Solve the problem. In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of $1050 \mathrm{kWh}$ and a standard deviation of $218 \mathrm{kWh}$. If 50 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than $1075 \mathrm{kWh}$.

Step 1 ：We are given a problem of probability involving normal distribution. The mean and standard deviation of the energy consumption levels are given as \(1050 \mathrm{kWh}\) and \(218 \mathrm{kWh}\) respectively. We are asked to find the probability that the mean energy consumption level of 50 randomly selected homes is greater than \(1075 \mathrm{kWh}\).

Step 2 ：The first step is to calculate the standard error of the mean (SEM), which is the standard deviation divided by the square root of the sample size. In this case, the standard deviation is \(218 \mathrm{kWh}\) and the sample size is 50. Therefore, the SEM is \(\frac{218}{\sqrt{50}} \approx 30.83\).

Step 3 ：Next, we calculate the z-score, which is the difference between the sample mean and the population mean, divided by the SEM. The sample mean is \(1075 \mathrm{kWh}\), the population mean is \(1050 \mathrm{kWh}\), and the SEM is \(30.83\). Therefore, the z-score is \(\frac{1075 - 1050}{30.83} \approx 0.81\).

Step 4 ：Finally, we use the z-score to find the probability. The z-score tells us how many standard deviations away from the mean our data point is. We can use a z-table or a statistical function to find the probability associated with this z-score. The probability that the mean energy consumption level for September is greater than \(1075 \mathrm{kWh}\) for 50 randomly selected homes is approximately 0.209.

Step 5 ：Final Answer: The probability that the mean energy consumption level for September is greater than \(1075 \mathrm{kWh}\) for 50 randomly selected homes is approximately \(\boxed{0.209}\).