Question

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.

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