Let $Y_{1},Y_{2},...,Y_{n}$ be a random sample from a normal distribution where the mean is $2$ and the variance is $4$. How large must $n$ be in order that $P(1.9 \leq \bar{Y}\leq 2.1) >= 0.99$.
Solution: Let $P(1.9 \leq \bar{Y}\leq 2.1) = P(\frac{1.9 - \mu}{\sigma/\sqrt{n}} \leq \bar{Y}\leq \frac{2.1 - \mu}{\sigma/\sqrt{n}}) = P(\frac{1.9 - 2}{2/\sqrt{n}} \leq Z\leq \frac{2.1 - 2}{2/\sqrt{n}}) >= 0.99 $ .
Then we look for a probability that is equal to about $0.99$ in the normal table. Then $\frac{2.1 - 2}{2/\sqrt{n}} = 2.3$.
So solving for n we have $n = 2116$.
Can anyone please verify this?, I don't know if I have to take in consideration $\frac{1.9 - 2}{2/\sqrt{n}}$, the other part of the inequality. In addition, there are so many approximations that are almost equal to 0.99. So I am not sure if $2.3$ works.
Any feedback/help is welcome. Thank you.
For a two-sided confidence interval $[1.9,2.1]$ with coverage probability $0.99$, the corresponding quantile is $0.995$, thus the correct $z$-score is $2.57583$, not $2.3$, and the desired $n$ satisfies $$\frac{2.1 - 2}{2/\sqrt{n}} = 2.57583.$$