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.

Step-by-step explanation:

1) Previous concepts

Normal 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:

And let represent the sample mean, the distribution for the sample mean is given by:

On this case  

2) Calculate the probability

We want this probability:

The best way to solve this problem is using the normal standard distribution and the z score given by:

If we apply this formula to our probability we got this:

 and rounded would be 0.928

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %