To solve this problem using the Central Limit Theorem (CLT), we follow these steps:

1. Define the parameters:

• Population mean $\mu = 137$
• Population standard deviation $\sigma = 40$
• Sample size $n = 238$

2. Find the standard error of the mean (SE):

The standard error of the mean (SE) is given by:

$SE = \frac{\sigma}{\sqrt{n}}$

Substitute the given values:

$SE = \frac{40}{\sqrt{238}} \approx \frac{40}{15.44} \approx 2.591$

3. Convert the sample means to z-scores:

The z-score for a value $x$ is given by:

$z = \frac{x - \mu}{SE}$

Now, calculate the z-scores for the bounds of the interval [135, 139]:

• For $x = 135$:

$z = \frac{135 - 137}{2.591} \approx \frac{-2}{2.591} \approx -0.772$

• For $x = 139$:

$z = \frac{139 - 137}{2.591} \approx \frac{2}{2.591} \approx 0.772$

4. Find the cumulative probabilities:

Using the standard normal distribution table (or a calculator), we find the cumulative probabilities for these z-scores:

• $P(z \leq -0.772) \approx 0.220$
• $P(z \leq 0.772) \approx 0.780$

5. Find the probability that the sample mean is between 135 and 139:

The probability that the sample mean is between 135 and 139 is:

$P(135 \leq \bar{x} \leq 139) = P(z \leq 0.772) - P(z \leq -0.772)$

$P(135 \leq \bar{x} \leq 139) \approx 0.780 - 0.220 = 0.560$

6. Final answer:

The probability that the sample mean is between 135 and 139 is approximately 0.560, rounded to the nearest thousandth.

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Would you like any further explanation or have questions?

Here are 5 related questions for practice:

1. What is the probability that a sample mean is greater than 140?
2. How does the sample size affect the standard error of the mean?
3. If the population standard deviation were 50, how would the probability change?
4. What is the z-score for a sample mean of 134?
5. How does the Central Limit Theorem apply when the population is not normally distributed?

Tip: The larger the sample size, the smaller the standard error, which increases the accuracy of the sample mean estimates.