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?

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+17

13.24 = X

13.24 minutes is the cutoff.