What Is the Probability of Zero Cracks in 5 Miles of Highway?

AI Thread Summary
The discussion centers on calculating the probability of zero cracks requiring repair in 5 miles of highway, assuming a Poisson distribution with a mean of two cracks per mile. Some participants question the validity of using a Poisson model, arguing that the independence of observations may not hold due to factors like the uniformity of pavement conditions. They suggest that defects might follow a different distribution based on the method of pavement laying and environmental conditions. For those pursuing the Poisson approach, the probability of zero defects in one mile can be calculated and then raised to the fifth power for the total distance. The conversation highlights the complexities and assumptions involved in statistical modeling for highway maintenance.
bartowski
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The number of cracks in a section of interstate highway that are significant enough to require repair is assumed
to follow a Poisson distribution with a mean of two cracks per mile. What is the probability that there are no cracks that require repair in 5 miles of highway?

any help guys? :)
 
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bartowski said:
The number of cracks in a section of interstate highway that are significant enough to require repair is assumed
to follow a Poisson distribution with a mean of two cracks per mile. What is the probability that there are no cracks that require repair in 5 miles of highway?

any help guys? :)

Just because the mean is a small number doesn't mean it's a Poisson process. In addition it's hard to make the argument that the observations are independent. The stretch of pavement was probably laid at the same time by the same method. If there are defects, I would expect the probability of defects has a uniform distribution along the length of the road under constant conditions. For inconstant conditions, you can't assume a constant distribution.

As an a academic exercise in irrelevant statistics, you would find the probability of 0 defects in one mile under the Poisson distribution with a mean of 2 and raise that value to the fifth power. It has nothing to do with the specific problem you described.
 
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