Andy is a restaurant owner: He believes 72% of his customers are…

Question

Andy is a restaurant owner: He believes 72% of his customers are satisfied with the food quality of his restaurant: From a random sample of 82 customers, what are the following probabilities? (Round to 2 decimal places for all z-values and round all other answers to four decimal places, if needed.)

a) What is the probability that less than 59 customers are satisfied with the food quality?

b)What is the probability that at least 59 customers are satisfied with the food quality?

c) What is the probability that the sample proportion of customers who are satisfied with the food quality is between 69% and 77%?

Answer 1

proportion 59/82 = 0.72, so sample proportion phat is equal to population proportion p, so 50% would be below or probability of 0.5 (a) and 50% would be above for probability of 0.5 (b).

for c) use z = (phat - p)/sqrt(p * (1-p)/n) = (phat - 0.72)/0.05)

P(0.69 < P < 0.77) = P((0.69 - 0.72)/0.05 < z < (0.77 - 0.72)/0.05) = P(-0.6 < z < 1) = P(z < 1) - P(z < -0.6) = 0.8413 - 0.2743 = 0.567