Sample mean has to differ from the population mean by no more than 10% of the population standard deviation.

Question

I found this exercise in my probability book. I will type it all below:

"A research worker intends to draw a sample from a large population and estimate the population mean by the sample mean. She wishes to guarantee a probability of at least 0.99 that the sample mean differs from the population mean by no more than 10% of the population standard deviation. Use the central limit theorem to find (approximately) the smallest sample size she should use."

I thought about it for quite a bit but I think I'm missing some knowledge.

For example I think I have to go from something like this:

$$P(\mid \bar{X}-\mu \mid \leq \sigma*0.1)\geq 0.99$$

to something like this:

$$f(n) \geq \Phi^{-1}(0.99)$$

I tried a bunch of things but nothing really worked.

Answer 1

Hint: You want a 99% confidence interval for $\mu$ to have margin of error of $.1\sigma.$

The margin of error for a 99% CI is $z^*\sigma/\sqrt{n},$ where $z^*$ cuts $.5\%$ of the area from the upper tail of a standard normal distribution.

What does that say about $n$?