How big of a sample size is necessary to be sufficiently confident in predictions? 
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? 
 A: 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. 
A: 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 $n=17$.
