Answered

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 1717 minutes and a standard deviation of 2.52.5 minutes. ​(a) the automotive center guarantees customers that the service will take no longer than 2020 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 77​% of its​ customers, how long should it make the guaranteed time​ limit?

Answer :

shinmin
Let X be the time or period needed for an automobile center to finish an oil change.

X ∼ N (17, 2.5) 

a) 

P(X ≥ 20) = P((X - 17)/5 ≥ (20- 17)/2.5) = P(Z ≥ 1.2) = .1151 *100 = 11.51% is the answer

b) 
P(X ≥ x) = 0.07 

P (X - 17)/2.5 ≥ (x - 17)/2.5) = 0.07

P (Z ≥ z) = 0.07
look at the z table, 0.07 lies between 1.47 and 1.48, add and then divide you'll get:z = 1.475

1.475 = (x - 17)/2.5 

x = 20.6875 ≈ 21 minutes

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