Answer:
C. Probability is 0.90, which is inconsistent with the Empirical Rule.
Step-by-step explanation:
We have been given that on average, the parts from a supplier have a mean of 97.5 inches and a standard deviation of 6.1 inches.
First of all, we will find z-score corresponding to 87.5 and 107.5 respectively as:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
![z=\frac{87.5-97.5}{6.1}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B87.5-97.5%7D%7B6.1%7D)
![z=\frac{-10}{6.1}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B-10%7D%7B6.1%7D)
![z=-1.6393](https://tex.z-dn.net/?f=z%3D-1.6393)
![z\approx-1.64](https://tex.z-dn.net/?f=z%5Capprox-1.64)
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
![z=\frac{107.5-97.5}{6.1}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B107.5-97.5%7D%7B6.1%7D)
![z=\frac{10}{6.1}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B10%7D%7B6.1%7D)
![z=1.6393](https://tex.z-dn.net/?f=z%3D1.6393)
![z\approx 1.64](https://tex.z-dn.net/?f=z%5Capprox%201.64)
Now, we need to find the probability
.
Using property
, we will get:
![P(-1.64](https://tex.z-dn.net/?f=P%28-1.64%3Cz%3C1.64%29%3DP%28z%3C1.64%29-P%28z%3C-1.64%29)
From normal distribution table, we will get:
![P(-1.64](https://tex.z-dn.net/?f=P%28-1.64%3Cz%3C1.64%29%3D0.94950-0.05050%20)
![P(-1.64](https://tex.z-dn.net/?f=P%28-1.64%3Cz%3C1.64%29%3D0.899)
![P(-1.64](https://tex.z-dn.net/?f=P%28-1.64%3Cz%3C1.64%29%5Capprox%200.90)
Since the probability is 0.90, which is inconsistent with the Empirical Rule, therefore, option C is the correct choice.