We conduct a simulation to mimic randomly sampling from a population with of college graduates. In the population 62% had student loans. Each sample has 50 graduates in it. What will be the mean of the distribution of sample proportions? Enter a number in decimal form. For example, you would enter 0.50, not 50 or 50%.

The correct answer is 0.069

Explanation:

Solution

Let recall that,

In the population, the number of student that took loans where = 62%

Each samples has graduates of = 50

The next step is to enter a number in decimal form.

Given that,

p = 62% = 0.62

1 - p = 1 - 0.62 = 0.38

Thus,

n = 50

The mean = μ p = p = 0.62

Then,

The standard deviation = б p =  √{p( 1 - p ) / n]

= √[(0.62 * 0.38 ) / 50 ] = 0.069

Therefore, the number form is 0.069

answer; importance to people; or acceptance;

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