Question

A quality control expert at bobcat computers want to test their new laptop monitors. The production manager claims the monitors have a mean life of 89 months with a population standard deviation of 5 months. If the claim is true, what is the probability that the mean monitor life would be greater than 87.7 months in a sample of 123 monitors? Round answer to four decimal places.

A).0001

B).5948

C).0039

D).9961

E).5000

Answer 1

Solution :

Given that ,

mean = = 89

standard deviation = = 5

n = 123

= = 89 and

= / n = 5 / 123 = 0.4508

P( > 87.7) = 1 - P( < 87.7)

= 1 - P(( - ) / < (87.7 - 89) / 0.4508)

= 1 - P(z < -2.88)

= 0.9961 Using standard normal table.

Probability = 0.9961

Option D is correct.