Mathematics, 25.03.2020 05:26 kykylove1126

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 36,309 miles, with a standard deviation of 4693 miles. What is the probability that the sample mean would differ from the population mean by less than 170 miles in a sample of 211 tires if the manager is correct? Round your answer to four decimal places.

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Answer from: MogTaee

0.4938 = 49.38% probability that the sample mean would differ from the population mean by less than 170 miles in a sample of 211 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 are 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:

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

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 problem, we have that:

$\mu = 36309, \sigma = 4693, n = 211, s = \frac{4693}{\sqrt{211}} = 323.1$

What is the probability that the sample mean would differ from the population mean by less than 170 miles in a sample of 211 tires if the manager is correct?

This is the pvalue of Z when X = 36309 + 170 = 36479 subtracted by the pvalue of Z when X = 36309 - 170 = 36139. So

X = 36479

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

By the Central Limit Theorem

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

$Z = \frac{36479 - 36309}{323.1}$

Z = 0.53

Z = 0.53 has a pvalue of 0.7019

X = 36139

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

$Z = \frac{36139 - 36309}{323.1}$

Z = -0.53

Z = -0.53 has a pvalue of 0.2981

0.7019 - 0.2081 = 0.4938

0.4938 = 49.38% probability that the sample mean would differ from the population mean by less than 170 miles in a sample of 211 tires if the manager is correct

Answer from: Quest

answer: ok

step-by-step explanation:

Answer from: Quest

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step-by-step explanation:

What is the best window on a graphing calculator to see the graph?

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