Answer
$16.26\lt\mu\lt20.16$
Work Step by Step
Given $\bar X=18.21,\sigma=5.92,n=50$, with a 98%
confidence, the critical z-value is given by $z_{\alpha/2}=2.326$
and the error is $E=2.326\times\frac{5.92}{\sqrt {50}}=1.95$
Thus the interval for the mean bill for all cars from the drive-thru with 98%
confidence is $\bar X-E\lt\mu\lt\bar X+E$ which gives
$16.26\lt\mu\lt20.16$