Central limit theorem questions Suppose that the proportion of defective items in a
large manufactured lot is 0.1. What is the smallest random
sample of items that must be taken from the lot in order
for the probability to be at least 0.99 that the proportion
of defective items in the sample will be less than 0.13?
So I know the proportion of n defective items = 0.1
and I have to determine the value of n such that $Pr(\bar{X}_n<0.13)\ge 0.99$
I was looking at an answer key and they said this:
p=0.1 ok, that makes sense
variance $\frac{p(1-p)}{n}=\frac{(0.1)(0.9)}{n}$
but I thought variance = np(1-p)? What am I missing?
Then they used this: 
$$Z=\frac{\bar{X}-\mu}{{\sigma}/\sqrt{n}}$$
$$\frac{Z\sigma}{\sqrt{n}}=\bar{X}-\mu$$
$$\frac{0.3Z}{\sqrt{n}}=\bar{X}-0.1$$
So $\mu=p$? but why does $\sigma=0.3$??
Any help clearing up these questions would be greatly appreciated


 A: This is a typical probability computation in 'acceptance
sampling'. The idea is to randomly select and inspect the
minimum number of items necessary to meet a certain quality
criterion.
I believe your question about the variance has been answered
in the comments.
Here is an exact computation in R, based directly on binomial probabilities
without using the CLT. The answer is very close to the one you got
using the normal approximation. The method is to evaluate the
relevant binomial probability for all $n$ from 10 through 100,
and then pick the smallest $n$ meeting the criterion.
 n=10:1000;  bp = pbinom(.13*n, n, .1)   # pbinom is CDF
 min(n[bp >= .99])   # brackets can be read 'such that'
 ## 554

 # Double check
 pbinom(.13*553, 553, .1)   # 553 not quite enough
 ## 0.9869744
 pbinom(.13*554, 554, .1)   # 554 just barely enough
 ## 0.9903957
 pbinom(.13*542, 542, .1)   # value from CLT--pretty close
 ## 0.9880745

Of course, this method provides no practice with the means,
variances, using normal tables, and so on. But it seems you
have gotten about as close to the correct answer as possible
using a normal approximation. (But you might want to check
for rounding errors.) 
Note: There is a little more complexity here than may
be immediately obvious. Because of discreteness, 'bp' is not exactly
monotone increasing with 'n'. Also, 'bp' comes very near .99 for
some smaller values of n: If '.99' is interpreted
to mean '>.985', then n = 477 is enough.
Nowadays, I believe something like this kind of 'brute
force' computer search is the way such a problem
would be solved in practice.
