To find the probability that the length of a randomly selected steel rod is greater than 180.4 cm, we need to calculate the z-score for the length of 180.4 cm using the given mean and standard deviation.
The formula for the z-score is:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
- \(X\) is the value we are interested in (180.4 cm),
- \(\mu\) is the mean (175.6 cm),
- \(\sigma\) is the standard deviation (1.7 cm).
Now, substituting the values:
\[ z = \frac{180.4 - 175.6}{1.7} = \frac{4.8}{1.7} \approx 2.8235 \]
Next, we will find the probability \(P(X > 180.4)\). This is equivalent to finding \(P(Z > 2.8235)\) in the standard normal distribution.
Using a standard normal distribution table or a calculator, we find \(P(Z > 2.8235)\). This is equal to \(1 - P(Z < 2.8235)\).
From the standard normal distribution table or using a calculator, we find:
\[ P(Z < 2.8235) \approx 0.9975 \]
Thus,
\[ P(Z > 2.8235) = 1 - P(Z < 2.8235) \approx 1 - 0.9975 = 0.0025 \]
Finally, the probability that the length of a randomly selected steel rod is greater than 180.4 cm is approximately:
\[ P(X > 180.4) \approx 0.0025 \]
Therefore, the final answer, rounded to 4 decimal places, is:
\[ \boxed{0.0025} \]