A quality inspector at a glass manufacturing company inspect

A quality inspector at a glass manufacturing company inspects sheets of glass to check for any imperfections. The number of flaws occur according to a Poisson distribution with an average of 0.05 flaws per sheet. A shipment of 10 sheets will be rejected if 2 or more flaws are found. Define the variable of interest for a shipment and give the parameter (lambda). What is the probability a shipment will be rejected? How many flaws would you expect to find in a shipment?

Solution

A)

Variable of interest: number of flaws in a shipment of 10 sheets [ANSWER]

lambda = 0.05*10 = 0.5 [ANSWER]

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b)

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 =    0.5      
          
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.90979599
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.09020401 [answer]

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c)

The expectation value is the parameter, lambda.

Thus,

E(x) = lambda = 0.5 [ANSWER]