Question

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 95 months with a standard deviation of 5 months. If the claim is true, what is the probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

Answer 1

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

Step(ii):-

The Empirical rule

         $Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

        $Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }$

       Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

    P(X⁻ ≥ 96.3) = P(Z≥2.383)

                      =  1- P( Z<2.383)

                      =  1-( 0.5 -+A(2.38))

                      = 0.5 - A(2.38)

                     = 0.5 -0.4913

                    = 0.0087

Final answer:-

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087