Due to random variations in the operation of an automatic coffee machine, not every cup is filled with the same amount of coffee. Assume that the mean amount of coffee dispensed is 6 ounces and the standard deviation is 0.5 ounce. Use the 68-95-99.7 rule to complete the following. a. What percent of the cups should have less than 6 ounces of coffee? b. What percent of the cups should have less than 5.5 ounces of coffee?

Step 1 ：The problem is about the distribution of coffee dispensed by an automatic coffee machine. The amount of coffee dispensed follows a normal distribution with a mean of 6 ounces and a standard deviation of 0.5 ounce.

Step 2 ：We are asked to find the percentage of cups that should have less than 6 ounces and less than 5.5 ounces of coffee respectively.

Step 3 ：According to the properties of a normal distribution, 50% of the data lies below the mean and 50% lies above it. Therefore, \(\boxed{50\%}\) of the cups should have less than 6 ounces of coffee.

Step 4 ：To find the percentage of cups that should have less than 5.5 ounces of coffee, we need to calculate the percentage of data that falls within 1 standard deviation below the mean.

Step 5 ：According to the 68-95-99.7 rule, 68% of the data falls within 1 standard deviation of the mean. Since this is symmetric about the mean, 34% of the data falls within 1 standard deviation below the mean.

Step 6 ：Therefore, the percentage of cups that should have less than 5.5 ounces of coffee is 50% - 34% = \(\boxed{16\%}\).