July 6, 2015

Homework Help: statistics

Posted by Karinna on Saturday, February 14, 2009 at 1:18am.

Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is 2.4 minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a 50% discount on the charges. The company wants to limit this discount to at most 5% of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.
I worked out the problem and found out that the z-value that corresponds to the required x-value. I found the z value from the normal distribution table for .0500. I used the standard normal distribution table to find out that the z-value is - 1.645. Substituting the values of , standard deviation(), and z in the formula x= +z, we obtain
x = +z = 15 +(-1.64)(2.4)=
15-3.936 = 11.064
I looked in the back of the book and the answer said approximately 19. I understand how they got to approximately 19. They figured out that the z-value was 1.64 and then they plugged it into the formula and that is how they got approx. 19. But I'm having trouble understanding why its 1.64 and not -1.64. If anyone could help me figure out where I'm going wrong I'd appreciate it. Thank you!

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