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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
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
.
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
subtracted by the pvalue of Z when ![X = 33208 - 633 = 32575](https://tex.z-dn.net/?f=X%20%3D%2033208%20-%20633%20%3D%2032575)
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](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cfrac%7B2503%7D%7B%5Csqrt%7B49%7D%7D%20%3D%20357.57)
So
X = 33841
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{33841 - 33208}{357.57}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B33841%20-%2033208%7D%7B357.57%7D)
![Z = 1.77](https://tex.z-dn.net/?f=Z%20%3D%201.77)
has a pvalue of 0.9616
X = 32575
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{32575- 33208}{357.57}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B32575-%2033208%7D%7B357.57%7D)
![Z = -1.77](https://tex.z-dn.net/?f=Z%20%3D%20-1.77)
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.