Samples are drawn from a population with mean 80 and standard deviation 32. Each sample has 369 randomly and independently chosen elements.
Use the Central Limit Theorem to estimate the probability that a sample mean is between 79 and 81.
Write your answer as a decimal rounded to the nearest thousandth.

We are asked to estimate the probability that the sample mean of a population falls between 79 and 81 using the **Central Limit Theorem (CLT)**.

The key facts provided are:
- Population mean (\(\mu\)) = 80
- Population standard deviation (\(\sigma\)) = 32
- Sample size (\(n\)) = 369

### Step 1: Applying the Central Limit Theorem
The **Central Limit Theorem** tells us that the sampling distribution of the sample mean will be approximately normal if the sample size is large (which it is, here \(n = 369\)). The mean of this sampling distribution will be the same as the population mean, but its standard deviation (also known as the **standard error** of the mean) is given by:

\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]

where:
- \(\sigma_{\bar{x}}\) is the standard error of the mean,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.

Substituting the given values:

\[
\sigma_{\bar{x}} = \frac{32}{\sqrt{369}} \approx \frac{32}{19.21} \approx 1.665
\]

### Step 2: Convert the given range to z-scores
Next, we calculate the **z-scores** for the sample means 79 and 81, which represent how many standard errors away these values are from the population mean (\(\mu = 80\)):

\[
z_1 = \frac{79 - 80}{1.665} = \frac{-1}{1.665} \approx -0.601
\]
\[
z_2 = \frac{81 - 80}{1.665} = \frac{1}{1.665} \approx 0.601
\]

### Step 3: Find the probabilities corresponding to these z-scores
Using standard normal distribution tables (or a calculator), we find the probabilities corresponding to these z-scores:

- For \(z_1 = -0.601\), the cumulative probability is approximately 0.274.
- For \(z_2 = 0.601\), the cumulative probability is approximately 0.726.

### Step 4: Calculate the probability that the sample mean is between 79 and 81
To find the probability that the sample mean is between 79 and 81, we subtract the cumulative probability at \(z_1\) from the cumulative probability at \(z_2\):

\[
P(79 \leq \bar{x} \leq 81) = P(z_2) - P(z_1) = 0.726 - 0.274 = 0.452
\]

### Final Answer:
The probability that the sample mean is between 79 and 81 is approximately **0.452**.

Would you like more details on any part of the process?

### Related questions:
1. How does the Central Limit Theorem apply to small sample sizes?
2. What would happen if the population distribution were not normal?
3. How do you calculate the z-score for other values of the sample mean?
4. Why is the standard error smaller than the population standard deviation?
5. Can the Central Limit Theorem be applied for proportions as well?

**Tip:** Always check whether the sample size is large enough to apply the Central Limit Theorem, typically \(n > 30\).

Samples are drawn from a population with mean 80 and standard deviation 32. Each sample has 369 randomly and independently chosen elements. Use the Central Limit Theorem to estimate the probability that a sample mean is between 79 and 81. Write your answer as a decimal rounded to the nearest thousandth.

Learn how to apply the Central Limit Theorem to estimate the probability that the sample mean lies between 79 and 81. This problem involves a population with a mean of 80 and a standard deviation of 32, using a sample size of 369. Explore the concepts of normal distribution, z-scores, and standard error.

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