A radio station claims that the amount of advertising per hour of broadcast time has an average of 15 minutes and standard deviation equal to 2.5 minutes. You listen to the radio station for 1 hour, at a randomly selected time, and carefully observe that the amount of advertising time is equal to 7 minutes. Calculate the $z$-score for this amount of advertising time. Round your answer to two decimal places as needed. A. -3.20 B. -20.00 C. 0.30 D. 3.20

Step 1 ：The problem provides the following information: the average amount of advertising per hour of broadcast time is 15 minutes, the standard deviation is 2.5 minutes, and the observed amount of advertising time is 7 minutes.

Step 2 ：We are asked to calculate the $z$-score for the observed advertising time. The $z$-score is a measure of how many standard deviations an element is from the mean.

Step 3 ：The formula for the $z$-score is: \( z = \frac{X - \mu}{\sigma} \), where $X$ is the observed value, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step 4 ：Substitute the given values into the formula: $X = 7$ minutes, $\mu = 15$ minutes, and $\sigma = 2.5$ minutes.

Step 5 ：Calculate the $z$-score: \( z = \frac{7 - 15}{2.5} = -3.2 \)

Step 6 ：Final Answer: The $z$-score for the observed advertising time of 7 minutes is \(\boxed{-3.20}\).