Question

Goop Inc. needs to order a raw material to make a special polymer. The demand for the polymer is forecasted to be normally distributed with a mean of 250 gallons and a standard deviation of 125 gallons. Goop sells the polymer for $21 per gallon. Goop purchases raw material for $13 per gallon and Goop must spend $3 per gallon to dispose of unused raw material due to government regulations. (One gallon of raw material yields one gallon of polymer.) If demand is more than Goop can make, then Goop sells only what they made, and the rest of demand is lost. Suppose Goop purchases 150 gallons of raw material. What is the probability that they will run out of raw material?

Answer 1

Answer:

There is a 78.81% probability that they will run out of raw material.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

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

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The demand for the polymer is forecasted to be normally distributed with a mean of 250 gallons and a standard deviation of 125 gallons. So $\mu = 250, \sigma = 125$.

Suppose Goop purchases 150 gallons of raw material. What is the probability that they will run out of raw material? That is $P(X > 150)$.

So

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

$Z = \frac{150 - 250}{125}$

$Z = -0.8$

$Z = -0.8$ has a pvalue of 0.2119. This means that $P(X \leq 150) = 0.2119$

Also

$P(X \leq 150) + P(X > 150) = 1$

$P(X > 150) = 1 - 0.2119 = 0.7881$

There is a 78.81% probability that they will run out of raw material.