Question 479410
p=15/80=.1875, q=.8125, 
{{{sigma=sqrt((.1875*.8125)/80)=.0436}}}
(a,b)=.1875 +-(2.576*.0436) [2.576 is the area under the normal curve for the 99% confidence level.]
=.1875 +-.1123
=(.2998, .0752)  the 99% confidence interval of the true proportion of customers who order only cheese pizza.