We need to find the probability that a rod's length is within 0.5 cm of the mean, i.e., between 24.7 cm and 25.7 cm.

Using the standardization formula, we calculate:

\(P\left( \frac{25.2 - (25.2 + 0.5)}{0.4} < z < \frac{25.2 - (25.2 - 0.5)}{0.4} \right)\)

This simplifies to:

\(P\left( \frac{-0.5}{0.4} < z < \frac{0.5}{0.4} \right) = P(-1.25 < z < 1.25)\)

Using the standard normal distribution table, we find:

\(P(-1.25 < z < 1.25) = 2 \Phi(1.25) - 1\)

\(\Phi(1.25) \approx 0.8944\)

Thus, \(2 \times 0.8944 - 1 = 0.7888\)

The expected number of rods is:

\(0.7888 \times 500 = 394.4\)

Rounding gives 394 or 395 rods.