Statistics question: What is the approximate probability that more than 4% of the guests are pescatarian and that he will not have enough pescatarian meals?

Question

An event planner does research and finds that approximately 2.75% of the people in the area where a large event is being held are pescatarian. Treat the 250 guests expected at the event as a simple random sample from the local population of about 150,000. Assume the sampling distribution is approximately Normal.

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?

Answer 1

population proportion: p = 0.0275

 

assumed sample proportion: pˆ = 0.04

 

standard deviation: σ = √(p(1-p)/n) = √(0.0275(1-0.0275)/250)

 

z-score: z = (pˆ - p)/σ = (0.04 - 0.0275)/(that thing)

 

P(sample proportion > 0.04) = P(Z > that z-score) = the appropriate entry in the appropriate standard normal table