Given the following parameter:
[tex]\begin{gathered} \mu=93 \\ \sigma=9 \\ \bar{x}=91.4 \\ n=66 \end{gathered}[/tex]
Using z-score formula
[tex]z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Substitute the parameter provided in the formula above
[tex]z=\frac{91.4-93}{\frac{9}{\sqrt{66}}}[/tex][tex]z=-1.4443[/tex]
The probability that the mean monitor life will be greater than 91.4 is given as
[tex]\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}[/tex]
Hence, the probability that the mean monitor life will be greater than 91.4 months is 0.9257
