Questions: A vending machine is designed to dispense a mean of 7.5 oz of coffee into an 8 -oz cup. If the standard deviation of the amount of coffee dispensed is 0.2 oz and the amount is normally distributed, find the percent of times the machine will dispense from 7.4 oz to 7.8 oz. Suggestion Hint: Use your calculator's Normal Cdf feature. % of the time the machine will dispense from 7.4 oz to 7.8 oz . (Type an integer or decimal rounded to one decimal place as needed.)

Solution Steps

We need to find the probability that a vending machine dispenses between \(7.4\) oz and \(7.8\) oz of coffee, given that the amount dispensed is normally distributed with a mean (\(\mu\)) of \(7.5\) oz and a standard deviation (\(\sigma\)) of \(0.2\) oz.

To find the probability, we first convert the values \(7.4\) oz and \(7.8\) oz into Z-scores using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

For the lower bound (\(X = 7.4\) oz):

\[ Z_{start} = \frac{7.4 - 7.5}{0.2} = -0.5 \]

For the upper bound (\(X = 7.8\) oz):

\[ Z_{end} = \frac{7.8 - 7.5}{0.2} = 1.5 \]

Using the Z-scores, we can find the probability that the machine dispenses between \(7.4\) oz and \(7.8\) oz:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.5) - \Phi(-0.5) \]

From the calculations, we find:

\[ P \approx 0.6247 \]

To express the probability as a percentage:

\[ P \times 100 \approx 62.47\% \]

Rounding to one decimal place gives us:

\[ P \approx 62.5\% \]

Final Answer