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Answer:The probability that the sample mean would be greater than 133.5 is 0.9926Explanation:Given:Mean = 137Standard deviation = 8Sample = 31To find the probability that the sample mean would be greater than 133.5, we have:[tex]\begin{gathered} z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}} \\ \\ =\frac{133.5-137}{\frac{8}{\sqrt{31}}}=-2.4359 \\ \\ P(X>133.5)=1-P(z<-2.4359) \\ =1-0.0074274 \\ \approx0.9926 \end{gathered}[/tex]