Answer:
52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct
Step-by-step explanation:
To solve this question, we have 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
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
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:
![\mu = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81](https://tex.z-dn.net/?f=%5Cmu%20%3D%2030393%2C%20%5Csigma%20%3D%202876%2C%20n%20%3D%2037%2C%20s%20%3D%20%5Cfrac%7B2876%7D%7B%5Csqrt%7B37%7D%7D%20%3D%20472.81)
What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct
This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So
X = 30732
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{30732 - 30393}{472.81}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B30732%20-%2030393%7D%7B472.81%7D)
![Z = 0.72](https://tex.z-dn.net/?f=Z%20%3D%200.72)
has a pvalue of 0.7642.
X = 30054
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{30054 - 30393}{472.81}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B30054%20-%2030393%7D%7B472.81%7D)
![Z = -0.72](https://tex.z-dn.net/?f=Z%20%3D%20-0.72)
has a pvalue of 0.2358
0.7642 - 0.2358 = 0.5284
52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct