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## Homework Statement

Two types of coins are produced at a factory: a fair coin and a biased one that comes up heads 55 percent of the time. We have one of these coins but do not know whether it is a fair coin or a biased one. In order to ascertain which type of coin we have, we shall perform the following statistical test: We shall toss the coin 1000 times. If the coin lands on heads 525 or more times, then we shall conclude that it is a biased coin, whereas, if it lands heads less than 525 times, then we shall conclude that it is the fair coin. If the coin is actually fair, what is the probability that we shall reach a false conclusion? What would it be if the coin were biased?

## Homework Equations

The DeMoivre-Laplace limit theorem

## The Attempt at a Solution

Since n is large, we can use the normal approximation to the binomial distribution.

We first assume that the coin is fair, that is p = .5

Our mean is np = 1000*.5 = 500

The variance is np(1-p) = 250, so the standard deviation is sqrt(250).

We would reach a false conclusion if the number of heads we observe is greater than or equal to 525.

[tex]P(S_{n} \geq 525) = 1 - P(S_{n} < 525)[/tex]

We now put into form to apply the DeMoivre-Laplace limit theorem:

[tex]1 -P(\frac{S_{n} - np}{\sqrt{np(1-p)}} < \frac{525-500}{\sqrt{250}})[/tex]

[tex]= 1 - \Phi(1.581)[/tex]

Using the table in the book, I look up the the value of the area beneath the standard normal curve to the left of 1.58. This value is .9429.

Hence 1-.9429 = .0571 is the probability of reaching a false conclusion.

The answer in the back of my book is .0606.

I think my method is sound, but I'm maybe acquiring phi incorrectly?

Rounding 1.581 up to 1.59 still doesn't get me to .0606.

Any ideas? I've run into a similar error in other problems, which makes me think I'm doing something systematically wrong.