Samples are drawn from a population with mean 112 and standard deviation 47. Each sample has 399 randomly and independently chosen elements.
Use the Central Limit Theorem to estimate the probability that a sample mean is between 111 and 119.
Write your answer as a decimal rounded to the nearest thousandth.

To estimate the probability that a sample mean is between 111 and 119 using the Central Limit Theorem (CLT), we proceed with the following steps:

### 1. Understanding the problem

Given:
- Population mean (\(\mu\)) = 112
- Population standard deviation (\(\sigma\)) = 47
- Sample size (\(n\)) = 399

We need to find:
- The probability that the sample mean (\(\bar{x}\)) lies between 111 and 119.

### 2. Apply the Central Limit Theorem (CLT)

According to the CLT, the sampling distribution of the sample mean \(\bar{x}\) is approximately normal, with the following parameters:
- Mean of the sampling distribution = \(\mu_{\bar{x}} = \mu = 112\)
- Standard deviation of the sampling distribution (Standard Error) = \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{399}}\)

### 3. Calculate the Standard Error (SE)

\[
\sigma_{\bar{x}} = \frac{47}{\sqrt{399}} \approx 2.354
\]

### 4. Convert to z-scores

To use the standard normal distribution, we convert the values 111 and 119 into z-scores using the formula:

\[
z = \frac{x - \mu_{\bar{x}}}{\sigma_{\bar{x}}}
\]

For \(x = 111\):

\[
z_{111} = \frac{111 - 112}{2.354} \approx -0.425
\]

For \(x = 119\):

\[
z_{119} = \frac{119 - 112}{2.354} \approx 2.973
\]

### 5. Find the probability using the standard normal distribution

We now look up the probabilities corresponding to these z-scores from the standard normal table:
- \(P(z_{111} \leq -0.425)\) ≈ 0.336
- \(P(z_{119} \leq 2.973)\) ≈ 0.9985

### 6. Compute the probability

The probability that the sample mean is between 111 and 119 is:

\[
P(111 \leq \bar{x} \leq 119) = P(z \leq 2.973) - P(z \leq -0.425) = 0.9985 - 0.336 = 0.6625
\]

### Final Answer:
The probability that the sample mean is between 111 and 119 is approximately **0.663**.

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Do you need more details on any part of this solution or have any questions?

Here are 5 related questions to further explore:
1. How does the Central Limit Theorem apply when the population is not normally distributed?
2. What happens to the standard error if the sample size is increased?
3. How do you find probabilities for values outside the range of the z-table?
4. Why is the Central Limit Theorem important in statistics?
5. How would the probability change if the population standard deviation was different?

**Tip:** Always ensure that the sample size is large enough when applying the Central Limit Theorem for a better approximation of normality.

Samples are drawn from a population with mean 112 and standard deviation 47. Each sample has 399 randomly and independently chosen elements. Use the Central Limit Theorem to estimate the probability that a sample mean is between 111 and 119. Write your answer as a decimal rounded to the nearest thousandth.

Estimate the probability that a sample mean is between 111 and 119 using the Central Limit Theorem. Given a population mean of 112, standard deviation of 47, and a sample size of 399, this problem demonstrates the application of z-scores and standard error calculation. Suitable for college-level statistics students.

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