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.
[tex]X\sim N(\mu=3.2\text{ hours},\ \sigma=0.4\text{ hours} )[/tex]
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, [tex]P(\dfrac{X-\mu}{\sigma}>\dfrac{x-\mu}{\sigma})=0.10[/tex]
[tex]P(z>\dfrac{x-3.2}{\sigma})=0.10\ \ \ [z=\dfrac{x-\mu}{\sigma}][/tex]
As, [tex]P(z>1.28)=0.10[/tex] [By z-table]
Then,
[tex]\dfrac{x-3.2}{0.4}=1.28\\\\\Rightarrow\ x=0.4\times1.28+3.2\\\\\Rightarrow\ x=3.712[/tex]
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.
