Question

A recent census revealed that the population mean and population standard deviation for rent paid for studio apartments a city was $1,090 and 30, respectively. This census was way too expensive, so the city is now exploring what it could learn from a random sample

a) What is the probability that a sample of 500 would produce a sample mean within $1 of the population mean?

b) What is the probability that a sample of 5,000 would produce a sample mean within $1 of the population mean?

Answer 1

Solution: Mean = u = 1090 Standard deviation = 0 = 30 Part-A: Sample size = n = 500 Probability that the sample mean will be within $1 of the population mean = P(u - 1<x<u + 1) Z-score for (x = ll – 1): - u Z =- o/vn | |-1-4 =z= 30/500 =z= -0.75 Z-score for (x = 1 + 1): X-u z= /vn ul +1-u 30/500 => z=0.75 > z=? Hence, P(u - 1<x<u + 1) = P(-0.75 <z < 0.75) = P(Z < 0.75) - P(Z < -0.75) From normal distribution table; P(Z < 0.75) - P(Z < -0.75) = 0.7734 -0.2266 = 0.5468 Hence, probability that the sample mean will be within $1 of the population mean = 0.5468

Part-B: Sample size = n = 5000 Probability that the sample mean will be within $1 of the population mean = P(u - 1<x<u+1) Z-score for (x = [1 – 1): X - u z= o/vn - 1-u 30/V5000 = -2.36 > INN = Z-score for (x = 1 + 1): X-u Z = 0/vn A+1-4 Z = 30/5000 > z= 2.36 Hence, P(u - 1<x<u + 1) = P(-2.36 <z < 2.36) = P(Z < 2.36) - P(Z < -2.36) From normal distribution table; P(Z < 2.36) - P(Z < -2.36) = 0.9909 - 0.0091 = 0.9818 Hence, probability that the sample mean will be within $1 of the population mean = 0.9818