Smallest n to align sample mean with population mean There's a question in my book that I just do not understand. This is it in its entirety:
Let $ \bar{X} $ be the sample mean of a random sample of size $ n $ from a normal distribution with a variance of 9. Find the smallest sample size such that the sample mean is within 0.5 units of the population mean with probability no less than (i) 0.9, (ii) 0.95
I really don't know where to begin. I've just started learning about confidence intervals, and usually the sample mean has been given. 
I think it would be setup something like this: 
$$P(\mu-0.5 \le \bar{x} \le \mu+0.5) = 0.9  $$
But how can I do that? And where does finding the smallest n come into play? 
I'm not given the sample mean, and I'm not given the population mean. 
I'd like to have more work to show, but I don't even understand how to start this problem. I read back through the chapter, and there's no comparable examples. I'm sure this is ultimately simple, but it has me thrown for a loop. 
If anyone could tell me how to set this up, I'd be very grateful. 

Hint:
If $X_i$~$N(\mu,\sigma)$ represent $n$ random variates, then
$$\frac1n \sum_{i=1}^{n}X_i\text{ has distribution } N(\mu, \frac{\sigma^2}{n})$$
