Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 8464 with a mean life of 886 minutes. If the claim is true, in a sample of 145 batteries, what is the probability that the mean battery life would be greater than 904.8 minutes? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 886

variance ^{2 =} 8464

standard deviation = = 92

n=145

= =886

= / n = 92 / 145 = 7.64

P( > 904.8) =

= 1 - P[( - ) / < (904.8-886) /7.64 ]

= 1 - P(z < 2.46)

Using z table

= 1 - 0.9931

= 0.0069

probability= 0.0069