Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer...

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 84648464 with a mean life of 886886 minutes.

If the claim is true, in a sample of 145145 batteries, what is the probability that the mean battery life would differ from the population mean by less than 13.413.4 minutes? Round your answer to four decimal places.

Population mean is 886

Population standard deviation is σ2 = 8464 , σ = 92

Let X is life of the computer. Here, X ~ N(  µ = 886,σ = 92 )

For this study a sample of 145 computers are selected at random.

So, the sample size is n =145

probability that the mean battery life would differ from the population mean by less than 13.4

P( X < 13.4 ) = P( Z < 13.4 -886 /92 / √ 145 )

= P( Z < -872.6 / 7.64)

= P( Z < -114.21)

P( Z < -114.21) = 1 - 0 = 1

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