Question

St. Andrew’s University receives 900 applications annually from prospective students. The application forms contain a variety of information including the individual’s scholastic aptitude test (SAT) score and whether or not the individual desires on-campus

housing.

What is the probability that a simple random sample of 30 applicants will provide an estimate of the population mean SAT score that is within plus or minus 10 (within 10 points) of the actual population mean m ? Given that the population Standard Deviation is 80.

Data show 648 applicants wanting On-Campus Housing. What is the probability that sample proportion exceeds 50%, when n =30?

What is the probability that a simple random sample of 30 applicants will provide an estimate of the population proportion of applicants desiring on-campus housing that is within plus or minus .05 of the actual population proportion?

If the University can provide to no more than 45% for the on –campus housing facilities, what would be the estimated number of accepted applicants desiring on-campus housing?

Answer 1

What is the probability that a simple random sample of 30 applicants will provide an estimate of the population mean SAT score that is within plus or minus 10 (within 10 points) of the actual population mean m? Given that the population Standard Deviation is 80.

Solution:

We are given

Population mean = m

Population standard deviation = σ = 80

Sample size = n = 30

We have to find P(m – 10 < Xbar < m + 10)

P(m – 10 < Xbar < m + 10) = P(Xbar < m + 10) – P(Xbar < m – 10)

First find P(Xbar < m + 10)

Z = (m + 10 – m)/[80/sqrt(30)]

Z = 10/ 14.60593

Z = 0.684653

P(Z< 0.684653) = P(Xbar < m + 10) = 0.753219

Now, find P(Xbar < m – 10)

Z = (m – 10 – m) /[80/sqrt(30)]

Z = -10/ 14.60593

Z = -0.684653

P(Z< -0.684653) = P(Xbar < m - 10) = 0.246781

P(m – 10 < Xbar < m + 10) = P(Xbar < m + 10) – P(Xbar < m – 10)

P(m – 10 < Xbar < m + 10) = 0.753219 – 0.246781

P(m – 10 < Xbar < m + 10) = 0.506437

Required probability = 0.506437

Data show 648 applicants wanting On-Campus Housing. What is the probability that sample proportion exceeds 50% when n =30?

Solution:

We have

Population proportion = p = X/N = 648/900 = 0.72

We have to find P(P>0.50)

P(P>0.50) = 1 – P(P<0.50)

Z = (P – p)/sqrt(pq/n)

q = 1 – p = 1 – 0.72 = 0.28, n=30

Z = (0.50 - 0.72) / sqrt(0.72*0.28/30)

Z = -2.6837252

P(Z<-2.6837252) = P(P<0.50) = 0.003640346

P(P>0.50) = 1 – P(P<0.50)

P(P>0.50) = 1 – 0.003640346

P(P>0.50) = 0.996359654

Required probability = 0.996359654

What is the probability that a simple random sample of 30 applicants will provide an estimate of the population proportion of applicants desiring on-campus housing that is within plus or minus .05 of the actual population proportion?

We have to find P(p - 0.05 < P < p + 0.05)

P(p - 0.05 < P < p + 0.05) = P(P<p+0.05) – P(P<p-0.05)

Z = 0.05/ sqrt(0.72*0.28/30) = 0.609938

P(Z<0.609938) = 0.729048

Z = - 0.05/ sqrt(0.72*0.28/30) = -0.60994

P(Z<-0.60994) = 0.270952

P(p - 0.05 < P < p + 0.05) = 0.729048 - 0.270952

P(p - 0.05 < P < p + 0.05) = 0.458097

Required probability = 0.458097

If the University can provide to no more than 45% for the on-campus housing facilities, what would be the estimated number of accepted applicants desiring on-campus housing?

Estimated number = n*p

We are given n=900, p=0.45

Estimated number = 900*0.45

Estimated number = 405

[***All probabilities are obtained by using z-table or excel.]