Question 1175343: Stafford Production, is concerned with the quality of the parts it purchases that will be used in the end items it assembles. Part number 34-78D is used in the company’s new laser printer. The parts are sensitive to dust and can easily be damaged in shipment even if they are acceptable when they leave the vendor’s plant. In a shipment of four parts, the purchasing agent has assessed the following probability distribution for the number of defective products:
x P(x)
0 0.20
1 0.20
2 0.20
3 0.20
4 0.20
a. What is the expected number of defectives in a shipment of four parts? Discuss what this value really means to Stafford Production. [2]
b. Compute and interpret the standard deviation of the number of defective parts in a shipment of four. [3]
c. Examine the probabilities as assessed and indicate why this probability distribution might be called a uniform distribution. Provide some reasons why the probabilities might all be equal, as they are in this case. [2] 5-83
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It is a uniform function with mean (4+0)/2=2 parts
or .2*0+.2*1+.2*2+.2*3+.2*4=.2+.4+.6+.8=2.0 parts
var is (n^2-1)/12 for a mass function that is uniform
=24/12=2
sd tis sqrt(var)=sqrt(2)
the mean and the variance are the same. While not fitting the definition of a Poisson distribution, it has the same relation between mean and variance as one.
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Also .2((2-0)^2+.2(2-1)^2+.2(2-2)^2+.2(2-3)^2+.2(2-4)^2
=.2(4+1+0+1+4)=2.0 parts^2 for the variance.
Maybe the the same probability in that they all are subject to the same type of damage over time (would see with Poisson) and that the less time spent in transit perhaps the less damage would occur.
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