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Suppose that a garage tells its customers that with the new synthetic oil, the average time people wait for an oil change is 5734.2 miles with a standard deviation of 1011.9 miles. What is the probability that a customer brings their car back between 4500 and 5500 miles?

User Ahmed Atia
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Final answer:

To find the probability of a customer returning their car between 4500 and 5500 miles for an oil change, we calculate the z-scores for each value, find the corresponding probabilities in the standard normal distribution, and subtract the smaller probability from the larger one.

Step-by-step explanation:

To determine the probability that a customer brings their car back for an oil change between 4500 and 5500 miles given a mean of 5734.2 miles and a standard deviation of 1011.9 miles, we can use the normal distribution. Since the question implies a normal distribution, we first standardize the values using the z-score formula: z = (X - μ) / σ, where X is the value in question, μ is the mean, and σ is the standard deviation.



The z-scores for 4500 and 5500 miles would be calculated as follows:



For X=4500, z = (4500 - 5734.2) / 1011.9

For X=5500, z= (5500 - 5734.2) / 1011.9



After calculating these z-scores, we can use a standard normal distribution table or a calculator to find the probability that corresponds to these z-scores. The probability of a value being between the two z-scores is the difference between the probabilities associated with each.



If the z-score for 4500 miles is negative and for 5500 miles is negative or less negative (closer to zero), the probabilities will increase as you move from the 4500-mile z-score to the 5500-mile z-score. The final answer will be the difference between the two values found in the z-table or calculator.