A vending machine is designed to dispense a mean of $7.8 \mathrm{oz}$ of coffee into an $8-0 \mathrm{z}$ cup. If the standard deviation the amount of coffee dispensed is $0.2 \mathrm{oz}$ and the amount is normally distributed, determine the percent of times th machine will dispense more than $7.3 \mathrm{oz}$. Click here to view page 1 of the standard normal distribution table Click here to view page 2 of the standard normal distribution table. $\square \%$ of the time the machine will dispense more than $7.3 \mathrm{oz}$ (Type an integer or decimal rounded to two decimal places as needed.)

Step 1 ：The problem is asking for the percentage of times the machine will dispense more than 7.3 oz of coffee. This is a problem of normal distribution. We know that the mean is 7.8 oz, the standard deviation is 0.2 oz, and we want to find the probability of getting more than 7.3 oz.

Step 2 ：To solve this, we need to convert the value 7.3 oz to a z-score, which is a measure of how many standard deviations an element is from the mean. The formula for the z-score is: \(z = \frac{X - \mu}{\sigma}\) where X is the value we are interested in (7.3 oz), \(\mu\) is the mean (7.8 oz), and \(\sigma\) is the standard deviation (0.2 oz).

Step 3 ：Let's calculate the z-score first. Using the values X = 7.3, \(\mu\) = 7.8, and \(\sigma\) = 0.2, we find that \(z = -2.5\). This means that 7.3 oz is 2.5 standard deviations below the mean.

Step 4 ：Now we need to find the probability associated with this z-score. However, the standard normal distribution table usually gives the probability for less than a certain z-score, and we want the probability for more than a certain z-score. So we need to subtract the probability we find from 1 to get the correct answer.

Step 5 ：The probability associated with a z-score of -2.5 is approximately 0.9938. This means that approximately 99.38% of the time, the machine will dispense more than 7.3 oz of coffee.

Step 6 ：Final Answer: \(\boxed{99.38\%}\) of the time the machine will dispense more than 7.3 oz.