A company produces optical-fiber cable with a mean of 0.7 flaws per 100 feet. What is the probability that there will be exactly 3 flaws in 1000 feet of cable? Round your answer to four decimal places.
+6248
\(\mbox{This is a Poisson distribution problem.}\\ \mbox{With 0.7 flaws per 100 feet we will have 7 flaws per 1000 feet.}\)
\(\mbox{So the parameter }\lambda = 7\\ \mbox{We evaluate the Poisson distribution PDF at 3} \\ P[\mbox{3 flaws in 1000 feet}]=\dfrac {7^3} {3!}e^{-7} = 0.05213\)
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