## The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard deviation of

Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard deviation of 2876 miles. What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

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2 months 2021-10-15T16:25:32+00:00 1 Answer 0 views 0

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem:

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by: 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 theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation In this problem, we have that: What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 – 339 = 30054. So

X = 30732 By the Central Limit Theorem    has a pvalue of 0.7642.

X = 30054    has a pvalue of 0.2358

0.7642 – 0.2358 = 0.5284

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct