A university wants to estimate the average distance that commuter students travel to get to class with an error of ±3 miles and 90 percent confidence. What sample size would be needed, assuming that distances are normally distributed with a range of X = 0 to X = 50 miles, and using the Empirical Rule ±3 to estimate ?

21

47

28

21

30 To estimate the sample size needed, we can use the formula:

n = (Z * σ / E)^2

where:
n = sample size
Z = Z-score corresponding to the desired confidence level
σ = standard deviation of the population
E = maximum acceptable error

In this case, we are given the range of distances from 0 to 50 miles, but we need to determine the standard deviation. Since we do not have access to the individual data points, we can use the Empirical Rule to estimate the standard deviation.

According to the Empirical Rule, for a normal distribution, approximately 68% of the data falls within one standard deviation (σ), approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations.

Given that the range is 50 miles, which is approximately 3 standard deviations, we can estimate the standard deviation as:

σ = Range / 3

σ = 50 miles / 3 ≈ 16.67 miles

Next, we need to determine the Z-score for a 90% confidence level. The Z-score for a 90% confidence level is obtained by subtracting the confidence level from 1 and dividing by 2. In this case:

Z = (1 - 0.90) / 2 = 0.10 / 2 = 0.05

Using a standard normal distribution table or calculator, we find that the Z-score corresponding to 0.05 is approximately 1.645.

Now, we can substitute the values into the formula:

n = (Z * σ / E)^2
= (1.645 * 16.67 / 3)^2
≈ (27.37835 / 3)^2
≈ 9.126^2
≈ 83.32

Therefore, a sample size of approximately 83.32 (which we will round up to 84) would be needed to estimate the average distance traveled by commuter students with an error of ±3 miles and 90% confidence, under the assumption that distances are normally distributed.