Question

The mean output of a certain type of amplifier is 496 watts with a variance of 144. If 40 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 1.2 watts? Round your answer to four decimal places.

Answer 1

µ = 496

sd = sqrt(144) = 12

n = 40

sd of sample mean = 12 / sqrt(40) = 1.8974

P(mean of the sample would differ from the population mean by less than 1.2 watts) = P(-1.2/1.8974 < Z < 1.2/1.8974)

                                    = P(-0.63 < Z < 0.63)

                                    = P(Z < 0.63) - P(Z < -0.63)

                                    = 0.7357 - 0.2643

                                    = 0.4714