Question

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.

Answer 1

Explanation:

The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 3639 miles, with a variance of 145,161.

If he is correct, what is the probability that the mean of a sample of 41 cars would differ from the population mean by less than 126 miles? Round your answer to four

Answer 2

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

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
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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
`\mu` and standard deviation
`\sigma`, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 4639, \sigma = 437, n = 32, s = (437)/(√(32)) = 77.25`

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

`Z = (X - \mu)/(\sigma)`

By the Central limit theorem

`Z = (X - \mu)/(s)`

`Z = (4820 - 4639)/(77.25)`

`Z = 2.34`

`Z = 2.34` has a pvalue of 0.9904

X = 4458

`Z = (X - \mu)/(s)`

`Z = (4458 - 4639)/(77.25)`

`Z = -2.34`

`Z = -2.34` 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