Question

3. The average price of a meal in restaurants at one city is 50 AED with a standard deviation of 10 AED. (8 marks) a. What is the probability of selecting a sample of 100 restaurants from this city and finding the sample mean is at least Dh 52 b. What is the probability that the mean for a sample of 100 restaurants is between Dh 52 and Dh 54 c. What is the standard error?

Answer 1

The price of the mean in a restaurant has

u= 50,0 = 10

Let X be the random variable for the meal price in the restaurant

a) Here sample size is n = 100, which is large enough to use central limit theorem, hence

The probability of selecting a sample of 100 restaurants from this city and finding the sample meal at least 52 is

P(X > 52) = P(Z > 52 – 50 10 ) = P(Z > 2) 100

P(Z > 2) = 1- P(Z < 2) = 1-0.9772 = 0.0228

b) The probability that the sample mean is between 52 and 58 is

54 - 50 P(52 < X < 54) = 52 – 50 10 100 a <7< 10 100

P(52 < X <54) = P(2<z <4)

P(2<Z < 4) = P(Z < 4) - P(Z <2) = 1-0.9772 = 0.0228

c) Standard error =

SE = lo = 10 100 -