Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard deviation of 86 minutes with a mean life of 505 minutes. If the claim is true, in a sample of 120 batteries, what is the probability that the mean battery life would differ from the population mean by greater than 16.6 minutes? Round your answer to four decimal places.

Answer 1

Solution :

Given that ,

mean = = 505

standard deviation = = 86

n = 120

=   = 505

= / n = 86 / 120 = 7.85

505 ± 16.6 = 488.4, 521.6

P(488.4 < < 521.6)  

= 1 - P[(488.4 - 505) / 7.85 < ( - ) / < (521.6 - 505) / 7.85)]

= 1 - P( -2.11 < Z < 2.11 )

= 1 - P(Z < 2.11) - P(Z < - 2.11)

Using z table,  

= 1 - P( 0.9826 - 0.0174)   

= 1 - 0.9652

= 0.0348