Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 120 months, with a standard deviation of 10 months.

If he is correct, what is the probability that the mean of a sample of 90 computers would be greater than 117.13 months? Round your answer to four decimal places

Answer 1

Solution :

Given that ,

mean = = 120

standard deviation = = 10

= / n = 10 / 90 = 1.0541

P( > 117.13) = 1 - P( < 117.13)

= 1 - P[( - ) / < (117.13 - 120) / 1.0541]

= 1 - P(z < -2.72)

= 1 - 0.0033

= 0.9967

probability = 0.9967