The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 4639 miles, with a standard deviation of 437 miles. If he is correct, what is the probability that the mean of a sample of 32 cars would differ from the population mean by less than 181 miles? Round your answer to four decimal places.

0.9808 = 98.08% probability that the mean of a sample of 32 cars would differ from the population mean by less than 181 miles

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:

If he is correct, what is the probability that the mean of a sample of 32 cars would differ from the population mean by less than 181 miles?

This is the pvalue of Z when X = 4639 + 181 = 4820 subtracted by the pvalue of Z when X = 4639 - 181 = 4458. So

X = 4820

By the Central limit theorem

has a pvalue of 0.9904

X = 4458

has a pvalue of 0.0096

0.9904 - 0.0096 = 0.9808

0.9808 = 98.08% probability that the mean of a sample of 32 cars would differ from the population mean by less than 181 miles

10000

step-by-step explanation: divide by 2,

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