Question

A drug company believes that the annual demand for a drug will follow a normal random variable with a mean of 900 pounds and a standard deviation of 60 pounds. If the company produces 1000 pounds of the drug, what is the chance (rounded to the nearest hundredth) that it will run out of the drug? Assume that the only way to meet the demand for the drug is to use this year's production number.

Answer 1

Answer:

There is a 4.75% chance that the company will run out of the drug

Step-by-step explanation:

Normal model problems can be solved by the zscore formula.

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$, the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

$\mu = 900, \sigma = 60$.

If the company produces 1000 pounds of the drug, what is the chance (rounded to the nearest hundredth) that it will run out of the drug?

This chance is 100% subtracted by the pvalue of the Z-score of $X = 1000$.

So:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1000 - 900}{60}$

$Z = 1.67$

$Z = 1.67$ has a pvalue of .95254. This means that there is 95.254% probability that the company will sell less than 1000 pounds of the drug.

The probability that the company will run out of the drug is

$P = 100% - 95.254% = 4.746% = 4.75%$