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Math Help - Another stat prob

  1. #1
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    Another stat prob

    Thanks
    Last edited by Smokey007; August 14th 2007 at 09:41 PM.
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  2. #2
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    Need some help guys I really don't think I'm doing this right

    right now I have:

    a. p(x=3)

    20!/3!1! (.05)^3 (.95)^17 = .24

    b. p (x > 19) = p(x=19) + p(x=20) = .27

    c. Mx = 1 s.d.= .97
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  3. #3
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    Binomial

    p = 0.05

    q = 1-p = 0.95

    n = 20

    1) Pr(3)

    2) Pr(19) + Pr(20)

    3) np \sqrt{npq}

    These definitions must be in your text. Find them and you will be able to identify similar information for other distributions.
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  4. #4
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    Quote Originally Posted by Smokey007 View Post
    Need some help guys I really don't think I'm doing this right

    right now I have:

    a. p(x=3)

    20!/3!1! (.05)^3 (.95)^17 = .24

    b. p (x > 19) = p(x=19) + p(x=20) = .27

    c. Mx = 1 s.d.= .97
    Sorry about notation differences. It happens.

    Pr(3) = p(x=3)

    You have the right idea, but something is going whacky.

    In part a, you have a 1! in the denominator where there should be a 17! Even so, what you have written comes nowhere near 0.24. Must be more arithmetic errors going on. [(20*19*18)/(3*2)]*(0.05^3)*(0.95^17) = 0.0596 or so.

    In part b, that probability is WAY to big. I'm thinking maybe you made the same kind of denominator error.

    20*(.05^19)*0.95 + 0.05^20 is a very small number.

    Good call on Part c.
    Last edited by TKHunny; August 14th 2007 at 09:05 PM.
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