Data from the U.S. Department of Education indicates that 47% of business graduate students from private universities had student loans. Suppose you randomly survey a sample of graduate business students from private universities.

Answers

The mean of this distribution is 99.

The standard deviation of this sampling distribution is 0.03399.

We are asked to find the mean and the standard deviation for the sampling distribution for this sample with a sample size, n = 215.

The sampling distribution of x has a mean μₓ = μ and standard deviation σₓ = σ/√n.

The true proportion of the sample is given to be p = 46% = 0.46.

Thus, the mean μ of the population = p*n = 0.46*215 = 98.8 ≈ 99.

Thus, the mean of the sample μₓ = μ = 99.

The standard deviation of the population, σ = √((p)(1 – p)) = √((0.46)(1 – 0.46)) = √0.46*.54 = √0.2484 = 0.498397431.

Thus, the standard deviation of the sample σₓ = σ/√n = 0.498397431/√215 = 0.03399.

Thus, The mean of this distribution is 99.

The standard deviation of this sampling distribution is 0.03399.

The provided question is incomplete. The complete question is:

“Data from the U.S. Department of Education indicates that 46% of business graduate students from private universities had student loans. Suppose you randomly survey a sample of graduate business students from private universities. Consider the sampling distribution (sample size n = 215) for the proportion of these students who have loans.

What is the mean of this distribution?

What is the standard deviation of this sampling distribution (i.e., the standard error)?”

meanof thisdistributionis99.standard deviationof thissampling distributionis0.03399.meanand thestandard deviationfor thesampling distributionfor this sample with a sample size, n = 215.sampling distributionof x has ameanμₓ = μ andstandard deviationσₓ = σ/√n.proportionof thesampleis given to be p = 46% = 0.46.meanμ of thepopulation= p*n = 0.46*215 = 98.8 ≈ 99.meanof thesampleμₓ = μ = 99.standard deviationof thepopulation, σ = √((p)(1 – p)) = √((0.46)(1 – 0.46)) = √0.46*.54 = √0.2484 = 0.498397431.standard deviationof thesampleσₓ = σ/√n = 0.498397431/√215 = 0.03399.meanof thisdistributionis99.standard deviationof thissampling distributionis0.03399.sampling distributionat