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 be less than 106.35 months? Round your answer to four decimal places

Answer 1

Solution :

Given that,

mean = = 105

variance = 2 = 81

standard deviation = = 9

n = 70

= 105

= / n = 9 70 = 1.0757

P( < 106.35 )

P ( - / ) < ( 106.35 - 105 / 1.0757)

P ( z < 1.35 / 1.0757 )

P ( z < 1.25 )

= 0.8944

Probability = 0.8944