Question

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 82 months with a standard deviation of 7 months.

If the claim is true, what is the probability that the mean monitor life would differ from the population mean by less than 1.8 months in a sample of 71 monitors? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

= / n = 7 / 71 = 0.8307

= P[(-1.8) / 0.8307< ( - ) / < (1.8) / 0.8307)]

= P(-2.17 < Z < 2.17)

= P(Z < 2.17) - P(Z < -2.17)

= 0.985 - 0.015

= 0.9700

Probability = 0.9700