Wiki Page, Math 13 - Perez Sandbox : Module 17 - Distribution of Sample Proportions (6 of 6)

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2026-01-17T21:15:22-08:00

Two-tailed Probabilities

The previous examples all had answers that came from the area in one-tail of the normal distribution. Why? Because all of the questions involved "less than" or "greater than" statements.

For example,

What is the probability that a random sample of 200 white male voters had less than 35% voting for Obama in 2008?

What is the probability that in a random sample of 100 young adults 15% or more enrolled in a community college?

Two-tailed probabilities involve questions that do not specify "less than" or "greater than."

EXAMPLE

Recall the use of data from the Centers for Disease Control and Prevention's (CDC) National Health Interview Survey to estimate behaviors such as alcohol consumption, cigarette smoking, and hours of sleep for adults in the United States. In the 2005–2007 report, the CDC estimated that 68% of men in the United States are overweight.

Suppose we select a random sample of 40 men and find that only 58% are overweight. If 68% of U.S. men are overweight, this sample percentage is off by 10%. Is this much error surprising? What is the probability that a sample proportion will an overestimate or underestimate of the parameter by more than 10%?

Check normality conditions:

Yes, the conditions are met. The number of expected successes and failures in a sample of 40 are at least 10.

Here is our work:

n is 40 because there are 40 men in the sample.

p is 0.68 because CDC estimated that 68% of population (men in the U.S.) are overweight.

Because 68% of the population is overweight, we expect 68% of the sample of 40 to be overweight; .

Because the remaining 32% (100% - 68% = 32%) is not overweight, we expect 32% of the sample of 40 to not be overweight; .

Therefore, we can use a normal model. This allows us to use a z-score and the Normal Distribution Calculator to find the probability.

Find the z-score:

We want the error to be more than 10% in either direction, so the sample proportion could be less than 0.58 (0.68 – 0.10 = 0.58) or greater than 0.78 (0.68 + 0.10 = 0.78).

It does not matter which sample proportion we use to find the z-score because of the symmetry in the normal distribution. We arbitrarily chose 0.58. We could also have used 0.78.

standard error =

Z =

Find the probability using the standard normal model:

Our probability estimate: We get a probability of about 18%.

Here is our work:

We want the probability described by the two tails. The probability for one tail is 0.0885, or about 0.09. So the probability for both tails is about 2 x 0.09 = 0.18.

Conclusion:

If it is true that 68% of U.S. men are overweight, then there is about an 18% chance that the percentage of overweight men in a random sample of 40 men is off by more than 10%. In other words, there is about an 18% chance that sample proportions will fall below 0.58 or above 0.78 if the true population proportion is 0.68.

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