Question

The length of time required for the periodic maintenance of an automobile or another machine usually has a mound-shaped probability distribution. Because some occasional long service times will occur, the distribution tends to be skewed to the right. Suppose that the length of time required to run a 5000-mile check and to service an automobile has mean 1.4 hours and standard deviation .7 hour. Suppose also that the service department plans to service 50 automobiles per 8-hour day and that, in order to do so, it can spend a maximum average service time of only 1.6 hours per automobile. On what proportion of all workdays will the service department have to work overtime?

Answer 1

Mean time for an automobile to run a 5000 mile check and service = 1.4 hours

Standard deviation = 0.7 hours

Maximum average service time = 1.6 hours for one automobile

The z - score for 1.6 hours =
`$(1.6-1.4)/(0.7 / √(50))$`

= 2.02

Now checking a normal curve table the percentage of z score over 2.02 is 0.0217

Therefore the overtime that will have to be worked on only 0.217 or 2.017% of all days.