Need help.

Question

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with the mean of 17 minutes and a standard deviation of 2 minutes.

a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price?

b) If the automotive center does not want to give the discount to more than 2% of its customers, how long should it make the guaranteed time limit?

Answer 1

Question a) is similar question to the other question you posted. Use the z-score formula to plug in X = 20, μ = 17, and σ = 2, and then, you get the z-score for X = 20 minutes. Again, it is longer than 20 minutes, the percentage (probability) of customers who receive the service for half-price is the proportion larger than the z-score that you computed.

Question b) asks you to think backward. It means, you are given percentage first (2%), and then, you can find a z-score that has the proportion (2%=0.02) larger than it as given in the question. Then, you can use the unit normal table to find first the closest proportion to 0.0200 in tail area of the distribution. Now, you can find the corresponding z-score for the proportion you just found. Plugging the z-score you found and μ = 17 and σ = 2 in the z-score formula, then, you have an equation with X is only value unknown. Then, you can solve the equation for X so that you can find the final answer for this question.