You are planning to collect a (simple random) sample to estimate the mean of a non-negative random variable. It is known that the population coefficient of variation is $1.2$. Use the central limit theorem to determine (approximately) the smallest sample size needed so that you will be at least $98\%$ confident that the sample mean is within $5\%$ of the population mean.
Just wondering if my steps below of right or not. First, I carry out this step to get the following:
$P\left(0.95\mu \le \bar{x}\le 1.05\mu \right)\:=\:1\:-\:2P\left(Z\le -0.041667\sqrt{n}\right)$.
After which, I solve for $n$ by this:
$1\:-\:2P\left(Z\le -0.041667\sqrt{n}\right)\ge 0.95$.
Am I doing it the right way?
Also, I would like to know if it is possible to just use the formula $n=\dfrac{\left(c^2\cdot z^2\right)}{ME^2}$, with $c=1.2$, $z=2.325$, and $ME=0.05$?
Thank you so much for your time and help!