A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of $258.2 \mathrm{~cm}$ and a standard deviation of $2.2 \mathrm{~cm}$. For shipment, 7 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is more than $259.1 \mathrm{~cm}$. Round your answer to at least three decimals.

Step 1 ：Given that the lengths of the steel rods are normally distributed with a mean of \(258.2 \mathrm{~cm}\) and a standard deviation of \(2.2 \mathrm{~cm}\).

Step 2 ：For shipment, 7 steel rods are bundled together. We are asked to find the probability that the average length of a randomly selected bundle of steel rods is more than \(259.1 \mathrm{~cm}\).

Step 3 ：First, we calculate the standard deviation of the average length, which is the standard deviation divided by the square root of the number of rods in a bundle. This gives us \( \frac{2.2}{\sqrt{7}} = 0.8315218406202999 \mathrm{~cm}\).

Step 4 ：Next, we calculate the z-score for \(259.1 \mathrm{~cm}\), which is the difference between this value and the mean, divided by the standard deviation of the average length. This gives us \( \frac{259.1 - 258.2}{0.8315218406202999} = 1.082352809071919 \).

Step 5 ：Finally, we find the probability that the z-score is more than the calculated z-score. This is \(1 - \) the cumulative distribution function (CDF) of the calculated z-score, which gives us \(1 - \) CDF(\(1.082352809071919\)) \(= 0.1395478944104951\).

Step 6 ：Rounding to three decimal places, the final answer is \(\boxed{0.140}\).