Question

Studies have shown that the average price for fish fry among all restaurants in Wisconsin is normally distributed with mean $10.20 and a standard deviation of 3.

(a) If we take a random sample of 50 Wisconsin restaurants, what is the probability that the average price of fish fry in this sample is less than $9.50? Show your work. (b) If we take a random sample of 35 Wisconsin restaurants, what is the probability that the average price of fish fry in this sample is between $9.20 and $11.00? Show your work. (c) What is the probability that Lake Side Bistro’s fish fry is less than $13.50? Show your work.

Answer 1

Mean, = $10.20

Standard deviation, = $3

(a) n = 50

Standard error = = 3/√50 = 0.4243

Probability that the average price is less than $9.50

= P( < 9.50) = P{Z < (9.50 - 10.20)/0.4243}

= P(Z < -1.65)

= 0.0495

(b) n = 35

Standard error = 3/√35 = 0.5071

The required probability = P(9.20 < < 11.00)

= P{(9.20 - 10.20)/0.5071 < Z < (11 - 10.20)/0.5071}

= P(-1.972 < Z < 1.5776)

= 0.9184

(c) The required probability = P(x < 13.50)

= P{Z < (13.50 - 10.20)/3}

= P(Z < 1.1)

= 0.8643