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the average wait time for an oil change at a popular tire and lube shop follows a normal distribution with a mean of 13 minutes and a standard deviation of 13 minutes. if a customer pulls into the tire and lube shop, determine the probability the probability that their wait time will be between 12 and 14 minutes longer than 15 minutes determine the wait time that corresponds to the 98th percentile determine the wait times that correspond to the boundaries of the middle 65% of wait times

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1. The probability that the customer's wait time will be between 12 and 14 minutes is 0.0304, or 3.04%.

2. The wait time that corresponds to the 98th percentile is a 39.715 minutes.

3. The wait times that correspond to the boundaries of the middle 65% of wait times are 0.722 minutes and 25.038 minutes.

The Breakdown

1. Determine the probability the probability that their wait time will be between 12 and 14 minutes longer than 15 minutes.

First, we need to standardize the values using the formula:

z = (x - μ) / σ

For 12 minutes:

z1 = (12 - 13) / 13 = -0.077

For 14 minutes:

z2 = (14 - 13) / 13 = 0.077

Using the standard normal distribution table or a calculator, we can find the probabilities associated with these z-scores. The probability between -0.077 and 0.077 is 0.0304.

Therefore, the probability that the customer's wait time will be between 12 and 14 minutes is 0.0304, or 3.04%.

2. Wait time longer than 15 minutes:

To determine the wait time that corresponds to the 98th percentile, we need to find the z-score associated with this percentile and then convert it back to the original wait time using the formula:

x = z × σ + μ

To find the z-score corresponding to the 98th percentile, we can use the standard normal distribution table or a calculator. The z-score for the 98th percentile is 2.055.

Using the formula, we can calculate the wait time:

x = 2.055 × 13 + 13 = 39.715

Therefore, the wait time that corresponds to the 98th percentile is 39.715 minutes.

3. Boundaries of the middle 65% of wait times:

To find the wait times that correspond to the boundaries of the middle 65% of wait times, we need to find the z-scores that enclose this percentage and then convert them back to the original wait times.

The middle 65% of the distribution corresponds to the area between the 17.5th and 82.5th percentiles. To find the z-scores for these percentiles, we can use the standard normal distribution table or a calculator. The z-score for the 17.5th percentile is -0.926, and the z-score for the 82.5th percentile is approximately 0.926.

Using the formula, we can calculate the wait times:

For the lower boundary:

x1 = -0.926 13 + 13 = 0.722 minutes

For the upper boundary:

x2 = 0.926 × 13 + 13 = 25.038 minutes

Therefore, the wait times that correspond to the boundaries of the middle 65% of wait times are 0.722 minutes and 25.038 minutes.

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