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 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?

Answer 1

Solution :

Given that ,

a) P(x < 20)

= P[(x - ) / < (20 - 19) / 3.5]

= P(z < 0.29)

Using z table,

= 0.6141

percent = 61.41%

b) Using standard normal table,

P(Z < z) = 2%

= P(Z < z ) = 0.02

= P(Z < -2.05 ) = 0.02  

z = -2.05

Using z-score formula,

x = z * +

x = -2.05 * 3.5 + 19

x = 11.8 minutes