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Math Help - Probability: Finding n and p.

  1. #1
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    Probability: Finding n and p.

    I am preparing for my finals and came across this question:

    \text{Let p be the probability that the vaccine is a success in any patient.}
    \text{Find the sample size n and p required to be 95 percent assured that when the} \text{vaccine is given to n infected patients, at least 1000 successful vaccinations will result}


    I am quite confused on this one since we have two unknowns. I tried using binomials distribution with P(X \geq 1000)\;=\;1-P(X=1000)\;=\;0.95 with n being the number of patients to be vacinated in order to get at least 1000 successes, but i cannot calculate both n and p from the same equation.

    is there any other approach to solve this problem? thank you.
    Last edited by chutiya; December 2nd 2011 at 11:16 AM.
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  2. #2
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    Re: Probability: Finding n and p.

    Quote Originally Posted by chutiya View Post
    I am preparing for my finals and came across this question:

    \text{Let p be the probability that the vaccine is a success in any patient.}
    \text{Find the sample size n and p required to be 95 percent assured that when the} \text{vaccine is given to n infected patients, at least 1000 successful vaccinations will result}


    I am quite confused on this one since we have two unknowns. I tried using binomials distribution with P(X \geq 1000)\;=\;1-P(X=1000)\;=\;0.95 with n being the number of patients to be vacinated in order to get at least 1000 successes, but i cannot calculate both n and p from the same equation.

    is there any other approach to solve this problem? thank you.
    X ~ Binomial(n, p)

    Pr(X < 1000) = 0.05

    You will now need to use the normal approximation to make further progress.
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  3. #3
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    Re: Probability: Finding n and p.

    Thank You. Thats what I was thinking about.

    Apparently, the value of p had been missing in the question, which was p=90%(which was the major source of my trouble) and I was supposed to find "n" only. That makes life easy. I just used normal approximation np-1.64\sqrt{np(1-p)}=1000 to find n.
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