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 3 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 2% 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 37.07%. (round to two decimal places as needed.)
(b) the guaranteed time limit is 13 minutes. (round up to the nearest integer as needed.)

Explanation:

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

For a normal distribution, if we want to find the percentage of customers that receive the service for half - price (i.e., the time is more than 20 minutes). The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 19\) (mean), \(\sigma=3\) (standard deviation), and \(x = 20\).

$$z=\frac{20 - 19}{3}=\frac{1}{3}\approx0.33$$

Using the standard normal table (or a calculator with a normal distribution function), \(P(X>20)=1 - P(X\leq20)\). From the standard normal table, \(P(Z\leq0.33) = 0.6293\), so \(P(X>20)=1 - 0.6293=0.3707 = 37.07\%\)

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

We want to find the value of \(x\) such that \(P(X>x)=0.02\). Then \(P(X\leq x)=1 - 0.02 = 0.98\). Looking up in the standard normal table, the z - score corresponding to a cumulative probability of \(0.98\) is approximately \(z = 2.05\) (using the formula \(z=\frac{x-\mu}{\sigma}\)).
We know \(\mu = 19\) and \(\sigma = 3\). Rearranging the formula \(z=\frac{x-\mu}{\sigma}\) for \(x\), we get \(x=\mu+z\sigma\)
Substitute \(\mu = 19\), \(z = 2.05\), and \(\sigma = 3\) into the formula:

$$x=19+2.05\times3=19 + 6.15=25.15\approx25$$

Answer:

(a) \(37.07\%\)
(b) \(25\) minutes