when a student attempt to log on to a computer time-sharing system, either all ports could be busy in which case the sudnet will fail to obtain access, or else there will be at least one port free, in which case the student wil be succesful in accessing the system. Suppose that 80% of studnets are successful in accessing the system, and consider observing a simple random sample (bernoulli sample) of n=20 students.

A)Calculate the mean and the variance of the sample proportion (sampling distribution mean and sampling distibution variance)

B)What is the probability that in a random sample of 200 studnets at least 85% would access the system?

My teacher said at point A) that the expected value is p=0.8 and that the variance is n*p*(1-p)/n^2 = p*(1-p)/n= 0.008

I do not understand why she did the way she did. Can you help please? Thank you.

Because what I do not understand is which reasoning a shall apply. Let me explain.

Reasoning 1) I have a bernoulli with p= 0.8 (is this correct or I am not getting what that 80% is?). I repeat the bernoulli experiment n times, where n =20. So I shall MULTIPLY p* n to get the expected value, and the variance is p*(1-p)*n

Reasoning 2) 80% is the data that we get from the population. So usually we see that in the all population 80% of times students have access. But then we want the mean of the SAMPLE so for inferential statistics the mean of the sample is equal to the mean of the population and var of sample is var_popultaion over n. But then why is the excercice talking about bernoulli and that 80%? Because I mean, Bernouilli cannot be the population.