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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?

Respuesta :

Answer:

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

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