Mathematics, 07.07.2021 02:00 DavidYSD46481

The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 4959 miles, with a standard deviation of 448 miles. If he is correct, what is the probability that the mean of a sample of 43 cars would differ from the population mean by less than 111 miles

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0.8948 = 89.48% probability that the mean of a sample of 43 cars would differ from the population mean by less than 111 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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean number of miles between services is 4959 miles, with a standard deviation of 448 miles

This means that $\mu = 4959, \sigma = 448$

Sample of 43:

This means that $n = 43, s = \frac{448}{\sqrt{43}}$

What is the probability that the mean of a sample of 43 cars would differ from the population mean by less than 111 miles?

p-value of Z when X = 4959 + 111 = 5070 subtracted by the p-value of Z when X = 4959 - 111 = 4848, that is, probability the sample mean is between these two values.

X = 5070

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{5070 - 4959}{\frac{448}{\sqrt{43}}}$

Z = 1.62

Z = 1.62 has a p-value of 0.9474

X = 4848

$Z = \frac{X - \mu}{s}$

$Z = \frac{4848 - 4959}{\frac{448}{\sqrt{43}}}$

Z = -1.62

Z = -1.62 has a p-value of 0.0526

0.9474 - 0.0526 = 0.8948

0.8948 = 89.48% probability that the mean of a sample of 43 cars would differ from the population mean by less than 111 miles

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one of each metals

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