Questions: This question: 1 point(s) possible 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 3.9 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 sample size needs to be greater than or equal to 30. B. The sample size needs to be less than or equal to 30. C. The normal model cannot be used if the shape of the distribution is skewed right. D. Any sample size could be used. (b) What is the probability that a random sample of n=35 oil changes results in a sample mean time less than 20 minutes? The probability is approximately . (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 35 oil changes between 10 A.M. and 12 PM. 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 minutes. (Round to one decimal place as needed.)

This question: 1 point(s) possible 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 3.9 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 sample size needs to be greater than or equal to 30. B. The sample size needs to be less than or equal to 30. C. The normal model cannot be used if the shape of the distribution is skewed right. D. Any sample size could be used. (b) What is the probability that a random sample of n=35 oil changes results in a sample mean time less than 20 minutes? The probability is approximately . (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 35 oil changes between 10 A.M. and 12 PM. 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 minutes. (Round to one decimal place as needed.)
Transcript text: This question: 1 point(s) possible 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 3.9 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 sample size needs to be greater than or equal to 30 . B. The sample size needs to be less than or equal to 30. C. The normal model cannot be used if the shape of the distribution is skewed right. D. Any sample size could be used. (b) What is the probability that a random sample of $n=35$ oil changes results in a sample mean time less than 20 minutes? The probability is approximately $\square$ . (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 35 oil changes between 10 A.M. and 12 PM. 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 $\square$ $\square$ minutes. (Round to one decimal place as needed.)
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Solution

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Solution Steps

Step 1: Determine Required Sample Size

To compute probabilities regarding the sample mean using the normal model, we need to consider the Central Limit Theorem. Since the distribution of the time required to get an oil change is skewed right, the sample size must be greater than or equal to 30 for the normal approximation to be valid.

Thus, the answer is: \[ \text{The sample size needs to be greater than or equal to } 30. \]

Step 2: Calculate Probability of Sample Mean Less Than 20 Minutes

We want to find the probability that a random sample of \( n = 35 \) oil changes results in a sample mean time less than 20 minutes. The calculated probability is given by:

\[ P(X < 20) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(-1.972) - \Phi(-\infty) = 0.0243. \]

Thus, the probability that the sample mean is less than 20 minutes is: \[ \boxed{P \approx 0.0243}. \]

Step 3: Determine Mean Oil-Change Time for 10% Chance

To find the mean oil-change time such that there is a 10% chance of being at or below this time, we first calculate the mean and standard deviation of the sampling distribution:

  • Mean of the sampling distribution: \[ \mu = 21.3. \]

  • Standard deviation of the sampling distribution: \[ \sigma = \frac{3.9}{\sqrt{35}} \approx 0.6592. \]

Next, we find the z-score corresponding to the 10th percentile: \[ z = \frac{0.1 - 0}{1} = 0.1. \]

Finally, we calculate the mean oil-change time for a 10% chance: \[ X = \mu + z \cdot \sigma = 21.3 + 0.1 \cdot 0.6592 \approx 21.4. \]

Thus, there is a 10% chance of being at or below a mean oil-change time of: \[ \boxed{21.4 \text{ minutes}}. \]

Final Answer

  • The sample size needs to be greater than or equal to 30.
  • The probability that the sample mean is less than 20 minutes is approximately \( \boxed{0.0243} \).
  • There is a 10% chance of being at or below a mean oil-change time of \( \boxed{21.4 \text{ minutes}} \).
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