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vfiekz [6]
2 years ago
6

According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, 62% of graduate

s 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.
Mathematics
1 answer:
arsen [322]2 years ago
7 0

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

<h3>What is Mean ?</h3>

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.

To know more about Probability

brainly.com/question/11234923

#SPJ1

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Step-by-step explanation:

From the question, we are informed that Allison earned $200 the first month she worked and $250 the second month she worked.

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mylen [45]

Answer:

My proof is in the explanation.  

Step-by-step explanation:

This is a two-column proof.

One column for statements and the other for the reason for that statement.

Hopefully it shows up well on your screen. Let me know if it doesn't.

Statement                                          |        Reason

1) CD is the perpendicular                   1) Given

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4) mAngleCDA=90                               4) Definition of perpendicular

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8) AC=CB                                               8) The two triangles are ............................................................................congruent so

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