Question

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.

1. 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.