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

Answer:

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

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

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

P(x < 20)

$P( x < 20) = P( z < \displaystyle\frac{20 - 19}{3}) = P(z < 0.3334)$

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)  

$P( X < x) = P( z < \displaystyle\frac{x - 19}{3})=0.77$  

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

$P(z < 0.739) = 0.77$

$\displaystyle\frac{x - 19}{3} = 0.739\\x = 21.28$  

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