Answer to Question #218504 in Statistics and Probability for Yasmin

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 2 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

(a) $P(x>20)=P(Z>\frac{20-19}{2})=P(Z>0.5)=1-P(Z<0.5)=0.3085=30.85\%.$

(b) $P(Z>z)=0.02\to P(Z<z)=0.98\to z=2.05.$

$\frac{x-19}{2}=2.05.$

$x=19+2.05*2=23.1\; min.$