Respuesta :
Answer:
Step-by-step explanation:
Given that the Office of Student Services at UNC would like to estimate the proportion of UNC's 34,000 students who are foreign students.
In their random sample of 50 students, 4 are foreign students.
Sample proportion = [tex]\frac{4}{50} =0.08[/tex]
the proportion of all UNC students that are foreign students is 0.061.
i.e. Population proportion= 0.061
Mean for 50 size sample = sample proportion = 0.08
Std deviation of sample proportion = [tex]\sqrt{0.061(1-0.061)/50} \\=0.0338[/tex]
Hence for the ampling distribution of the sample proportion for a sample of size 50, mean =0.08 and std dev = 0.0338
(a) The Mean of the students will be 3.05
(b) The standard deviation will be 1.69.
What will be the Mean and the standard deviation?
For every student, the possible outcomes are only two. Either they are foreign students or they are not.
Since this concept is based on Binomial probability distribution
So from Binomial probability distribution
The probability of x successes on n repeated trials, with p probability.
In this problem, we have a sample of 50 students, so [tex]n=50[/tex] The proportion of all UNC students that are foreign students is 0.061 [tex]P=0.061[/tex],
The mean is given by the following formula:
[tex]E(x)=n\times p=50\times0.061=3.25[/tex]
The standard deviation is given by the following formula
[tex]\sigma =\sqrt{n\times p\times(1-p)}[/tex]
[tex]\sigma =\sqrt{50\times0.061\times (1-0.061 )[/tex]
[tex]\sigma =1.69[/tex]
Thus
(a) The Mean of the students will be 3.05
(b) The standard deviation will be 1.69.
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