Answer:
0.8185
Step-by-step explanation:
To solve this question, we have to understand the normal probability ditribution 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 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 = 1.01, \sigma = 0.003, n = 9, s = \frac{0.003}{\sqrt{9}} = 0.001](https://tex.z-dn.net/?f=%5Cmu%20%3D%201.01%2C%20%5Csigma%20%3D%200.003%2C%20n%20%3D%209%2C%20s%20%3D%20%5Cfrac%7B0.003%7D%7B%5Csqrt%7B9%7D%7D%20%3D%200.001)
Find the probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.
This is the pvalue of Z when X = 1.012 subtracted by the pvalue of Z when X = 1.009. So
X = 1.012
![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{1.012 - 1.01}{0.001}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1.012%20-%201.01%7D%7B0.001%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a pvalue of 0.9772
X = 1.009
![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{1.009 - 1.01}{0.001}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1.009%20-%201.01%7D%7B0.001%7D)
![Z = -1](https://tex.z-dn.net/?f=Z%20%3D%20-1)
has a pvalue of 0.1587
0.9772 - 0.1587 = 0.8185
0.8185 = 81.85% probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.