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?

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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