The shape of the distribution of the time required to get an oil change at a 10 -minute oil-change facility is skewed right. However, records indicate that the mean time is 11.6 minutes. and the standard deviation is 4.3 minutes. Complete parts (a) through (c) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution tasle (page 2). B. The normal model cannot be used if the shape of the distribution is skewed right C. The sample size needs to be greater than 30 D. The sample size needs to be less than 30 . (b) What is the probability that a random sample of $n=40$ oil changes results in a sample mean time less than 10 minutes? The probability is approximately 0.0094 . (Round to four decimal places as needed.) (c) Suppose the manager agrees to pay each employee a $\$ 50$ bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between $10 \mathrm{A.M}$ and $12 \mathrm{PM}$. Treating this as a random sample, at what mean oil-change time would there be a $10 \%$ chance of being at or below? This will be the goal established by the manager. There would be a $10 \%$ chance of being at or below $\square$ minutes. (Round to one decimal place as needed) Clear all Check answer Search
Solution
Step 1 ：Given that the population mean \(\mu = 11.6\) minutes and the population standard deviation \(\sigma = 4.3\) minutes. The sample size is \(n = 40\).
Step 2 ：We first calculate the standard deviation of the sample means (standard error), \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4.3}{\sqrt{40}} \approx 0.68\) minutes.
Step 3 ：We want to find the probability that the sample mean is less than 10 minutes. To do this, we standardize the score to get a z-score, \(Z = \frac{X - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{10 - 11.6}{0.68} \approx -2.35\).
Step 4 ：Using a standard normal distribution table, we find that the probability of getting a z-score less than -2.35 is approximately 0.0094. So, the probability that a random sample of 40 oil changes results in a sample mean time less than 10 minutes is \(\boxed{0.0094}\).
Step 5 ：To find the mean oil-change time that corresponds to a 10% chance, we find the z-score that corresponds to a cumulative probability of 0.10, which is approximately -1.28.
Step 6 ：We then use the formula for the z-score to solve for X, \(X = Z\sigma_{\bar{x}} + \mu_{\bar{x}} = -1.28(0.68) + 11.6 \approx 10.7\) minutes.
Step 7 ：So, the manager should set the goal at \(\boxed{10.7}\) minutes. If the mean oil-change time is at or below this value, there would be a 10% chance of this occurring.
