a radio station claims that the amount of advertising per hour of broadcast time has an average of 17 minutes and a standard deviation equal to 2.7 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 11 minutes. Calculate the z-score for this amount of advertising time

Answer:z-score for 11 minutes of advertising time is [tex]z=\frac{-6}{2.7}\approx -2.222[/tex]Step-by-step explanation:Z-scores measure the distance of any data point from the mean in units of standard deviations and are useful because they allow us to compare the relative positions of data values in different samples.The z-score for any single data value can be found by the formula:[tex]z=\frac{data \:value- \:mean}{standard \:deviation}[/tex]From the information given we know:Data value = 11 minutesMean = 17 minutesStandard deviation = 2.7 minutesSo[tex]z=\frac{11-17}{2.7} = \frac{-6}{2.7}\approx -2.222[/tex]