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

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 7​% of its​ customers, how long should it make the guaranteed time​ limit?

Answer 1

A)

µ =    19                  
σ =    3                  
                      
P ( X ≥   20.00   ) = P( (X-µ)/σ ≥ (20-19) / 3)              
= P(Z ≥   0.333   ) = P( Z <   -0.333   ) =    0.3694  

= 36.94% (answer)

B)

µ=   19                  
σ =    3                  
proportion=   0.93                  
                      
Z value at    0.93   =   1.48   (excel formula =NORMSINV(   0.93   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.48   *   3   +   19  
X   =   23.43   (answer)          

Thanks in advance!

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