Answer:
The diameter that separates the top 3% is of 5.85 millimeters, and the one which separates the bottom 3% is of 5.55 millimeters.
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.
Mean of 5.7 millimeters and a standard deviation of 0.08 millimeters.
This means that ![\mu = 5.7, \sigma = 0.08](https://tex.z-dn.net/?f=%5Cmu%20%3D%205.7%2C%20%5Csigma%20%3D%200.08)
Top 3%
The 100 - 3 = 97th percentile, which is X when Z has a p-value of 0.97, so X when Z = 1.88.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.88 = \frac{X - 5.7}{0.08}](https://tex.z-dn.net/?f=1.88%20%3D%20%5Cfrac%7BX%20-%205.7%7D%7B0.08%7D)
![X - 5.7 = 1.88*0.08](https://tex.z-dn.net/?f=X%20-%205.7%20%3D%201.88%2A0.08)
![X = 5.85](https://tex.z-dn.net/?f=X%20%3D%205.85)
Bottom 3%
The 3rd percentile, which is X when Z has a p-value of 0.03, so X when Z = -1.88.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-1.88 = \frac{X - 5.7}{0.08}](https://tex.z-dn.net/?f=-1.88%20%3D%20%5Cfrac%7BX%20-%205.7%7D%7B0.08%7D)
![X - 5.7 = -1.88*0.08](https://tex.z-dn.net/?f=X%20-%205.7%20%3D%20-1.88%2A0.08)
![X = 5.55](https://tex.z-dn.net/?f=X%20%3D%205.55)
The diameter that separates the top 3% is of 5.85 millimeters, and the one which separates the bottom 3% is of 5.55 millimeters.