The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 19 minutes and a standard deviation of 3.5 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?

a. 61.41%

b. 27 minutes

Step-by-step explanation:

a:  Find the z-score for the situation.  

µ = 19

x-bar = 20

σ = 3.5

  z = (20 - 19)/3.5 = 0.29

The p-value for z = 0.29 is 0.6141, so 61.41% of people will get this discount

b:  They want no more than 2% to get the discount, so they want less than 98% getting the discount.  The z-score for 98% (0.98 as a decimal) is 2.055

*You need to look at the chart and find where 0.98 would be.  It's between a z-score of 2.05 and 2.06.  

The z-score is    2.055 = (x - 19)/3.5      We are solving for the time for this one.  So solve for x...

              7.1925 = x - 19      (multiply both sides by 3.5)

             26.1925 = x    (add 19 to both sides)

So 26.1925 minutes, or about 26 minutes, 12 seconds, so round up to 27 minutes because they want less than 2%.   I chose 27 minutes because no places give odd wait times like 26 minutes 12 seconds.

2x3 – 6x2 – 14x + 24 square centimeters

if i'm wrong i'm sorry : )

other answer is correct. i had used the radius of the radius somehow. a website: