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 1919 minutes and a standard deviation of 33 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 1

a) 36.9% of customers receive the service for​ half-price

b) The automotive center should make 21.28 minutes, the guaranteed time​ limit.

Explanation:

We are given the following information in the question:

Mean, μ = 19 minutes

Standard Deviation, σ = 3 minutes.

We are given that the distribution of time required is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) P(service will take no longer than 20 minutes)

P(x < 20)

`P( x < 20) = P( z < \displaystyle(20 - 19)/(3)) = P(z < 0.3334)`

Calculation the value from standard normal z table, we have,

`P(x < 20) = 0.631 = 63.1\%`

If it does take​ longer than 20 minutes, the customer will receive the service for​ half-price.

Customers receive the service for​ half-price

=
`100\%-63.1\% = 36.9\%`

b) We have to find the value of x such that the probability is 0.77.

P(X < x)

`P( X < x) = P( z < \displaystyle(x - 19)/(3))=0.77`

Calculation the value from standard normal z table, we have,

`P(z < 0.739) = 0.77`

`\displaystyle(x - 19)/(3) = 0.739\\x = 21.28`

Hence, the automotive center should make 21.28 minutes, the guaranteed time​ limit.