Question 1022899
 
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<k<64;&lambda;)=P({59,60,61,62,63};&lambda;)
=0.04555+0.04783+0.04940+0.05020+0.05020
=0.2432