Questions: The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is skewed right. However, records indicate that the mean time is 16.4 minutes, and the standard deviation is 4.1 minutes. Complete parts (a) through (c) below. B. Any sample size could be used. C. The normal model cannot be used if the shape of the distribution is skewed right. D. The sample size needs to be greater than 30. (b) What is the probability that a random sample of n=40 oil changes results in a sample mean time less than 15 minutes? The probability is approximately 0.0001. (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 A.M. and 12 P.M. 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 minutes. (Round to one decimal place as needed.)

The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is skewed right. However, records indicate that the mean time is 16.4 minutes, and the standard deviation is 4.1 minutes. Complete parts (a) through (c) below. B. Any sample size could be used. C. The normal model cannot be used if the shape of the distribution is skewed right. D. The sample size needs to be greater than 30. (b) What is the probability that a random sample of n=40 oil changes results in a sample mean time less than 15 minutes? The probability is approximately 0.0001. (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 A.M. and 12 P.M. 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 minutes. (Round to one decimal place as needed.)
Transcript text: The shape of the distribution of the time required to get an oil change at a 15 -minute oil-change facility is skewed right. However, records indicate that the mean time is 16.4 minutes, and the standard deviation is 4.1 minutes. Complete parts (a) through (c) below. B. Any sample size could be used. C. The normal model cannot be used if the shape of the distribution is skewed right. D. The sample size needs to be greater than 30. (b) What is the probability that a random sample of $n=40$ oil changes results in a sample mean time less than 15 minutes? The probability is approximately $0.0001$. (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 A.M. and 12 P.M. 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.)
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Solution

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Solution Steps

Step 1: Calculate the Probability for Part (b)

To find the probability that a random sample of \( n = 40 \) oil changes results in a sample mean time less than \( 15 \) minutes, we first calculate the Z-score:

\[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} = \frac{15 - 16.4}{4.1 / \sqrt{40}} \approx -2.1596 \]

Using the standard normal distribution, we find:

\[ P(X < 15) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(-2.1596) - \Phi(-\infty) \approx 0.0154 \]

Thus, the probability that the sample mean time is less than \( 15 \) minutes is approximately \( 0.0154 \).

Step 2: Calculate the Mean Time for Part (c)

To determine the mean oil-change time at which there is a \( 10\% \) chance of being at or below, we use the Z-score corresponding to the \( 10\% \) percentile, which is approximately \( -1.2816 \).

We calculate the mean time as follows:

\[ \text{Mean Time} = \mu + Z \cdot \left(\frac{\sigma}{\sqrt{n}}\right) = 16.4 + (-1.2816) \cdot \left(\frac{4.1}{\sqrt{40}}\right) \]

Calculating this gives:

\[ \text{Mean Time} \approx 15.6 \]

Thus, the mean oil-change time with a \( 10\% \) chance of being at or below is \( 15.6 \) minutes.

Final Answer

The answers to the questions are:

  • Part (b): The probability that the sample mean time is less than \( 15 \) minutes is \( \boxed{0.0154} \).
  • Part (c): The mean oil-change time with a \( 10\% \) chance of being at or below is \( \boxed{15.6} \).
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