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.
I'm not very happy with this problem statement. The independence of the samples is dubious, at best. Plus, is the sample 20 or 200? That will make a difference. Further, we know nothing of the size of the population, excepting that it is likely to be lass than 6 billion. It would be unwise to make conclusions that tell us a sample of 20 or 200 gives THE mean or variance.
In any case. Each experiment is Bernoulli. The entire experiment is Binomial with p = 0.8 and n = 20.
The Population data are given (or are they?):
Population Mean: 20*0.8
Population Variance 20*0.8*0.2
The Sample data are a little sketchy:
Sample Mean: ??? I don't see where we are told this.
Sample Variance This either...
If the problem MEANS (it certainly doesn't SAY so) that p = 0.80 was determined by the sample of 20, then we have an entirely different problem and I have to go with your teacher's response.
The Population data are NOT given:
Population Mean: ??
Population Variance ??
The Sample data are given:
n = 20
p = 0.80
Sample Mean: 20*0.80
Sample Variance: (20*0.8*0.20)/(20*20)
You should be able to calculate the probability in b. Is 200 sufficiently large for a normal approximation?
I checked if I mistook reportig the excercice but I wrote it Okay. It says 20 before and 200 later.
All these problems I have been given are all written in this way, and I feel so stupid, because it is hard to understand what they mean. At least, reading your words, I feel a little bit less idiot.
Why are you doing the sample variance the way you are doing? that is dividing by n^2 ? because I know sample variance= variance population/ n
Thank you so much.
I copied your teacher's correct formula: n*p*(1-p)/n^2 = p*(1-p)/n
and why is the formula like this? the book says Var sample= var pop/ n
Originally Posted by TKHunny
so n*p(1-p)/n= p*(1-p) :(