Central Limit Theorem

**feiyingx** · October 21st 2007, 11:54 AM

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!

**CaptainBlack** · October 21st 2007, 12:22 PM

Originally Posted by feiyingx

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:

$ n=\sum_{i=1}^{100} x_i $

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:

$ z=\frac{49.5-45}{\sqrt{24.75}} \approx 0.9045 $

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

**feiyingx** · October 21st 2007, 12:50 PM

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!

**CaptainBlack** · October 21st 2007, 01:27 PM

Originally Posted by feiyingx

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

**feiyingx** · October 21st 2007, 01:54 PM

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
$ z=\frac{49.5-45}{\sqrt{24.75}} \approx .9045 $
Using the table and calculating for the probability, I got 0.183.

Is the book's solution incorrect?

Thanks!

**CaptainBlack** · October 21st 2007, 02:01 PM

Originally Posted by feiyingx

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
$ z=\frac{49.5-45}{\sqrt{24.75}} \approx .9045 $
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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