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vovikov84 [41]
3 years ago
10

A warehouse distributor of carpet faces a normally distributed demand for its carpet. The average demand for carpet from the sto

res 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.)
Mathematics
1 answer:
dimulka [17.4K]3 years ago
3 0

Answer:

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 \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. 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:

\mu = 4500, \sigma = 900.

The order will not be met if X > 6000. So we find the pvalue of Z when X = 6000, and subtract 1 by this value.

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

Z = \frac{6000 - 4500}{900}

Z = 1.67

Z = 1.67 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.

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