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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?

Respuesta :

Solution :

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 = [tex]$\frac{1.6-1.4}{0.7 / \sqrt{50}}$[/tex]

                                           = 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.

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