Question

Assume the time required to complete a product is normally distributed with a mean 3.2 hours and standard deviation .4 hours. How long should it take to complete a random unit in order to be in the top 10% (right tail) of the time distribution?

Answer 1

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.

$X\sim N(\mu=3.2\text{ hours},\ \sigma=0.4\text{ hours} )$

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, $P(\dfrac{X-\mu}{\sigma}>\dfrac{x-\mu}{\sigma})=0.10$

$P(z>\dfrac{x-3.2}{\sigma})=0.10\ \ \ [z=\dfrac{x-\mu}{\sigma}]$

As, $P(z>1.28)=0.10$  [By z-table]

Then,

$\dfrac{x-3.2}{0.4}=1.28\\\\\Rightarrow\ x=0.4\times1.28+3.2\\\\\Rightarrow\ x=3.712$

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.