Sampling distributions are probability distributions.

If a normal curve is a good fit for a sampling distribution, then we can apply the Empirical Rule or use Z-scores and the standard normal distribution to determine probabilities, as we have done previously.

EXAMPLE

A 2007 report by the Pew Research Center stated that about 10% of the 3.1 million 18- to 24-year-olds in the United States were enrolled in a community college.

If we randomly selected 100 U.S. adults in this age group, what is the probability that less than 7% are enrolled in a community college.

Answer: 16%

How do we know?

First, we need to determine that a normal model is a good fit for the distribution of sample proportions.

Check normality conditions: Are $LaTeX: n\cdot p$ and $LaTeX: n\left(1-p\right)$ both equal to or greater than 10? YES. Therefore, we can use a normal curve to represent the probability distribution.

Here is our work: n is the sample size (100 students) and p is the population proportion. In this case, p=0.10 because the problem states that 10% of the population of 18- to 24-year-olds in the U.S. were enrolled in a community college.

$LaTeX: np\:=\:\left(100\right)\left(0.10\right)=10$ and $LaTeX: n\left(1-p\right)\:=\:\left(100\right)\left(1-0.10\right)\:=\:90$

Second, sketch the normal probability curve and mark the mean and several standard errors above and below the mean.

What is the standard error? We need to calculate it. (Calculator Keystrokes)
$s t a n d a r d e r r o r = \sqrt{\frac{p (1 - p)}{n}} = \sqrt{\frac{0.10 (1 - 0.10)}{100}} = 0.03$
Mark the several standard errors above and below the mean. For example, above the mean: 0.10+0.03=0.13, 0.10+2(0.03)=0.16, etc. Similarly below the mean: 0.10-0.03 = 0.07, 0.10-2(0.03)=0.04, etc.

$bell curve; norma distribution of sample means; shaded right tail area = 0.16$

Third, use the Empirical Rule because the sample proportion of 0.07 is one standard error below the mean:

The central area for values within one standard error is 0.68.
The area in the two tails is 1 – 0.68 = 0.32.
The area in the left tail is half of 0.32 = 0.16 $End of example$

What if we cannot use the Empirical Rule? Then we find a z-score and use the Standard Normal Curve as we did before.

What is the z-score? As before, the z-score is a standardized unit; it is the number of standard errors the sample proportion is from the mean. We can say this another way: the z-score is the error in the statistic divided by the standard error.

For sample proportions, we have the following formulas:

Z = $\frac{statistic - parameter}{standard error} = \frac{\hat{p} - p}{standard error}$

where the standard error is $\sqrt{\frac{p (1 - p)}{n}}$

We can also write this as one formula:

Z = $\frac{\hat{p} - p}{\sqrt{\frac{p (1 - p)}{n}}}$

As before, a positive z-score indicates that the sample proportion is larger than the parameter. A negative z-score indicates that the sample proportion is smaller than the parameter.

EXAMPLE Using the z-score and the Standard Normal Curve to find a probability

Let’s return to the example of community college enrollment: a 2007 report by the Pew Research Center stated that about 10% of the 3.1 million 18- to 24-year-olds in the United States were enrolled in a community college.

Let’s again suppose we randomly selected 100 young adults in this age group.

What is the probability that a random sample of this size has 15% or more enrolled in a community college?

To answer this question, we first determine if a normal model is a good fit for the sampling distribution.

Check normality conditions:
Yes, the conditions are met (duh!, we already did this above.) Therefore, we can use a normal model, which allows us to use a z-score to find the probability.
But before we move on, let's look at the normality conditions from a different point of view:
- When 10% of the population is enrolled in a community college, we expect 10% of the sample of 100 students to be enrolled in a community college. This expected count is $n p = 100 (0.10) = 10$ .
- Similarly, when 10% of the population is enrolled in a community college, 90% of the population is not enrolled; therefore, we expect 90% of the 100 is not enrolled, $n (1 - p) = 100 (0.90) = 90$ .
- In other words, another way to describe the normality conditions is to say: the count of expected successes (np) and failures (n(1-p)) are both at least 10.
  Find the z-score:
We already found the standard error previously. It is 0.03.

$Z = \frac{s t a t i s t i c - p a r a m e t e r}{s t a n d a r d e r r o r} = \frac{0.15 - 0.10}{0.03} \approx 1.67$

Find the probability using the standard normal model using the Normal Distribution Calculator

Here are our results: We want the probability that the sample proportion is greater than 15%; therefore, we want the probability that the z-score is greater than 1.67. The probability is about 0.0475.

$Standard normal model$

Conclusion: If it is true that 10% of the population of 18- to 24-year-olds are enrolled at a community college, then it is unusual to see a random sample of 100 with 15% or more enrolled. The probability is about 0.0475. $End of example$

Comment

This probability is a conditional probability. Recall that we write a conditional probability P(A given B) as P(A | B).

We write the probability that a sample proportion is greater than or equal to 0.15 given the population proportion is 0.10. as

P (\hat{p} \geq 0.15 ∣ p = 0.10) \approx 0.0475

As we now know, we can use a normal model only when certain conditions are met. Whenever we want to use a normal model, we must check the conditions to make sure a normal model is a good fit.

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