Andy is a restaurant owner…

Question

Andy is a restaurant owner: He believes 85% of his customers are satisfied with the food quality of his restaurant: From a random sample of 96 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 82 customers are satisfied with the food quality?

b)What is the probability that at least 82 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 82% and 90%?

Answer 1

Can use normal approximation to the binomial:

mean is n * p = 96 * 0.85 = 81.6

standard error (SE) = sqrt(n * p * (1-p)) = 3.5

x = number of customers

standardized variable z = (x - mean)/SE = (x - 81.6)/3.5

P(x < 82) = P(x < 81.5) (applying continuity correction)

Replacing limit with standardized variable:

P(x < 81.5) = P(z < (81.5 - 81.6)/3.5) = P(z < -0.03) = 0.488

P(x >= 82) = 1 - P(x < 82) = 1 - 0.488 = 0.512

P(82 < x < 90) = P(82.5 < x < 89.5) (applying continuity correction)

Replacing limits with standardized variable:

P(82.5 < x < 89.5)= =P((82.5 - 81.6)/3.5 < z < (89.5 - 81.6)/3.5) = P(0.26 < z < 2.26) = P(z < 2.26) - P(z < 0.26) = 0.9881 - 0.6026 = 0.3855