Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles. What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct? Round your answer to four decimal places.

Answer 1

Answer:

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

Step-by-step explanation:

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 $\sigma$, a large sample size 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 operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that $\mu = 33208, \sigma = 2503$.

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when $X = 33208+633 = 33841$ subtracted by the pvalue of Z when $X = 33208 - 633 = 32575$

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57$

So

X = 33841

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

$Z = \frac{33841 - 33208}{357.57}$

$Z = 1.77$

$Z = 1.77$ has a pvalue of 0.9616

X = 32575

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

$Z = \frac{32575- 33208}{357.57}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.