1. ## normal distribution apprx

find the size n, the number of times an unbiased coin must be tossed if the probability will be at least 0.75 that proportion of heads is btw 0.4 and 0.6.

i know that i have to approx normal distribution but how do i find?

2. The equation for finding sample size with a given proportion is given by (see this link for details):

$n = \frac{Z^2 p(1-p)}{e^2}$

where p is the true proportion, Z is the z-value for a given confidence level, and e is the sampling error. So consider the information you're given. We know that since the coin is unbiased that half of the tosses should end up on one side or the other. Thus, p = 0.5, and p(1-p) = 0.25. Your error is the difference between the observed and the true proportion. The question states that we want at least 75% of heads to be within 0.1 of the true proportion. Thus, the error is just that difference. The last piece of information you require for your calculation is Z for an $\alpha = 0.25$. I get that because the confidence level just is $1-\alpha$. For instance, standard confidence levels are to find the observed at least 95% of the time with an $\alpha = 0.05$, which gives a Z = 1.96. In our case, we should get Z = 1.15.

I get a value of n just slightly above 33. Since these are discrete, we have to round up to whole numbers.

3. although the OP asked for a normal approximation its worth mentioning that it doesn't land on the right answer in this case. The distribution is discrete - if the sample size is 33 to get a sample proportion in (0.4,0.6) you need successes in the interval of [14,19]. The probability of that is 0.703. You need 34 in your sample to get a probability over 0.75.

It may be more appropriate to use the normal approximation to get a first guess then use trial and error to get your final answer.

4. My calculations produced a number slightly above 33. As I said these are discrete, so the rounding has to go to the ceiling of n, and that is 34. It might be more apt to say we want n*, where n* = ceiling(n), with n defined as above. I did wonder, though, if a binomial distribution might be used instead (bearing in mind that by the CLT it is approximately normal for large populations).

5. ah sorry i misread, i thought you had rounded up to 33