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This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 70 customer orders to fill. Each order requires one component part that is purchased from a supplier. However, typically, 3% of the components are identified as defective, and the components can be assumed to be independent.

Required:
a. If the manufacturer stocks 70 components, what is the probability that the 70 orders can be filled without reordering components?
b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?
c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

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Answer:

Step-by-step explanation:

a) This is a binomial distribution.

p = probability it is defective = 3% = 0.03, q = 1 - p = 1 - 0.03 = 0.97

This is a binomial distribution is given by the formula:

[tex]P(x)=C(n,x)*p^xq^{n-x}[/tex]

But n = 70

[tex]P(x=0)=C(70,0)*0.03^0*(0.97)^{70}=0.1186[/tex]

b) n = 102. Hence there can be 2 defective components, hence:

[tex]P(X\leq2)=P(X=0)+P(X=1)+P(X=2)=C(102,0)*0.03^0*(0.97)^{102}+C(102,1)*0.03^1*(0.97)^{102-1}+C(102,2)*0.03^2*(0.97)^{102-2}=0.0447+0.1441+0.2204=0.4062[/tex]

c) n = 105, there can be 5 defective products:

[tex]P(X\leq2)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=C(102,0)*0.03^0*(0.97)^{102}+C(102,1)*0.03^1*(0.97)^{102-1}+C(102,2)*0.03^2*(0.97)^{102-2}+C(102,3)*0.03^3 *(0.97)^{102-3}+C(102,4)*0.03^4*(0.97)^{102-4}+C(102,5)*0.03^5*(0.97)^{102-5}=0.0447+0.1411+0.2204+0.2273+0.174+0.1054=0.913[/tex]

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