Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 109109 months, with a variance of 100100.

If he is correct, what is the probability that the mean of a sample of 7171 computers would differ from the population mean by less than 3.373.37 months? Round your answer to four decimal places.

Answer 1

Solution :

Given that ,

mean = = 109

standard deviation = = 100 = 10

n = 71

=   = 109

= / n = 10 / 71 = 1.187

109 ± 3.37 = 105.63, 112.37

P(105.63 < < 112.37)  

= P[(105.63 -109) / 1.187 < ( - ) / < (112.37 - 109) / 1.187)]

= P( -2.84 < Z < 2.84 )

= P(Z < 2.84) - P(Z < -2.84)

Using z table,  

= 0.9977 - 0.0023

= 0.9954