Let X be the number of material anomalies occurring in a particular region of an

Question

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodology for Probabilistic Life Prediction of Multiple-Anomaly Materials"† proposes a Poisson distribution for X. Suppose that = 4. (Round your answers to three decimal places.)

(a) Compute both P(X 4)and P(X < 4).

(b) Compute P(4 X 5).

(c) Compute P(5 X).

(d) What is the probability that the number of anomalies does not exceed the mean value by more than one standard deviation?

Explanation / Answer

a)

Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    4      
          
x = the maximum number of successes =    4      
          
Then the cumulative probability is          
          
P(x<=4) =    0.628836935 [ANSWER]

**********

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 =    4      
          
x = our critical value of successes =    4      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   3   ) =    0.43347012
          
Which is also          
          
P(x<4) =    0.43347012 [ANSWER]

******************************************

b)

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 =    4      
          
          
Then          
          
P(at most    3   ) =    0.43347012
P(at most    5   ) =    0.785130387
          
Thus,          
          
P(between 4 and 5) =    0.351660267   [ANSWER]

*************************  

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 =    4      
          
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   ) =    0.628836935
          
Thus, the probability of at least   5   successes is  
          
P(5<=x) =    0.371163065 [ANSWER]

**************************

d)

Here,

sigma = sqrt(u) = sqrt(4) = 2

Hence, we need the probability of between 2 and 6.

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    2      
x2 =    6      
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    4      
          
          
Then          
          
P(at most    1   ) =    0.091578194
P(at most    6   ) =    0.889326022
          
Thus,          
          
P(between x1 and x2) =    0.797747827   [ANSWER]  

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