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 2.5 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 7% of its customers, how long should it make the guaranteed time limit?
(a) the percent of customers that receive the service for half - price is
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 19\), \(\sigma=2.5\), and \(x = 20\).
\(z=\frac{20 - 19}{2.5}=\frac{1}{2.5}=0.4\)
Using the standard normal table (or a calculator with a normal - distribution function), \(P(Z\leq0.4)\) is the cumulative probability.
\(P(Z\leq0.4)=0.6554\)
The percentage of customers that receive the service for half - price is \(1 - 0.6554=0.3446\) or \(34.46\%\)

Step2: Find the z - score for part (b)

We want to find \(x\) such that \(P(X>x)=0.07\), then \(P(X\leq x)=1 - 0.07 = 0.93\)
Looking up the z - score in the standard normal table, the z - score \(z\) corresponding to a cumulative probability of \(0.93\) is approximately \(z = 1.48\)
Using the z - score formula \(z=\frac{x-\mu}{\sigma}\), and solving for \(x\) (where \(\mu = 19\) and \(\sigma = 2.5\))
\(1.48=\frac{x - 19}{2.5}\)
\(x-19=1.48\times2.5\)
\(x-19 = 3.7\)
\(x=19 + 3.7=22.7\)

Answer:

a. \(34.46\%\)
b. \(22.7\) minutes