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Math Help - Central Limit Theorem

  1. #1
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    Central Limit Theorem

    Candidates A and B are running for office and 55% of the electorate favor candidate B. What is the probability that in a sample of size 100 at least one-half of those sampled will favor candidate A?

    Here's what I did:
    Let X_i = 1 if the ith person voted for A.
    S_100 = X_1 + X_2 + X_3 + ... + X_100

    P(S_100 >= 50) = 1 - P(S_100 <= 49)
    = 1 - P([S_100 - 100(.45)]/sqrt(100*.45*.55) <= [49 + 1/2 - 100(.45)]/sqrt(100*.45*.55)) #transform into standard norm rv
    = 1-P(Z<=0.9045)
    = 1-z(0.9045) #z is the standard normal function phi
    = 1-0.817
    = 0.183

    However, the book's solution is 0.133. I don't know how to arrive at that solution. Can someone go through this problem step by step. Thanks!
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  2. #2
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    Quote Originally Posted by feiyingx View Post
    Candidates A and B are running for office and 55% of the electorate favor candidate B. What is the probability that in a sample of size 100 at least one-half of those sampled will favor candidate A?
    This is a central limit theorem question, so let X_i=1 if the i-th samped voter
    votes A, and 0 otherwise. then the mean for X_i is

    \mu = 0.45

    and the variance is

    \sigma ^2=0.45(1-0.45)^2+0.55(-0.45)^2 = 0.2475.

    So the number of votes for A in the sample is:

    <br />
n=\sum_{i=1}^{100} x_i<br />

    which approximatly has a normal distribution with mean 100\mu=45, and variance 100 \sigma^2 = 24.75.

    So now we want the probability that 50 or more will vote A. As we have
    a continuous distribution modelling a discrete we ask what is the probability
    of a value greater than 49.5 occuring from a normal distribution with mean 45 and variance 24.75.

    The z-score for this problem is:

    <br />
z=\frac{49.5-45}{\sqrt{24.75}} \approx 0.9045<br />

    which we look up in a standard normal table to get a probability of 0.183.

    Note if we had been asked for the probability of more than 50 voted for A this would drop to 0.134

    (If this were not a CLT question I would have used a binomial distribution to
    model the distribution of the number of votes for A in the sample, but when the
    normal approximation is used for the binomial the answer is exactly the same
    as we get with the above argument)

    RonL
    Last edited by CaptainBlack; October 21st 2007 at 02:59 PM.
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  3. #3
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    Thanks for the response RonL.
    I have a question that I'm confused about.

    Shouldn't the mean for Xi be 0.45 since Xi is a Bernoulli RV with probability of people voting for A of 45%?
    Therefore, the total mean is .45*100 = 45

    Thanks!
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  4. #4
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    Quote Originally Posted by feiyingx View Post
    Thanks for the response RonL.
    I have a question that I'm confused about.

    Shouldn't the mean for Xi be 0.45 since Xi is a Bernoulli RV with probability of people voting for A of 45%?
    Therefore, the total mean is .45*100 = 45

    Thanks!
    Opps.. I had the probabilities the wrong way around. You should be able to fix that yourself.

    RonL
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  5. #5
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    So the only difference would be the mean of Xi which is \mu = 0.45
    This translates to the normal mean of .45*100 = 45.
    Then by using the normal approximation, it gives us
    <br />
z=\frac{49.5-45}{\sqrt{24.75}} \approx .9045<br />
    Using the table and calculating for the probability, I got 0.183.

    Is the book's solution incorrect?

    Thanks!
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  6. #6
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    Quote Originally Posted by feiyingx View Post
    So the only difference would be the mean of Xi which is \mu = 0.45
    This translates to the normal mean of .45*100 = 45.
    Then by using the normal approximation, it gives us
    <br />
z=\frac{49.5-45}{\sqrt{24.75}} \approx .9045<br />
    Using the table and calculating for the probability, I got 0.183.

    Is the book's solution incorrect?

    Thanks!
    Well with the wording you give I agree with you (I have now corrected the
    earlier post), but if the wording had been:

    "What is the probability that in a sample of size 100 at more than one-half of those sampled will favor candidate A"

    the answer would be 0.134.

    RonL
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