I need to find the probability of 320 or more flaws in a 10km length of fibre optic cable. There is an average of 1.5 flaws per 50m.
I could solve this using:
P(X≥320) = 1 - [P(X=0) + P(X=1) + ... + P(X=319)]
But this would be extremely long and tedious to do, plus it wouldn't be an appropriate method.
Is there any other way I could get to my answer in a appropriate method?
(post links if necessary)
An average of 1.5 flaw per 50m is 3 flaws per 100 m so 30 flaws per km and 300 flaws per 10 km.
With a Poisson distribution with parameter [itex]\lambda=300[/itex] has mean and standard deviation [itex]300[/itex] so, like any distribution with finite mean and standard deviation, can be approximated by a normal distribution with that same mean and standard deviation. That is, we can approximate this by the normal distribution with mean and standard deviation 300. The probability x is greater than 320 can be changed to the probability that x is greater than 319.5 (since the normal distribution is continuous, we use the "half integer correction" treating any number from 319.5 to 320.5 as "320"). That converts to the standard normal distribution . Now look it up in a table of the normal distribution, such as the one at http://www.stat.duke.edu/~banks/111-...mDistTable.pdf. Since that gives the probability a value is less than or equal to z, subtract the value in the table from 1.