Question: Numeric 5 points The average one-way commuting time in Richmond, Virginia is 21.9 minutes with a…

Numeric 5 points

The average one-way commuting time in Richmond, Virginia is 21.9 minutes with a standard deviation of 6.4 minutes.

If commuting times are normally distributed, what is the probability that the average of 8 commuters in one Richmond neighborhood would have a commuting time of less than 20 minutes?

0.11

Solution

To solve this problem, we use the Central Limit Theorem which states that the distribution of the sample mean will be approximately normal if the sample size is large enough. Given: - Mean (\(\mu\)) = 21.9 minutes - Standard deviation (\(\sigma\)) = 6.4 minutes - Sample size (\(n\)) = 8 - We’re finding the probability that the sample mean is less than 20 minutes. First, calculate the standard error of the mean (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{6.4}{\sqrt{8}} \] Calculate \(SE\): \[ SE = \frac{6.4}{2.828} \approx 2.26 \] Next, calculate the z-score for the sample mean of 20 minutes: \[ z = \frac{\bar{x} - \mu}{SE} = \frac{20 - 21.9}{2.26} \] Calculate \(z\): \[ z = \frac{-1.9}{2.26} \approx -0.841 \] Now, use the z-score table to find the probability corresponding to a z-score of \(-0.841\). The probability \(P(Z < -0.841) \approx 0.20045\). Therefore, the probability that the average of 8 commuters would have a commuting time of less than 20 minutes is approximately 0.200.