Question

A quality control expert at glow tech computers wants to test their new monitors . The production manager claims that have a mean life of 93 months with the standard deviation of nine months. If the claim is true what is the probability that the mean monitor life will be greater than 91.4 months and a sample of 66 monitors? Round your answers to four decimal places

Answer 1

Given the following parameter:

`\begin{gathered} \mu=93 \\ \sigma=9 \\ \bar{x}=91.4 \\ n=66 \end{gathered}`

Using z-score formula

`z=\frac{\bar{x}-\mu}{(\sigma)/(√(n))}`

Substitute the parameter provided in the formula above

`z=(91.4-93)/((9)/(√(66)))`
`z=-1.4443`

The probability that the mean monitor life will be greater than 91.4 is given as

`\begin{gathered} P(z>-1.4443)=P(0\leq z)+P(0-1.4443)=0.5+0.4257 \\ P(z>-1.4443)=0.9257 \end{gathered}`

Hence, the probability that the mean monitor life will be greater than 91.4 months is 0.9257