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This question is about the Central Limit Theorem and involves calculating the probability of the average length of steel rods in a randomly selected bundle falling within a certain range.

### Problem breakdown:
- **Population mean (μ)** = 136.7 cm
- **Population standard deviation (σ)** = 2.2 cm
- **Sample size (n)** = 21 rods
- **Target range for the mean (M)** = between 136.6 cm and 136.8 cm

We can apply the Central Limit Theorem (CLT) because the sample size is reasonably large. According to the CLT:
- The **sampling distribution** of the sample mean (M) is normally distributed with:
  - **Mean of the sample distribution (μM)** = Population mean (μ) = 136.7 cm
  - **Standard deviation of the sample mean (σM)** = \( \frac{σ}{\sqrt{n}} \)

### Steps to solve:
1. **Calculate the standard deviation of the sample mean**:
   \[
   σ_M = \frac{σ}{\sqrt{n}} = \frac{2.2}{\sqrt{21}} \approx 0.480
   \]

2. **Convert the target values (136.6 cm and 136.8 cm) to z-scores**:
   The z-score formula is:
   \[
   z = \frac{M - μ}{σ_M}
   \]
   - For 136.6 cm:
     \[
     z_1 = \frac{136.6 - 136.7}{0.480} \approx -0.2083
     \]
   - For 136.8 cm:
     \[
     z_2 = \frac{136.8 - 136.7}{0.480} \approx 0.2083
     \]

3. **Find the probability using a standard normal distribution table or a calculator**:
   - The probability corresponding to \( z_1 = -0.2083 \) is approximately 0.4182.
   - The probability corresponding to \( z_2 = 0.2083 \) is approximately 0.5818.

4. **Final probability**:
   The probability that the mean falls between 136.6 cm and 136.8 cm is:
   \[
   P(136.6 < M < 136.8) = P(z_2) - P(z_1) = 0.5818 - 0.4182 = 0.1636
   \]

Thus, the answer is approximately **0.1636**.

Would you like more details or have any questions?

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Here are 5 related questions:
1. How does the Central Limit Theorem apply to non-normal distributions?
2. What happens to the standard deviation of the sample mean as the sample size increases?
3. How would the probability change if the sample size were larger?
4. What is the difference between population standard deviation and sample standard deviation?
5. How do you interpret z-scores in normal distribution tables?

**Tip:** Always check that the sample size is sufficiently large when applying the Central Limit Theorem, especially if the population distribution is not normal.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 136.7 cm and a standard deviation of 2.2 cm. For shipment, 21 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 136.6 cm and 136.8 cm.

Learn how to calculate the probability that the average length of steel rods is within a certain range using the Central Limit Theorem. This problem involves calculating z-scores and applying normal distribution principles, suitable for statistics students.

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