Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes. Suppose that in an effort to provide better service to the public, the director of the local office is permitted to provide discounts to those individuals whose waiting time exceeds a predetermined time. The director decides that 15 percent of the customers should receive this discount. What number of minutes do they need to wait to receive the discount?

The top 15% of the customer will receive a discount. We need to find the number of minutes above which only 15% of the customers wait.In other words we can say, we need to find the time below which 85% of the customers wait.

Since, 85% of the values will be below this point, this point will be the 85th percentile.

In order to solve this problem we can use the concept of z scores. We can find the z score which represents the 85th percentile for a Normal Distribution and using that z score we can find an equivalent time in minutes.

From the z table, the z score associated with 85th percentile or a probability of 0.85 under the standard normal curve is z = 1.04

The formula to calculate the z score is:

[tex]z=\frac{x-u}{\sigma}[/tex]

Using the values, we can find x, the time that will separate the lower 85% from the top 15%.

[tex]1.04=\frac{x-30}{8}\\\\ 8.32=x-30\\\\ x=38.32\\\\ x=38.3[/tex]

This means, the customers need to wait 38.3 minutes to get a discount.