Question

A plant manager knows that the number of boxes of supplies received weekly is normally distributed with a mean of 200 and a standard deviation of 20. What percentage of the time will the number of boxes of supplies that arrive be greater than 210 or less than 180?

Answer 1

Answer:

46.72%.

Step-by-step explanation:

We have been given that the number of boxes of supplies received weekly is normally distributed with a mean of 200 and a standard deviation of 20. We are asked to find the number of boxes of supplies that arrive will be greater than 210 or less than 180.  

We will use z-score formula and normal distribution table to solve our problem.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{180-200}{20}$

$z=\frac{-20}{20}$

$z=-1$

Let us find z-score corresponding to normal score 210.

$z=\frac{210-200}{20}$

$z=\frac{10}{20}$

$z=0.5$

$P(z0.5)=P(z$

Using normal distribution table, we will get:

$P(z0.5)=0.15866+1-0.69146$

$P(z0.5)=0.4672$

$0.4672\times 100\%=46.72\%$

Therefore, 46.72% of the time the number of boxes of supplies that arrive will be greater than 210 or less than 180.