A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be $90\%$ confident that her estimate is within $2$ ounces of the true mean? Assume that $\sigma =4.9\ $ounces and that birth weights are normally distributed.
Why did they get 17?
On
The sample mean $\bar{X}$ has mean $\mu$ and standard deviation $\frac{4.9}{\sqrt{n}}$.
We want $\Pr\left(|\bar{X}-\mu|\le 2\right)\ge 0.9$.
Note that $\frac{\bar{X}-\mu}{4.9/\sqrt{n}}$ is standard normal.
So we want $$\Pr\left(|Z|\le \frac{0.2}{4.9/\sqrt{n}}\right)\le 0.9.$$ The table for the standard normal now says that $\Pr(|Z|\le 1.645$ is about $0.9$. Thus we want $$\frac{2}{4.9/\sqrt{n}}\gt 1.645.$$ Thus we want $\sqrt{n}\gt 4.03025$. Square. We want $n$ to be bigger than $16$, and $17$ will do the job.
Consider the distribution of sample mean of birth weight from $n$ samples. This distribution is normal and has a same mean as birth weight, and has a standard deviation of $\sigma\over\sqrt n$.
We would like to find minimum $n$ such that $$P(\text{true mean}-2<\text{sample mean of birth weight}<\text{true mean}+2)\ge90\%$$ By normalising to standard normal distribution, $$\begin{align} P\left(\frac{-2}{\sigma/\sqrt n}<Z<\frac{2}{\sigma/\sqrt n}\right)\ge& 90\%\\ \frac{2}{\sigma/\sqrt n}>&1.64\\ n>&16.144 \end{align}$$ Therefore the answer takes $n=17$.