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1. The time required for servicing transmissions is normally distributed with   = 45 minutes and   = 8minutes. The service manager plans to have work begin on the transmission of a customer’s car 10 minutes after the car is dropped off  and  the customer is told that the car will be ready within 1 hour from drop-off. What is the probability that the service manager cannot  meet his commitment?

Ans- We have a normal distribution with   = 45 and   = 8.0. Let X  be the amount of time it takes to complete the repair on a customer's car. To finish in one hour you must have X ≤ 50 so the question is to find Pr(X > 50).
Pr(X > 50) = 1 - Pr(X ≤ 50).
Z = (X -  )/  = (X - 45)/8.0
Thus the question can be answered by using the normal table to find
Pr(X ≤ 50) = Pr(Z ≤ (50 - 45)/8.0) = Pr(Z ≤ 0.625)=73.4%
Probability that the service manager will not meet his demand will be = 100-73.4 = 26.6% or 0.2676

2. The current age (in years) of 400 clerical employees at an insurance claims processing center is normally distribute

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