To solve this problem, we need to calculate the proportion of steel rods that have lengths between $23.9$ cm and $24.1$ cm, given that the lengths are normally distributed with:

• Mean ($\mu$) = $24.24$ cm
• Standard deviation ($\sigma$) = $0.09$ cm

The number of rods within this range is then determined by multiplying the proportion by the total order size ($10,000$).

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Steps to Solve:

1. Convert the lengths to $z$-scores: The formula for the $z$-score is: $z = \frac{x - \mu}{\sigma}$ For $23.9$ cm: $z_1 = \frac{23.9 - 24.24}{0.09} = \frac{-0.34}{0.09} = -3.78$ For $24.1$ cm: $z_2 = \frac{24.1 - 24.24}{0.09} = \frac{-0.14}{0.09} = -1.56$

2. Find the cumulative probabilities: Using the standard normal distribution table:

   • $P(Z < -3.78) \approx 0.00008$
   • $P(Z < -1.56) \approx 0.0594$

   The probability of a rod being between $23.9$ cm and $24.1$ cm is: $P(-3.78 < Z < -1.56) = P(Z < -1.56) - P(Z < -3.78)$ $P(-3.78 < Z < -1.56) = 0.0594 - 0.00008 = 0.05932$

3. Calculate the number of rods: Multiply the proportion of rods in the range by the total number of rods: $\text{Number of rods} = 0.05932 \times 10,000 = 593.2$

   The plant manager should expect to manufacture approximately 593 rods that meet the specification.

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Final Answer:

The plant manager can expect approximately 593 rods to meet the specified length requirements.

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Do you have any questions or need further details?
Here are 5 related questions for deeper exploration:

1. How do we use a standard normal distribution table effectively?
2. What happens to the probability if the standard deviation changes?
3. How is the concept of a $z$-score used in quality control processes?
4. Can we use software to compute these probabilities more accurately?
5. How does increasing the sample size affect the accuracy of the prediction?

Tip: Always double-check $z$-score calculations, especially when working with extremely small probabilities, to ensure accuracy.