Melissa Burkhalter $\quad 10 / 18 / 238: 41 \mathrm{PM}$ ons of the Normal Question 18, 7.2.51 HW Score: $71.67 \%, 14.33$ of 20 points Part 1 of 2 Points: 0.5 of 1 Save 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 15 minutes and a standa deviation of 4 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? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The percent of customers that receive the service for half-price is $\square \%$ (Round to two decimal places as needed) example Get more help. Clear all Chick answer
Solution
Step 1 ：The problem is asking for the percentage of customers who receive the service for half-price, which means the service took longer than 20 minutes. This is a problem of finding the area under the curve of a normal distribution to the right of a certain value. In this case, the value is 20 minutes, the mean is 15 minutes, and the standard deviation is 4 minutes.
Step 2 ：We need to convert the value to a z-score. The formula for calculating the z-score is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 3 ：Substituting the given values into the formula, we get \(Z = \frac{20 - 15}{4} = 1.25\).
Step 4 ：The z-score of 1.25 corresponds to the percentage of 10.56 in the standard normal distribution table, which means that approximately 10.56% of customers receive the service for half-price.
Step 5 ：Final Answer: The percent of customers that receive the service for half-price is approximately \(\boxed{10.56\%}\).
