Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 97 months, with a variance of 81.

If he is correct, what is the probability that the mean of a sample of 50 computers would differ from the population mean by more than 1.35 months? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 97

standard deviation = = 9

n = 50

= = 97

= / n = 9 / 50 = 1.2728

P(95.65 < < 98.35) = 1 - (P((95.65 - 97) /1.2728 <( - ) / < (98.35 - 97) / 1.2728)))

= 1 - (P(-1.06 < Z < 1.06))

= 1 - ( P(Z < 1.06) - P(Z < -1.06) ) Using standard normal table,  

= 1 - 0.7109

= 0.2891

Probability = 0.2891