Answer:
0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 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
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
, the sample means with size n of at least 30 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 = 48564, \sigma = 3293, n = 281, s = \frac{3293}{\sqrt{281}} = 196.44](https://tex.z-dn.net/?f=%5Cmu%20%3D%2048564%2C%20%5Csigma%20%3D%203293%2C%20n%20%3D%20281%2C%20s%20%3D%20%5Cfrac%7B3293%7D%7B%5Csqrt%7B281%7D%7D%20%3D%20196.44)
What is the probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct?
This is the pvalue of Z when X = 48101. So
![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{48101 - 48564}{196.44}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B48101%20-%2048564%7D%7B196.44%7D)
![Z = -2.36](https://tex.z-dn.net/?f=Z%20%3D%20-2.36)
has a pvalue of 0.0091
0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct