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 2.5 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: about 66%

b: 25 minutes (24 minutes, 8.25 seconds, but we round up)

Step-by-step explanation:

Part a:

We will find the z-score if the situation.  

µ = 19, σ = 2.5

x = 20.

The z-score is:  z = (20 - 19)/2.5 = 0.40

P(z < 0.40) = 0.6554, so about 66% of customers would receive that discount

Part b:

We want to find the z-score that corresponds to 0.980, (top 2%).  Looking of the chart we have to estimate since the exact value of 0.9800 isn't on the chart.  z = 2.05 gives us 0.9798, and z = 2.06 gives  us 0.9803.  0.9800 is just about in between them, so we average 2.05 and 2.06 giving us 2.055.  We need to find the x-value that gives us a z-score of 2.055

2.055 = (x - 19)/2.5

    5.1375 = x - 19   (multiply both sides by 2.5)

    24.1375 = x     (add 19 to both sides)

   24.1375 minutes is 24 minutes 8.25 seconds.  Rounding  down to 24 minutes would make a little more than 2% get the discount, we need no more than 2%, so rounding up to 25 minutes would satisfy this requirement

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: