A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 183.5 cm and a standard deviation of 2.1 cm. For shipment, 16 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is between 182.5 cm and 184.5 cm.
P(182.5 cm < M < 184.5 cm) =

To find the probability that the average length of a randomly selected bundle of steel rods is between 182.5 cm and 184.5 cm, we need to use the properties of the sampling distribution of the sample mean.

Given:
- Mean length of steel rods (\(\mu\)) = 183.5 cm
- Standard deviation of lengths (\(\sigma\)) = 2.1 cm
- Sample size (\(n\)) = 16

The mean of the sampling distribution of the sample mean (\(\mu_{\bar{X}}\)) is the same as the population mean:
\[
\mu_{\bar{X}} = \mu = 183.5 \, \text{cm}
\]

The standard deviation of the sampling distribution of the sample mean (standard error, \(\sigma_{\bar{X}}\)) is given by:
\[
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{16}} = \frac{2.1}{4} = 0.525 \, \text{cm}
\]

We want to find:
\[
P(182.5 < \bar{X} < 184.5)
\]

First, we convert the lengths to \(z\)-scores using the formula:
\[
z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}}
\]

For \(\bar{X} = 182.5\):
\[
z_{1} = \frac{182.5 - 183.5}{0.525} = \frac{-1}{0.525} \approx -1.90
\]

For \(\bar{X} = 184.5\):
\[
z_{2} = \frac{184.5 - 183.5}{0.525} = \frac{1}{0.525} \approx 1.90
\]

Next, we find the probability corresponding to these \(z\)-scores using the standard normal distribution table or a calculator.

\[
P(-1.90 < Z < 1.90)
\]

Looking up the values in the standard normal distribution table:
\[
P(Z < 1.90) \approx 0.9713
\]
\[
P(Z < -1.90) \approx 0.0287
\]

Thus:
\[
P(-1.90 < Z < 1.90) = P(Z < 1.90) - P(Z < -1.90) \approx 0.9713 - 0.0287 = 0.9426
\]

Therefore, the probability that the average length of a randomly selected bundle of steel rods is between 182.5 cm and 184.5 cm is approximately 0.9426 or 94.26%.

Would you like more details or have any questions? Here are some more questions you might consider:

1. How do you interpret a \(z\)-score in a normal distribution?
2. What is the significance of the Central Limit Theorem in this context?
3. How would the probability change if the sample size increased?
4. What is the impact of a larger standard deviation on the probability?
5. How can you use a calculator to find the \(z\)-scores and probabilities?

**Tip:** When working with normal distributions, always remember to check if the problem refers to individual values or sample means, as the standard deviation will be adjusted accordingly.

Calculate the probability that the average length of a randomly selected bundle of steel rods falls between 182.5 cm and 184.5 cm. Understand the properties of normal and sampling distributions with a step-by-step solution. Suitable for undergraduate students and professionals.

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