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 differ from the population mean by greater than 13.4 minutes? Round your answer to four decimal places.

Answer 1

The population statistics are:

u = 886
σ2 = 8464 OR  σ = 92

Now, we have a sample of 145 batteries, the sample statistics will be approximately equal to:

x-bar = 886

s = σ/√n
= 92/√145
s = 7.6402

We need to find: P(X > 886 + 13.4) + P(X < 886 - 13.4)

= P(X > 899.4) + P(X < 872.6)

= P(z > (899.4 - 886)/7.6402) + P(X < (872.6-886)/7.6402)

= P(z > 1.75) + P(z < -1.75)

= 1 - P(z < 1.75) + P(z < -1.75)

From the z-table,

= 1 - 0.9599 + 0.0401

Final answer = 0.0802