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Math Help - pneumonia and normal distribution

  1. #1
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    pneumonia and normal distribution

    I am having trouble solving the following question after reading my textbook. Thank you for any help you can give me.

    In adults the pneumococcus bacterium cause 70% of the pneumonia cases. In a random sample of n=84 adults who have pneumonia, let X equal the number whose pneumonia was caused by the pneumococcus bacterium. Use the normal distribution to find, approximately, P(X \le 52)
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  2. #2
    MHF Contributor matheagle's Avatar
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    X is a binomial rv with n=84 and p=.7.

    The exact probability is \sum_{k=0}^{52}{84\choose k}(.7)^k(.3)^{84-k} (or you can use the complement).

    The normal approximation is...

     P(X_B\le 52) \sim  P(X_N\le 52.5)

    where  X_B is the original binomial rv, while X_N is the overlapping normal approximation to that binomial.
    Next we center the normal...

    P(X_N\le 52.5)=P\biggl(Z\le {52.5-(84)(.7)\over \sqrt{(84)(.7)(.3)}}\biggr)
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  3. #3
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    Thank you so much for clearing that all up! I just finished your work to get P(Z \le -1.5)=P(Z \ge 1.5)=1-0.9332=0.0668. Hopefully I didn't mess that little part up!
    Last edited by logitech; May 28th 2009 at 08:49 PM.
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  4. #4
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    Smile

    Can someone help me figure out where I went wrong? I got 0.1530328 for the exact probability but 0.0668 using normal distribution, but I think these numbers are too far apart. Thank you!
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  5. #5
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    I think I got it! I calculated the real probability wrong. It should have been 0.068987
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  6. #6
    MHF Contributor matheagle's Avatar
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    With n=84, they should be quite close.
    I just checked my approximation and I also P(Z \le -1.5).
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