Question

Suppose that the number of defects in the production of a length of copper wire is modeled by a Poisson process with an average of 8 defects per 500 feet of wire. a. What is the probability that there will be no defects in 50 feet of wire? (.4493) b. What is the probability that the first defect occurs between the 25th and 50th foot of wire produced? (.2210) c. What is the probability that the 6th defect comes before the 350th foot of wire produced? Use fnInt. (.4881)

Answer 1

inter-defect distance =500/8 =62.5 feet = $\beta$

1) hence   probability that there will be no defects in 50 feet of wire =P(X>50) =e^{-x/ $\beta$} =e^-50/62.5 =0.4493

2)probability that the first defect occurs between the 25th and 50th foot of wire produced=P(25<X<50)

=e^-25/62.5 -e^-50/62.5) =0.2210

3) expected number of failure in 350 foot =8*350/500=5.6 = $\lambda$

hence  probability that the 6th defect comes before the 350th foot of wire produced

=1-P(at most 5 defect in 350 foot) =1- $\sum_{x=0}^{5}e^{-5.6}5.6^{x}/x!$ =1-0.5119 =0.4881