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?

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.

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:

Multiply both sides by 2:

Add 17 to each side:

-3.76+17 = X-17+17

13.24 = X

13.24 minutes is the cutoff.

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629