SOLUTION: THE MEAN NUMBER OF DEFECTIVE PRODUCTS PRODUCED IN A FACTORY IN ONE DAY IS 21.WHAT IS THE PROBABILITY THAT IN A SPAN OF 3 DAYS THERE WILL BE MORE THAN 58 BUT LESS THAN 64 DEFECTIVE
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Question 1022899: THE MEAN NUMBER OF DEFECTIVE PRODUCTS PRODUCED IN A FACTORY IN ONE DAY IS 21.WHAT IS THE PROBABILITY THAT IN A SPAN OF 3 DAYS THERE WILL BE MORE THAN 58 BUT LESS THAN 64 DEFECTIVE PRODUCTS
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
THE MEAN NUMBER OF DEFECTIVE PRODUCTS PRODUCED IN A FACTORY IN ONE DAY IS 21.WHAT IS THE PROBABILITY THAT IN A SPAN OF 3 DAYS THERE WILL BE MORE THAN 58 BUT LESS THAN 64 DEFECTIVE PRODUCTS
Solution:
Assuming the number of defective products of each day is independent of previous days, the Poisson distribution may be used to model this situation.
Average number of defects in 3 days = 3*21=63=λ.
P(k;λ)=λ^k*e^(-λ)/k!
We need to calculate
P(58
=0.04555+0.04783+0.04940+0.05020+0.05020
=0.2432
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