company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 241.4-cm and a standard deviation of 0.6-cm. For shipment, 5 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is between 241.7-cm and 241.9-cm.
P(241.7-cm < M < 241.9-cm) = 

To find the probability that the average length of a randomly selected bundle of 5 steel rods is between 241.7 cm and 241.9 cm, we can follow these steps:

### Given:
- Mean of the steel rod lengths, \( \mu = 241.4 \, \text{cm} \)
- Standard deviation of the steel rod lengths, \( \sigma = 0.6 \, \text{cm} \)
- Number of rods in a bundle, \( n = 5 \)

We are looking for the probability that the average length of the bundle is between 241.7 cm and 241.9 cm.

### Step 1: Standard error of the mean (SEM)
The standard error of the mean (SEM) for the sample mean is given by:

\[
\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{0.6}{\sqrt{5}} \approx 0.268
\]

### Step 2: Z-scores for 241.7 cm and 241.9 cm
To find the probability, we need to convert the given values (241.7 cm and 241.9 cm) into Z-scores using the following formula:

\[
Z = \frac{X - \mu}{\text{SEM}}
\]

- For \( X = 241.7 \, \text{cm} \):

\[
Z_1 = \frac{241.7 - 241.4}{0.268} \approx 1.12
\]

- For \( X = 241.9 \, \text{cm} \):

\[
Z_2 = \frac{241.9 - 241.4}{0.268} \approx 1.87
\]

### Step 3: Find the probabilities
Using the Z-table (or a calculator), find the probabilities corresponding to these Z-scores:

- \( P(Z_1 = 1.12) \approx 0.8686 \)
- \( P(Z_2 = 1.87) \approx 0.9693 \)

### Step 4: Find the probability between the Z-scores
The probability that the average length is between 241.7 cm and 241.9 cm is:

\[
P(241.7 < M < 241.9) = P(Z_2) - P(Z_1) = 0.9693 - 0.8686 = 0.1007
\]

### Final Answer:
The probability that the average length of a randomly selected bundle of 5 steel rods is between 241.7 cm and 241.9 cm is approximately **0.1007** or **10.07%**.

---

Would you like more details or have any questions?

### Related Questions:
1. How would the probability change if the standard deviation was larger?
2. What is the probability that the average length is greater than 241.8 cm?
3. How does the sample size affect the standard error of the mean?
4. How would you find the probability for only one steel rod instead of a bundle?
5. What is the Z-score if the average length of the bundle is 242.0 cm?

### Tip:
Increasing the sample size reduces the standard error of the mean, making the distribution of the sample mean narrower.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 241.4 cm and a standard deviation of 0.6 cm. For shipment, 5 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 241.7 cm and 241.9 cm. P(241.7 cm < M < 241.9 cm) = ?

Calculate the probability that the average length of a bundle of 5 steel rods, each normally distributed with a mean of 241.4 cm and a standard deviation of 0.6 cm, falls between 241.7 cm and 241.9 cm. This problem involves the use of normal distribution, standard error of the mean, and Z-scores.

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