A warehouse 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.)

Answer with Step-by-step explanation:Since the demand is normally distributed the required probability can be found from the area under the normal distribution curve as Part a)Given mean = 4500 yards per monthStandard deviation = 900 yardsThus area under the curve corresponding to 6000 yards is found from the standard variate factor Z as[tex]Z=\frac{X-\bar{X}}{\sigma }\\\\Z=\frac{6000-4500}{900}=1.67[/tex]Area for Z = 1.67 = 95.22%Thus the probability that the demand will be met is 0.9522 hence the probability that the demand will not be met is [tex]P(E)=1-0.9522=0.0478[/tex]Part b)The reuired answer is area between 5000 and 7000 yards in the normal distribution curve thus we have[tex]Z_1=\frac{5000-4500}{900}=0.56[/tex].[tex]Z_2=\frac{7000-4500}{900}=2.78[/tex]The area between these 2 values is 49.73% hence the reuired probability is 0.4973.Part c)For 97% satisfaction of demand the Z factor corresponding to 97% of area is found to be 1.88thus we can write[tex]1.88=\frac{X-4500}{900}\\\\X=4500+1.88\times 1.88=6193[/tex]