Question

The time required for an automotive center to complete an oil change service on an automobile...

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 15 minutes and a standard deviation of 4 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) The percent of customers that receive the service for​ half-price is 1.06​%.

​(Round to two decimal places as​ needed.)

​(b) The guaranteed time limit is_minutes.

Homework Answers

Answer #1

Solution :

Given that ,

mean = = 15 minutes

standard deviation = = 4 minutes

a) P(x < 20) = P[(x - ) / < (20 - 15) / 4]

= P(z < 1.25)

Using z table,

= 0.8944

The percentage is = 89.44%

b) Using standard normal table,

P(Z > z) = 3%

= 1 - P(Z < z) = 0.03  

= P(Z < z) = 1 - 0.03

= P(Z < z ) = 0.97

= P(Z < 1.88 ) = 0.97  

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 4 + 15

x = 22.52 minutes

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