Problem: Let $\overline X$ be the mean of a random sample of size $n$ from $N(\mu,9)$. Find $n$ so that $P(\overline X -1 < \mu < \overline X + 1) = .9$
This problem is given at the end of a chapter in my statistics textbook concerning confidence intervals for means.
Thoughts: I am not sure where to begin. First of all, is there a unique interval for which this probability holds? Secondly If i re-write the inequality dividing by $\sqrt {\sigma^2 / n }$ I get
$Pr ( \frac {-1}{\sqrt {9/n}} >\frac { \overline X - u } {\sqrt {9/n} } > \frac {1}{\sqrt {9/n}} ) = .9$ but this is as far as I get. Any insights/clarifications much appreciated.
By the way the answer is $25$.
If $W$ is a standard normal random variable, then notice that $ \mu - \bar{X}$ is i.i.d to $\dfrac{3}{\sqrt{n}}W$. Thus we need to find $n$ such that $$\mathbb{P}\left(|W| < \frac{\sqrt{n}}{3}\right) = 0.9.$$ A z table can then be used to find $$ \frac{\sqrt{n}}{3} \approx 1.645 \Rightarrow n \approx 24$$