Question

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?

Answer 1

Answer:

38.3 minutes

Step-by-step explanation:

Waiting time at the department is normally distributed.

Mean waiting time = u = 30 minutes

Standard Deviation = $\sigma$ = 8 minutes

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:

$z=\frac{x-u}{\sigma}$

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

$1.04=\frac{x-30}{8}\\\\ 8.32=x-30\\\\ x=38.32\\\\ x=38.3$

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