he time required for an automotive center to complete an oil change service on an automobile approximately follows a normal​ distribution, with a 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 3​% of its​ customers, how long should it make the guaranteed time​ limit?

Answer:A) 6.68%; B) No more than 13.24 minutes, or 13, to the nearest whole minute.Step-by-step explanation:For part A,We find the z score associated with 20 minutes to answer this.  Then we will look up the z score and find the area under the curve less than or equal to this.[tex]z=\frac{X-\mu}{\sigma}\\\\=\frac{20-17}{2}=\frac{3}{2}=1.5[/tex]From the z table, we see that the area under the curve less than z = 1.5 is 0.9332 = 93.32%.  This is the percentage of drivers whose oil change takes less than or equal to 20 minutes.  To find the number that go over and receive the half-price discount, we subtract from 1.00:1-0.9332 = 0.0668 = 6.68%.For part B,We work backward on this problem.  First we convert 3% to a decimal:  3% = 3/100 = 0.03.Next we find the closest value to 0.03 in the cells of our z table.  The closest is 0.0301; this corresponds with a z-score of -1.88.  We substitute this into our z-score formula:[tex]z=\frac{X-\mu}{\sigma}\\\\-1.88=\frac{X-17}{2}[/tex]Multiply both sides by 2:[tex]2(-1.88)=\frac{X-17}{2}\times 2\\\\-3.76 = X-17[/tex]Add 17 to each side:-3.76+17 = X-17+1713.24 = X13.24 minutes is the cutoff.