A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 228-cm and a standard deviation of 2.4-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 227.3-cm and 228.8-cm.
P(227.3-cm < M < 228.8-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.

To find the probability that the average length of a randomly selected bundle of steel rods is between 227.3 cm and 228.8 cm, we need to use the properties of the sampling distribution of the sample mean.

Given:
- Mean of the population, \(\mu = 228\) cm
- Standard deviation of the population, \(\sigma = 2.4\) cm
- Sample size, \(n = 21\)

The standard deviation of the sample mean (also known as the standard error) is given by:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{2.4}{\sqrt{21}} \approx 0.5238 \text{ cm}
\]

We need to find:
\[
P(227.3 < \bar{X} < 228.8)
\]

First, we convert the raw scores to z-scores using the formula:
\[
z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}
\]

For \( \bar{X} = 227.3 \):
\[
z = \frac{227.3 - 228}{0.5238} \approx -1.34
\]

For \( \bar{X} = 228.8 \):
\[
z = \frac{228.8 - 228}{0.5238} \approx 1.53
\]

Now, we look up these z-scores in the standard normal distribution table or use a calculator to find the corresponding probabilities.

The probability corresponding to \(z = -1.34\) is approximately 0.0901.
The probability corresponding to \(z = 1.53\) is approximately 0.9370.

To find the probability that the sample mean is between 227.3 and 228.8, we calculate:
\[
P(-1.34 < Z < 1.53) = P(Z < 1.53) - P(Z < -1.34)
\]
\[
P(-1.34 < Z < 1.53) = 0.9370 - 0.0901 = 0.8469
\]

Thus, the probability that the average length of a randomly selected bundle of steel rods is between 227.3 cm and 228.8 cm is:
\[
\boxed{0.8469}
\]

Would you like more details or have any questions?

Here are some related questions to consider:
1. What is the probability that the average length of a bundle is less than 227.5 cm?
2. What is the probability that the average length of a bundle is more than 229 cm?
3. How does changing the sample size affect the standard error?
4. What is the probability that a single steel rod is longer than 230 cm?
5. How would the probability change if the standard deviation were 3 cm instead of 2.4 cm?
6. What is the 95% confidence interval for the mean length of the steel rods?
7. How would the probability change if the mean length were 227 cm instead of 228 cm?
8. If the bundle size is increased to 30, what is the new probability for the same range of average lengths?

**Tip:** When working with normally distributed data, always check if the Central Limit Theorem applies, especially for small sample sizes.

Explore how to calculate the probability that the average length of a bundle of steel rods falls between 227.3 cm and 228.8 cm. Understand the application of the Central Limit Theorem, sampling distributions, and standard error of the mean in statistical analysis. Suitable for advanced high school students and beyond.

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