The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles. What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct? Round your answer to four decimal places.

Answer:There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.Step-by-step explanation:Problems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.Central Limit TheoremThe Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean \mu and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]In this problem, we have that:The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.This means that [tex]\mu = 33208, \sigma = 2503[/tex].What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?This is the pvalue of Z when [tex]X = 33208+633 = 33841[/tex] subtracted by the pvalue of Z when [tex]X = 33208 - 633 = 32575[/tex]By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57[/tex]SoX = 33841[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{33841 - 33208}{357.57}[/tex][tex]Z = 1.77[/tex][tex]Z = 1.77[/tex] has a pvalue of 0.9616X = 32575[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{32575- 33208}{357.57}[/tex][tex]Z = -1.77[/tex][tex]Z = -1.77[/tex] has a pvalue of 0.0384.This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.