Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 35,869 miles, with a variance of 12,194,060. What is the probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 tires if the manager is correct? Round your answer to four decimal places.

Answer 1

Answer:

0.9216 = 92.16% probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 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 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:

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

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 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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The standard deviation is the square root of the variance. So

$\mu = 35869, \sigma = \sqrt{12194060} = 3492, n = 269, s = \frac{3492}{\sqrt{269}} = 212.9$

What is the probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 tires if the manager is correct?

This is the pvalue of Z when X = 35869 + 375 = 36244 subtracted by the pvalue of Z when X = 35869 - 375 = 35494.

X = 36244

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

By the Central Limit Theorem

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

$Z = \frac{36244 - 38569}{212.9}[tez]Z = 1.76$

[tez]Z = 1.76[/tex] has a pvalue of 0.9608

X = 35494

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

$Z = \frac{35494 - 38569}{212.9}[tez]Z = -1.76$

[tez]Z = -1.76[/tex] has a pvalue of 0.0392

0.9608 - 0.0392 = 0.9216

0.9216 = 92.16% probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 tires if the manager is correct