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De Moivre-Laplace theorem help

  1. Apr 7, 2014 #1
    1. The problem statement, all variables and given/known data
    There are 10 000 000 people who are going to vote. 5 500 000 will vote for the party A, the rest will vote for the party B.
    20 000 voters are randomly chosen. What is the probability that there will be more party B voters than there will be party A voters (sorry for poor translation).

    2. The attempt at a solution

    P(party A voter) = 0.55
    P(party B voter) = 0.45

    There have to be at least 10 001 party B voters for the condition to be satisfied. This number can range from 10 001 to 20 000. Then, according to de Moivre-Laplace theorem the probability which I need to find:

    P = Θ(λ2) - Θ(λ1) ,

    where λ1 = (10 001 - 20 000*0.45)/√(20 000*0.45*(1-0.45))

    Unfortunately, λ1 ≈ 14.23. λ2 is even bigger. Both, when put under the Θ function asymptotically approximate to 1. Do I have to use a different approach here?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 7, 2014 #2

    jbunniii

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    You should expect the probability to be extremely small. The probability distribution for the number of B voters is approximately normal with mean ##9000## and standard deviation ##\sigma = \sqrt{npq} \approx 70.4##. The probability of exceeding ##3\sigma## above the mean is already only about one in a thousand, and you are looking at the probability that it exceeds around ##14\sigma## above the mean. So I don't necessarily think you are doing anything wrong, but the answer is going to be tiny.
     
    Last edited: Apr 7, 2014
  4. Apr 7, 2014 #3
    Thank you for the clarification. I guess it is just a bit counter-intuitive to me.
     
  5. Apr 7, 2014 #4

    jbunniii

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    To get some intuition, consider flipping a coin 20000 times. Surely if the coin is fair (##p = q = 0.5##), it should be extremely unlikely that you would see, say, 11000 heads or more. Probably the only surprise is exactly how tiny the probability is, but you would expect it to be quite small. Your problem is similar, but with slightly different ##p## and ##q##.
     
  6. Apr 7, 2014 #5

    Ray Vickson

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    If you want an actual numerical approximation, use the fact that for the standard normal pdf
    [tex] \phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-z^2/2}[/tex]
    we have, for large ##z > 0##:
    [tex] \Phi_{>}(z) \equiv \int_{z}^{\infty} \phi(t) \, dt \sim \frac{1}{z} \phi(z).[/tex]
    This follows from integration by parts: in ##\int \phi(t) \, dt,## take ##u = 1/t,
    : dv = t \phi(t) \, dt = -d (\phi(t))##. Thus
    [tex] \int_z^{\infty} \phi(t) \, dt = \phi(z)/z - \int_z^{\infty} \frac{1}{t^2} \phi(t) \, dt,[/tex]
    and for large ##z > 0## we can drop the second term (its magnitude is ##< \phi(z)/z^2##). If you want better accuracy, get another term using integration by parts again, etc.

    Anyway, for ##z = 14.23## we have ##\Phi_{>}(z) \approx 0.2999 \times 10^{-45}##.
     
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