A widget manufacturer knows that their widget lengths are approximately Normally distributed with mean 12.1 inches and standard deviation 1.8 inches. One widget is 14.7 inches long. That widget z-score is about...

answer:

Get a z-score chart

https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf

z = (X - μ) / σ, where X = date, μ = mean, σ = standard deviation

z = (12.8 - 15.2) / 1.6

z = -1.500

0.06681 is the area for this z.

6.681% shall be shorter than 12.8 inches.

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answer:

b

Step-by-step explanation:

this was already solved by me with a different person. if you look it up in the search bar at the top of the page, you will get the answer.

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