The shape of the distribution of the time required to get an oil change at a 

2020​-minute 

​oil-change facility is skewed right.​ However, records indicate that the mean time is 

21.3 minutes21.3 minutes​, 

and the standard deviation is 

3.8 minutes3.8 minutes.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 

4545 

oil changes between 10 A.M. and 12 P.M. Treating this as a random​ sample, there would be a​ 10% chance of the mean​ oil-change time being at or below what​ value? This will be the goal established by the manager.

There is a​ 10% chance of being at or below a mean​ oil-change time of 

enter your response here 

minutes.

​(Round to one decimal place as​ needed.)

To solve this problem, we need to determine the value of the mean oil-change time such that there is a 10% chance that the mean time for 45 oil changes is at or below that value. This involves finding the 10th percentile of the sampling distribution of the sample mean.

### Steps:

1. **Determine the population mean and standard deviation:**
   - Mean ($\mu$) = 21.3 minutes
   - Standard deviation ($\sigma$) = 3.8 minutes

2. **Determine the sample size:**
   - $n = 45$

3. **Calculate the standard error (SE) of the sample mean:**
   \[
   \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{3.8}{\sqrt{45}} \approx \frac{3.8}{6.708} \approx 0.5662
   \]

4. **Find the z-score that corresponds to the 10th percentile:**
   - The z-score associated with the 10th percentile is approximately $-1.28$.

5. **Use the z-score formula to find the sample mean that corresponds to this z-score:**
   \[
   z = \frac{X - \mu}{\text{SE}}
   \]
   Solving for $X$ (the mean time at the 10th percentile):
   \[
   X = \mu + z \times \text{SE}
   \]
   \[
   X = 21.3 + (-1.28) \times 0.5662 \approx 21.3 - 0.7247 \approx 20.6
   \]

### Conclusion:
There is a 10% chance of the mean oil-change time being at or below **20.6 minutes**.

Let me know if you'd like more details or have any questions! 

Here are 8 related questions:
1. What is the significance of the 10th percentile in this context?
2. How would changing the standard deviation affect the target mean time?
3. What does a z-score of -1.28 represent in a normal distribution?
4. How would the goal change if the sample size were increased?
5. Why is the shape of the original distribution important for this calculation?
6. What are some factors that could cause the actual mean time to deviate from the calculated goal?
7. How would you set a similar goal for the 25th percentile?
8. What is the impact of a skewed distribution on setting goals like this?

**Tip:** When using percentiles to set goals, it’s important to understand how the standard error affects the spread of the sample mean distribution.

Learn how to calculate the mean oil-change time goal based on a skewed distribution with specific data points: mean, standard deviation, and sample size. Understand the concept of sampling distribution, standard error of the mean, and z-score in statistics.

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