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:
\[ \boxed{0.0778} \]
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