Answer:
0.7888 = 78.88% probability that the first machine produces an acceptable cork.
0.9772 = 97.72% probability that the second machine produces an acceptable cork.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
First machine:
Mean 3 cm and standard deviation 0.08 cm, which means that ![\mu = 3, \sigma = 0.08](https://tex.z-dn.net/?f=%5Cmu%20%3D%203%2C%20%5Csigma%20%3D%200.08)
What is the probability that the first machine produces an acceptable cork?
This is the p-value of Z when X = 3.1 subtracted by the p-value of Z when X = 2.9. So
X = 3.1
![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{3.1 - 3}{0.08}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B3.1%20-%203%7D%7B0.08%7D)
![Z = 1.25](https://tex.z-dn.net/?f=Z%20%3D%201.25)
has a p-value of 0.8944
X = 2.9
![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{2.9 - 3}{0.08}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B2.9%20-%203%7D%7B0.08%7D)
![Z = -1.25](https://tex.z-dn.net/?f=Z%20%3D%20-1.25)
has a p-value of 0.1056
0.8944 - 0.1056 = 0.7888
0.7888 = 78.88% probability that the first machine produces an acceptable cork.
What is the probability that the second machine produces an acceptable cork?
For the second machine,
. Now to find the probability, same procedure.
X = 3.1
![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{3.1 - 3.04}{0.03}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B3.1%20-%203.04%7D%7B0.03%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a p-value of 0.9772
X = 2.9
![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{2.9 - 3.04}{0.03}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B2.9%20-%203.04%7D%7B0.03%7D)
![Z = -4.67](https://tex.z-dn.net/?f=Z%20%3D%20-4.67)
has a p-value of 0
0.9772 - 0 = 0.9772
0.9772 = 97.72% probability that the second machine produces an acceptable cork.