Question

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 8989 months with a standard deviation of 55 months.

If the claim is true, what is the probability that the mean monitor life would differ from the population mean by less than 1.21.2 months in a sample of 123123 monitors? Round your answer to four decimal places.

Answer 1

WE are given

µ = 89

σ = 5

n = 123

We have to find P(89 – 1.2 < Xbar < 89 + 1.2)

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = -1.2/(5/sqrt(123))

Z = -2.66172876

P(Z<-2.66172876) = P(Xbar<89 – 1.2) = 0.003887025

(by using z-table)

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = 1.2/(5/sqrt(123))

Z =2.66172876

P(Z<2.66172876) = P(Xbar<89 + 1.2) = 0.996112975

(by using z-table)

P(89 – 1.2 < Xbar < 89 + 1.2) = 0.996112975 - 0.003887025 = 0.99222595

Required probability = 0.9922