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 month
Standard deviation = 900 yards
Thus 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 1.88
thus we can write
[tex]1.88=\frac{X-4500}{900}\\\\X=4500+1.88\times 1.88=6193[/tex]
