Answer:
There is a 20% probability that the store selected does not violate the institute scanner accuracy standard
Step-by-step explanation:
70 company stores were investigared.
56 violated the institute's scanner accuracy standard.
70-56 = 14 did not violate the institute scanner accuracy standard.
If 1 of the 70 company stores is randomly selected, what is the probability that that store does not violate the institute scanner accuracy standard?
This is 14 divided by 70, so:
![P = \frac{14}{70} = 0.2](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B14%7D%7B70%7D%20%3D%200.2)
There is a 20% probability that the store selected does not violate the institute scanner accuracy standard