Question

the average student spends about 17 hours each week preparing for classes; preparation for classes include homework, reading and any other assignment. Assume the standard deviation if time spent preparing for class is 4 hours. if you select a random sample of 16 students. what is the probability that the mean time spent preparing for classes is at least 16 hours per week?

Answer 1

P(X^- greaterthanorequalto 16) Normal Distribution, mu = 17, sigma = 4, n = 16 Since this is continuous distribution, P(X6- greaterthanorequalto 16) = P(X^- > 16) we convert this to standard normal using z = X^- - mu_x^-/sigma_X^- = X^- - mu/sigma/squareroot n z = 16 - (17)/4 squareroot 16 = -1.00 P(Z > - 1.00) = Area to the right of -1.00 P(X^- > 16) = P(Z > - 1.00) = 1 - P(Z < - 1.00) = 1 - 0.1587 = 0.8413 P(Z < - 1): in a z - table having area to the left of z, locate -1 in the left most column. Move across the row to the right under column 0.00 and get value 0.1587 Using technology, answer is 0.841344746068543 P(X^- > 16) = 0.8413 = 84.13%