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

Answer 1

a)

Given: $\mu$ = 19, $\sigma$ = 2

To find the probability, we need to find the Z scores first.

Z = (X - $\mu$ )/ [ $\sigma$ /Sqrt(n)]. Since n = 1, Z = (X - $\mu$ )/ $\sigma$

(a) P(X > 20) = 1 - P(X < 20) ), as the normal tables give us the left tailed probability only.

For P( X < 20), Z = (20 – 19)/2 = 0.50

The probability for P(X < a) from the normal distribution tables is = 0.6915

Therefore the required probability = 1 – 0.6915 = 0.3085. Therefore 30.85% of the customers receive the service for half the price.

(b) P(X > x) = 0.03 (Since it wants to restrict it to not more than 3%)

Therefore P(X < x) = 1 - 0.03 = 0.97 and the corresponding Z value is 1.88

Therefore 1.88 = (X - 19)/2. Solving for X, we get X = 1.88*2 + 19 = 22.76 $\approx$ 23 minutes

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