The Question: Let $Y_{1}, Y_{2}, ..., Y_{n}$ be random samples from a normal distribution where the mean is 2 and the variance is 4. How large must n be in order that $P(1.9 \leq \overline{Y} \leq 2.1) \geq 0.99$?
The Attempt: We are computing for the sample mean of our random sample that was given in the problem. By definition, $z= \frac{\overline{Y} - \mu}{\frac{\sigma}{\sqrt{n}}}$. So rewrite the equation so we can transform the data to make the mean 0 and the standard deviation 1. If I do that, I get,
$P(\frac{1.9 - 2}{\frac{2}{\sqrt{n}}} \leq z \leq \frac{2.1-2}{\frac{2}{\sqrt{n}}}) \geq 0.99.$ This means that
$P(\frac{-0.1}{\frac{2}{\sqrt{n}}} \leq z \leq \frac{0.1}{\frac{2}{\sqrt{n}}}) \geq 0.99.$ This also means that
$P(-0.05\sqrt{n} \leq z \leq 0.05\sqrt{n}) \geq 0.99.$ I am not really sure what to do after this step. I am trying to use the definition of the normal distribution, however that was too difficult to do.
Do you guys know what to do after this step? Thank you for all of your help.
You just have to set $$0.05\sqrt{n}\geq\Phi^{-1}(0.995)$$ and solve for $n$