Question

Moving to another question will save this response. · Question 3 2 points Save Ans The mean television viewing time for Americans is 20 hours per week. Suppose a sample of 30 Americans is taken to further investigate viewing habits Assume the population standard deviation for weekly viewing time is o- 4 hours What is the probability that the sample mean will be within 1 hour of the population mean? 0.4687 0.9784 。0.9147 0.8294

Answer 1

Given, \mu = 20, \sigma = 4, n = 30 Using central limit theorem, P(\bar {x} < x) = P(z < x - \mu/ (\sigma/sqrt(n)) So, Probability that sample mean will be with in 1 hour of population mean = P(\mu - 1 < \bar {x} < \mu + 1) = ? = P(20 - 1 < \bar {x} < 20 + 1) = (19 < \bar{x} < 21) = P(\bar {x} < 21) - P(\bar {x} < 19) = P(Z < 21 - 20/(4/sqrt(30)) - P(2 < 19 - 20 /(4/sqrt(30)) = P(Z < 1.37) - P(Z < -1.37) = P(Z < 1.37) - (1 - p(Z < 1/37)) = 0.9147 - (1 - 0.9147) = 0.8294