Assume the time required to complete a product is normally distributed with a mean 3.2 hours and standard deviation .4 hours. Ho
w long should it take to complete a random unit in order to be in the top 10% (right tail) of the time distribution?
1 answer:
Answer: 3.712 hours or more
Step-by-step explanation:
Let X be the random variable that denotes the time required to complete a product.
X is normally distributed.

Let x be the times it takes to complete a random unit in order to be in the top 10% (right tail) of the time distribution.
Then, 
![P(z>\dfrac{x-3.2}{\sigma})=0.10\ \ \ [z=\dfrac{x-\mu}{\sigma}]](https://tex.z-dn.net/?f=P%28z%3E%5Cdfrac%7Bx-3.2%7D%7B%5Csigma%7D%29%3D0.10%5C%20%5C%20%5C%20%5Bz%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D)
As,
[By z-table]
Then,

So, it will take 3.712 hours or more to complete a random unit in order to be in the top 10% (right tail) of the time distribution.
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