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 17 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 3% of its customers, how long should it make the guaranteed time limit?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 17

standard deviation = $\sigma$ =2.5

P(x <20 ) = P(( x -$\mu$) / $\sigma$ (20 - 17) /2.5 )

= P(z < 1.2)

Using z table

= 0.8849

=88.49%

(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 $\sigma$ + $\mu$

x =1.88 *2.5+17

x=21.7