A study was conducted of college students that stated the true average number of units that a college student enrolls in each semester is 12.7 with a true standard deviation of 6.8. Suppose you randomly sampled 40 college students and recorded the number of units that each of them were enrolled in. Use this information to answer the following question.

What is the probability that the average number of units those students from your sample were enrolled in was more than 11.3.? Make sure to round your answer to 2 decimal places; i.e. if your answer was 0.54321 then you would type in 0.54 .

Step 1 ：Given values are: population mean (\(\mu\)) is 12.7, population standard deviation (\(\sigma\)) is 6.8, sample size (\(n\)) is 40, and sample mean (\(\bar{x}\)) is 11.3.

Step 2 ：Calculate the z-score using the formula \(z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\).

Step 3 ：Substitute the given values into the formula to get \(z = \frac{11.3 - 12.7}{6.8 / \sqrt{40}}\), which simplifies to \(z = -1.3021143306575667\).

Step 4 ：Calculate the probability using the cumulative distribution function (CDF) of the normal distribution. The probability that the z-score is less than or equal to the calculated z-score is given by the CDF, denoted as \(P(Z \leq z)\).

Step 5 ：However, we want the probability that the z-score is more than the calculated z-score, which is given by \(P(Z > z) = 1 - P(Z \leq z)\).

Step 6 ：Substitute the calculated z-score into the formula to get \(P(Z > -1.3021143306575667) = 1 - P(Z \leq -1.3021143306575667)\).

Step 7 ：Using the CDF of the normal distribution, we find that \(P(Z \leq -1.3021143306575667)\) is approximately 0.1.

Step 8 ：Substitute this value into the formula to get \(P(Z > -1.3021143306575667) = 1 - 0.1\), which simplifies to \(P(Z > -1.3021143306575667) = 0.9\).

Step 9 ：Round the probability to 2 decimal places to get 0.90.

Step 10 ：Final Answer: The probability that the average number of units those students from your sample were enrolled in was more than 11.3 is \(\boxed{0.90}\).