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 \(\boxed{0.90}\).