Question

According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, 62% of graduates from public universities had student loans.
We randomly select 50 students at a time.
What is the mean of the distribution of sampling proportions? Do not round, and enter your answer as a proportion (decimal number) not a percentage.

Answer 1

The mean of the distribution sample is 0.62 and the standard error of the distribution sample is 0.069

What is Mean ?

Mean is the study of central tendency of a data set. The average value of the data points is the mean.

It is given that

The probability that the students have the student loan is  = 62% = 0.62

The probability that the students do not have student loan is

   1 - p

= 1 - 0.62

= 0.38

Total number of students is

n = 50

1) $\rm \mu \;\hat p$ = p = 0.62

2) $\rm \sigma\hat p$  = $\rm \sqrt {p ( 1 - p ) / n}$

=  $\rm \sqrt {(0.62 * 0.38) / 50}$

= 0.069

Therefore the mean of the distribution sample is 0.62 and the standard error of the distribution sample is 0.069.

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