The auto parts department of an automotive dealership sends out a mean of 5.45.4 special orders daily. What is the probability that, for any day, the number of special orders sent out will be less than 33? Round your answer to four decimal places.
Answer:
Given,
Mean
= 5.4
Consider,
Poisson distribution
P(X = x) = e^-
*
^x/x!
P(X < 3) = P(0) + P(1) + P(2)
= e^-5.4*5.4^0/0! + e^-5.4*5.4^1/1! + e^-5.4*5.4^2/2!
= 0.0045 + 0.0244 + 0.0659
= 0.0948
The auto parts department of an automotive dealership sends out a mean of 5.45.4 special orders...
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