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?

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:

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

P(x < 20)

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

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

Customers receive the service for half-price

=

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

P(X < x)  

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

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

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %

answer: 100

step-by-step explanation: