The shape of the distribution of the time required to get an oil change at a 20 -minute oil-change facility is skewed right. However, records indicate that the mean time is 21.3 minutes, and the standard deviation is 3.8 minutes. Complete parts (a) through (c). (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? A. The normal model cannot be used if the shape of the distribution is skewed right. B. The sample size needs to be greater than or equal to 30 . The sample size needs to be less than or equal to 30 . D. Any sample size could be used. (b) What is the probability that a random sample of \( n=40 \) oil changes results in a sample mean time less than 20 minutes? The probability is approximately (Round to four decimal places as needed.)

To appropriately use the normal model for probability calculations involving sample means, you'll want to remember the Central Limit Theorem, which tells us that as the sample size increases (specifically beyond 30), the distribution of the sample mean will approach normality even if the underlying distribution is skewed. So, in this case, the answer is that the sample size needs to be greater than or equal to 30. Now for the probability calculation, you will use the formula for the sampling distribution of the sample mean, which is \( \mu_{\bar{x}} = \mu = 21.3 \) and the standard deviation of the sample mean \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3.8}{\sqrt{40}} \approx 0.6 \). To find the probability of getting a sample mean less than 20 minutes, you would compute the z-score: \( z = \frac{(20 - 21.3)}{0.6} \approx -2.167 \). Consulting a z-table or calculator, the probability is approximately 0.0150.