What is the minimum number of random samples that should be taken so that with probability at least 0.95, the proportion of smokers in the sample will not differ from the unknown population of smokers by more than $\pm 0.04$?
I am given the added information that: If $Z$ is $N(0,1)$, then $P(Z > 1.96) = 0.025$
This is exam revision. I can't seem to find the procedure for this in my notes so I don't know how to approach it.
If the sample consists of just one person, then the number of smokers is $0$ or $1$, with respective probabilities $1-p$ and $p$, so you have a random variable whose expected value is $p$ and whose variance is $p(1-p)$. If the sample size is $n$, then the number of smokers is the sum of $n$ random variables with that distribution, so the expected number is $np$. If they're independent, then the variance is $np(1-p)$. The proportion of smokers in the sample is the number of smokers in the sample divided by $n$. Dividing the random variable by $n$ has the effect of dividing its expected value by $n$ and its variance by $n^2$. So the expected proportion is $p$ and the variance is $p(1-p)/n$.
It looks as if you want the probability that the sample proportion differs from the population proportion by more than $0.04$ to be no more than $0.05$; that would explain why $\Pr(Z>1.96)=0.05/2$ is relevant. You have $\Pr(-1.96<Z<1.96)=0.05$. $$ \Pr\left(p-1.96\sqrt\frac{p(1-p)}{n} < X < p+1.96\sqrt\frac{p(1-p)}{n}\,\right) = 0.05. $$ So you want $$ 1.96\sqrt\frac{p(1-p)}n \le 0.04. $$ Since $p(1-p)\le\frac 1 4$ (with equality when $p=1/2$), you it is enough to make $n$ big enough so that $$ 1.96\frac1{\sqrt{4n}} \le0.04, \text{ or }\frac{1.96}{2\sqrt n}\le0.04. $$
You may not always find a procedure for something in your notes, but if you can find all the information needed in order to figure out the procedure yourself, you'll have enough.