Question

An air conditioning contractor has kept detailed records on the lifetimes of a random sample of 300 compressors. She plans to use the sample mean​ lifetime, x overbar​, of those 300 compressors as her estimate for the population mean​ lifetime, mu​, of all such compressors. If the lifetimes of this brand of compressor have a standard deviation of 50 ​months, what is the probability that the​ contractor's estimate will be within 5 months of the true​ mean?

Answer 1

Here we have

$n=300,sigma=50$

The z-score for $ar{x}=mu-5$ is

$z=rac{ar{x}-mu}{sigma/sqrt{n}}=rac{mu-5-mu}{50/sqrt{300}}=-1.73$

The z-score for $ar{x}=mu+5$ is

The probability that the​ contractor's estimate will be within 5 months of the true​ mean is

P(μ-5 < < μ + 5)-P(-1.73 < z < 1.73) = 0.9164