Determining $n$ for which the confidence interval will have level of confidence >= 0.9

Question

I have come to the point where there is an $X\text{~}N(a,\frac{9}{n})$ and $\psi$ is its CDF, I have to determine the smallest n for which $\psi(a+1)-\psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!

Answer 1

$\psi(a+1)-\psi(a-1)=\Phi(1/\sqrt{(9/n)})-\Phi(-1/\sqrt{(9/n)})=2\Phi(1/\sqrt{(9/n)})-1\geq 0.9\Rightarrow \Phi(1/\sqrt{(9/n)})\geq (1+0.9)/2=0.95 \Rightarrow \sqrt{n}/3\geq \Phi^{-1}(0.95)\approx 1.64 \Rightarrow n\geq (3\times 1.64)^2=24.20\Rightarrow n\geq 25.$

Here $\Phi$ is cdf of $N(0,1).$

Answer 2

Hint. $\psi(a+1)-\psi(a-1)=\mathbf{P}(a - 1 < X < a + 1) = \mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.