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​ 2.75% of the people in the area where a large event is to be held are pescatarian. Treat the 250 guests expected at the event as a simple random sample from the local population of about​ 150,000. Suppose the event planner assumes that​ 4% of the guests will be pescatarian so he orders 10 pescatarian meals. What is the approximate probability that more than​ 4% of the guests are pescatarian and that he will not have enough pescatarian​ meals? Round to three decimal places.

Answer 1

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.0275
sample size       =n=	250
std error of proportion=σp=√(p*(1-p)/n)=	0.0103

approximate probability that more than​ 4% of the guests are pescatarian and that he will not have enough pescatarian​ meals :

probability =P(X>0.04)=P(Z>(0.04-0.0275)/0.01)=P(Z>1.21)=1-P(Z<1.21)=1-0.8869=0.113