Suppose the quality control manager at Milestone Optics Inc is inspecting 5.000

Question

Suppose the quality control manager at Milestone Optics Inc is inspecting 5.000 foot reels of mono-mode fiber optic cable produced for installation into high speed communication systems throughout the East Coast If the production process is in control, the mean number of defects per unit length of cable is 6 0 What is the probability that in any particular reel being inspected: Less than 5 defects will be found? Exactly 5 defects will be found? Five or more defects will be found? Either four or five defects will be found?

Explanation / Answer

a)

Note that P(fewer than x) = P(at most x - 1).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    6      
          
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   4   ) =    0.2850565
          
Which is also          
          
P(fewer than   5   ) =    0.2850565 [ANSWER]

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b)

Note that the probability of x successes is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    6      
          
x = the number of successes =    5      
          
Thus, the probability is          
          
P (    5   ) =    0.160623141 [ANSWER]

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c)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    6      
          
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   4   ) =    0.2850565
          
Thus, the probability of at least   5   successes is  
          
P(at least   5   ) =    0.7149435 [ANSWER]

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d)

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    4      
x2 =    5      
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    6      
          
          
Then          
          
P(at most    3   ) =    0.151203883
P(at most    5   ) =    0.445679641
          
Thus,          
          
P(between x1 and x2) =    0.294475759   [ANSWER]  

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