Question

A pescatarian is a person who eats fish and seafood but no other animal. An event planner does some research and finds that approximately​ 4% of the people in the area where a large event is to be held are pescatarian. She expects 300 guests at the event and the chef is deciding upon the menu. Treat the guests as a simple random sample from the local population of about​ 150,000.

If the event planner decides to make​ 5% of the meals​ (or 15)​ pescatarian, what is the probability that there will not be enough pescatarian meals​ -- that is what is the probability that more than​ 5% of the guests will be​ pescatarians? Assume the pescatarians are independent and there are no families of pescatarians.

Answer 1

We use normal approximation to binomial

Mean = n*p = 300*0.04 = 12

Standard dev = (n*p*q)^0.5 = (300*0.04*0.96)^0.5 = 3.394

P(X>=16)