Awarehouse distributor of carpet faces a normally distributed demand for its carpet. the average demand for carpet from the stores that purchase from the distributor is 4,500 yards per month, with a standard deviation of 900 yards. a. suppose the distributor keeps 6,000 yards of carpet in stock during a month. what is the probability that a customer’s order will not be met during a month? (this situation is referred to as a stockout.)

There is a 25.14% probability that the order will not be met during a month.

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 , the zscore of a measure X is given by

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

.

The order will not be met if . So we find the pvalue of Z when , and subtract 1 by this value.

has a pvalue of 0.7486.

So there is a 1 - 0.7486 = 0.2514 = 25.14% probability that the order will not be met during a month.

table 3 represents a direct variation function. : )

answer: 5

step-by-step explanation: