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natta225 [31]
3 years ago
12

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

iance of 145,161 145,161 . If he is correct, what is the probability that the mean of a sample of 41 41 cars would differ from the population mean by less than 126 126 miles
Mathematics
1 answer:
Andrei [34K]3 years ago
3 0

Answer:

96.6% probability that the mean of a sample would differ from the population mean by less than 126 miles

Step-by-step explanation:

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

A reminder is that the standard deviation is the square root of the variance.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 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 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.

In this question:

\mu = 3639, \sigma = \sqrt{145161} = 381, n = 41, s = \frac{381}{\sqrt{41}} = 59.5

Probability that the mean of the sample would differ from the population mean by less than 126 miles

This is the pvalue of Z when X = 3639 + 126 = 3765 subtracted by the pvalue of Z when X = 3639 - 126 = 3513. So

X = 3765

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

By the Central Limit Theorem

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

Z = \frac{3765 - 3639}{59.5}

Z = 2.12

Z = 2.12 has a pvalue of 0.983

X = 3513

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

Z = \frac{3513 - 3639}{59.5}

Z = -2.12

Z = -2.12 has a pvalue of 0.017

0.983 - 0.017 = 0.966

96.6% probability that the mean of a sample would differ from the population mean by less than 126 miles

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