That leaves 5% outside this range.

    The normal distribution is symmetric about the mean so there must be 2.5% on either end.

    Hence P(Y<10-a) = 2.5% = 0.0025


    EDIT: Didn't see previous reply.
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