A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 161.9-cm and a standard deviation of 1.8-cm.

To find the probability that the length of a randomly selected steel rod lies between 164.4 cm and 166.6 cm, we will use the properties of the normal distribution and the concept of z-scores.

First, we will convert the lengths to z-scores using the formula:

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

where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Given:
- Mean (\(\mu\)) = 161.9 cm
- Standard deviation (\(\sigma\)) = 1.8 cm

### Step 1: Calculate the z-score for 164.4 cm

\[
z_1 = \frac{(164.4 - 161.9)}{1.8} = \frac{2.5}{1.8} \approx 1.3889
\]

### Step 2: Calculate the z-score for 166.6 cm

\[
z_2 = \frac{(166.6 - 161.9)}{1.8} = \frac{4.7}{1.8} \approx 2.6111
\]

### Step 3: Use a standard normal distribution table (or calculator) to find the probabilities

We will find the area to the left of both z-scores.

Using standard normal distribution tools (such as a z-table or a calculator):

- For \(z_1 \approx 1.389\), the probability \(P(Z < 1.389) \approx 0.9177\)
- For \(z_2 \approx 2.611\), the probability \(P(Z < 2.611) \approx 0.9955\)

### Step 4: Calculate the probability that the length is between 164.4 cm and 166.6 cm

\[
P(164.4 < X < 166.6) = P(Z < 2.611) - P(Z < 1.389)
\]

\[
P(164.4 < X < 166.6) \approx 0.9955 - 0.9177 = 0.0778
\]

Thus, the probability that the length of a randomly selected steel rod is between 164.4 cm and 166.6 cm is:

\[
\boxed{0.0778}
\]