A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 195-cm and a standard deviation of 0.6-cm. For shipment, 25 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is between 194.7-cm and 194.9-cm.
P(194.7-cm < M < 194.9-cm) = 


Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

The problem involves calculating the probability that the average length \( M \) of a randomly selected bundle of 25 steel rods is between 194.7 cm and 194.9 cm. The lengths of the steel rods are normally distributed with a mean \( \mu = 195 \) cm and a standard deviation \( \sigma = 0.6 \) cm.

Since we're dealing with the average length of 25 rods, we use the **Central Limit Theorem**. This theorem tells us that the distribution of the sample mean \( M \) will be approximately normal with:

- Mean \( \mu_M = \mu = 195 \) cm (the same as the population mean).
- Standard deviation \( \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{0.6}{\sqrt{25}} = \frac{0.6}{5} = 0.12 \) cm.

We now need to calculate the probability \( P(194.7 < M < 194.9) \).

### Step 1: Convert the raw scores to z-scores.
The z-score for a value \( M \) is given by:
\[
z = \frac{M - \mu_M}{\sigma_M}
\]
- For \( M = 194.7 \):
\[
z_1 = \frac{194.7 - 195}{0.12} = \frac{-0.3}{0.12} = -2.5
\]
- For \( M = 194.9 \):
\[
z_2 = \frac{194.9 - 195}{0.12} = \frac{-0.1}{0.12} = -0.8333
\]

### Step 2: Use the standard normal distribution to find the probabilities associated with these z-scores.

- The probability corresponding to \( z_1 = -2.5 \) (from standard normal tables or a calculator) is approximately \( P(z_1) = 0.0062 \).
- The probability corresponding to \( z_2 = -0.8333 \) is approximately \( P(z_2) = 0.2023 \).

### Step 3: Calculate the probability between the two z-scores.
The probability that \( M \) is between 194.7 and 194.9 is the difference between the two probabilities:
\[
P(194.7 < M < 194.9) = P(z_2) - P(z_1) = 0.2023 - 0.0062 = 0.1961
\]

Thus, the probability is approximately \( \boxed{0.1961} \).

Would you like further clarification or details? 

Here are 5 related questions:
1. How do you interpret z-scores in the context of normal distributions?
2. What is the Central Limit Theorem and why is it important in statistics?
3. How would the result change if the sample size increased to 50 rods?
4. How can we approximate probabilities using z-scores in a non-normal distribution?
5. How does the standard deviation of a sample mean differ from the population standard deviation?

**Tip:** Always double-check your z-score calculations, especially when rounding decimals, to ensure the final probability is as accurate as possible.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 195-cm and a standard deviation of 0.6-cm. For shipment, 25 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 194.7-cm and 194.9-cm. P(194.7-cm < M < 194.9-cm) = 

Learn how to calculate the probability that the average length of a bundle of 25 steel rods is between 194.7-cm and 194.9-cm using the Central Limit Theorem. This problem involves normal distribution, z-scores, and the application of probability concepts, perfect for students studying advanced statistics.

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