The average GRE score at the University of Pennsylvania for the incoming class of 2016-2017 was 311. Assume that the standard deviation was 13. If you select a random sample of 40 students, what is the probability that the sample mean will be greater than 308? Round your answer to three decimal places.

Answer:[tex]P(\bar X>308)=1-0.0721=0.928[/tex]Step-by-step explanation:1) Previous conceptsNormal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  Let X the random variable that represent the GRE score at the University of Pennsylvania for the incoming class of 2016-2017, and for this case we know the distribution for X is given by:[tex]X \sim N(\mu=311,\sigma=13)[/tex]  And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]On this case  [tex]\bar X \sim N(311,\frac{13}{\sqrt{40}})[/tex]2) Calculate the probabilityWe want this probability:[tex]P(\bar X>308)=1-P(\bar X<308)[/tex]The best way to solve this problem is using the normal standard distribution and the z score given by:[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]If we apply this formula to our probability we got this:[tex]P(\bar X >308)=1-P(Z<\frac{308-311}{\frac{13}{\sqrt{40}}})=1-P(Z<-1.46)[/tex][tex]P(\bar X>308)=1-0.0721=0.9279[/tex]  and rounded would be 0.928