A factory worker productivity is normally distributed. one worker produces an average of 75 units per day with a standard deviat
ion of 20. another worker produces at an average rate of 65 per day with a standard deviation of 21. what is the probability that in 1 week (5 working days), worker 1 will outproduce worker 2?
Let Xi be the random variable representing the number of units the first worker produces in day i. Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5. Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html. Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546 .
