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To solve this problem, we need to find the probability that the average length \( M \) of a randomly selected bundle of 8 steel rods falls between 206.9 cm and 207.7 cm. We are given the following information:

- Mean of the distribution, \( \mu = 208 \) cm
- Standard deviation of the distribution, \( \sigma = 2 \) cm
- Sample size, \( n = 8 \)

Since the lengths of the steel rods are normally distributed, the sampling distribution of the sample mean will also be normally distributed with:

- Mean \( \mu_M = \mu = 208 \) cm
- Standard deviation (standard error), \( \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{8}} \)

Let's calculate \( \sigma_M \):
\[
\sigma_M = \frac{2}{\sqrt{8}} = \frac{2}{2.828} \approx 0.707
\]

Now, we need to standardize the values 206.9 cm and 207.7 cm using the z-score formula:
\[
z = \frac{X - \mu_M}{\sigma_M}
\]

### Step 1: Calculate the z-scores for 206.9 and 207.7

1. For \( X = 206.9 \):
   \[
   z_1 = \frac{206.9 - 208}{0.707} \approx \frac{-1.1}{0.707} \approx -1.556
   \]

2. For \( X = 207.7 \):
   \[
   z_2 = \frac{207.7 - 208}{0.707} \approx \frac{-0.3}{0.707} \approx -0.424
   \]

### Step 2: Find the probability corresponding to these z-scores

We need to find the probability that \( z \) falls between -1.556 and -0.424. Using the standard normal distribution table, we find the corresponding probabilities:

- Probability for \( z_1 = -1.556 \approx 0.0606 \)
- Probability for \( z_2 = -0.424 \approx 0.3364 \)

### Step 3: Calculate the difference

The probability that \( M \) is between 206.9 cm and 207.7 cm is:
\[
P(206.9 < M < 207.7) = 0.3364 - 0.0606 = 0.2758
\]

Thus, the probability is approximately 0.2758.

Would you like further details or have any questions?

Here are 5 related questions:
1. How would the probability change if the standard deviation were larger?
2. What if the sample size increased to 16 steel rods? How would that affect the standard error?
3. How do we interpret the z-scores in the context of a normal distribution?
4. What is the significance of using a standard normal table for these calculations?
5. How does changing the range of \( M \) affect the probability calculation?

**Tip:** Always ensure that when calculating probabilities involving the mean of a sample, you adjust for the sample size by using the standard error.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 208 cm and a standard deviation of 2 cm. For shipment, 8 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 206.9 cm and 207.7 cm.

Learn how to calculate the probability that the average length of steel rods is within a specified range using normal distribution. This example involves a company producing steel rods with a mean length of 208 cm and a standard deviation of 2 cm, considering a sample of 8 rods.

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