1. ## 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!

2. 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

3. 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!

4. 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

5. 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!

6. 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"