Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 105 months, with a variance of 81

.

If he is correct, what is the probability that the mean of a sample of 70

computers would differ from the population mean by less than 1.9

months? Round your answer to four decimal places.

Answer 1

Solution :

variance = 81

= 9

= / n = 6 / 70

= P[ -1.9 / 6 / 70 < ( - ) / < 1.9 / 6 / 70)]

= P(-2.65 < Z < 2.65)

= P(Z < 2.65) - P(Z < -2.65)

= 0.992

Probability = 0.9920