The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 35,869 miles, with a variance of 12,194,060. What is the probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 tires if the manager is correct? Round your answer to four decimal places.

0.9216 = 92.16% probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 tires if the manager is correct

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples can be 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 , the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

The standard deviation is the square root of the variance. So

What is the probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 tires if the manager is correct?

This is the pvalue of Z when X = 35869 + 375 = 36244 subtracted by the pvalue of Z when X = 35869 - 375 = 35494.

X = 36244

By the Central Limit Theorem

[tez]Z = 1.76[/tex] has a pvalue of 0.9608

X = 35494

[tez]Z = -1.76[/tex] has a pvalue of 0.0392

0.9608 - 0.0392 = 0.9216

0.9216 = 92.16% probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 tires if the manager is correct

a 90 degree rotation across the the origin followed by a reflection across the x-axis.

step-by-step explanation: