A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 124.7-cm and a standard deviation of 2.2-cm.

To find the probability that the length of a randomly selected steel rod is between 119.9 cm and 122.9 cm, we can use the properties of the normal distribution.

Given:
- Mean (\(\mu\)) = 124.7 cm
- Standard deviation (\(\sigma\)) = 2.2 cm

We will first calculate the z-scores for the values 119.9 cm and 122.9 cm using the following formula for z-scores:

\[
z = \frac{X - \mu}{\sigma}
\]

**Calculating the z-score for 119.9 cm:**

\[
z_{119.9} = \frac{119.9 - 124.7}{2.2} = \frac{-4.8}{2.2} \approx -2.1818
\]

**Calculating the z-score for 122.9 cm:**

\[
z_{122.9} = \frac{122.9 - 124.7}{2.2} = \frac{-1.8}{2.2} \approx -0.8182
\]

Next, we will find the probabilities corresponding to these z-scores using the standard normal distribution table (or a calculator).

1. **Find \(P(Z < -2.1818)\)**:

Using a z-table or calculator:
\[
P(Z < -2.1818) \approx 0.0144
\]

2. **Find \(P(Z < -0.8182)\)**:

Using a z-table or calculator:
\[
P(Z < -0.8182) \approx 0.2061
\]

Now, to find the probability that the length of a randomly selected steel rod is between 119.9 cm and 122.9 cm, we subtract the two probabilities:

\[
P(119.9 < X < 122.9) = P(Z < -0.8182) - P(Z < -2.1818)
\]
\[
P(119.9 < X < 122.9) = 0.2061 - 0.0144 = 0.1917
\]

Therefore, the probability that the length of a randomly selected steel rod is between 119.9 cm and 122.9 cm is approximately:

\[
\boxed{0.1917}
\]