A roll of plastic-coated wire has an average of 0.09 flaws per 5-meter length of

Question

A roll of plastic-coated wire has an average of 0.09 flaws per 5-meter length of wire. Suppose a quality control engineer will sample a 5-meter length of wire from a roll of wire 230 meters in length. If no flaws are found in the sample, the engineer will accept the entire roll of wire.

What is the probability that the roll will be rejected?

Before examining the sample, what is the probability that there are no flaws in the 230 meters of wire? What is the probability that there are exactly 3 flaws in the entire roll?

Based on your answers from parts a and b, what is the probability that if there is at least one flaw in the entire roll, a randomly sampled 5-meter length of wire from that roll will have at least one flaw? [Hint: It may behelpful to recognize that if the roll has no flaws, the 5-meter length of wire will have no flaws.]

Given that no flaws were found in the sample, what is the probability that the entire roll has no flaws? Is sampling 5 meters of wire sufficient for determining if the entire roll has flaws? Why or why not?

Now assume that if there is at least one flaw in the 230-meter roll of wire, then at least one flaw appears in a randomly sampled 5-meter length of wire with probability 0.99. With this new information, what is the probability that the entire roll has no flaws given that no flaws were found in the sample?

Comment briefly on the difference in answers between part d and part e. What important assumption regarding independence makes these answers different? Which assumption do you think is more realistic? Why?

Explanation / Answer

0.09 flaws for 5 m length

For 230 m flaws = 230(0.09)/5 = 4.14

Thus flaws average for 230 m is Poisson with mean 4.14

P(rejected) = 1-P(x=0) = 1-0.0159

= 0.9841

P(x=0) = 0.0159

P(X=3) = 0.1883

P(x=1 for 5 m length/x=1 for 230 m length)

=0.0860/0.9841=0.0874

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