To solve this, we apply the Central Limit Theorem (CLT). Here’s the step-by-step process:

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Step 1: Identify key parameters

• Population mean $\mu = 14$
• Population standard deviation $\sigma = 3.6$
• Sample size $n = 41$
• Sample mean of interest $\bar{x} = 15.6$

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Step 2: Compute the standard error (SE) of the sample mean

The standard error is given by:

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

Substitute the values:

$SE = \frac{3.6}{\sqrt{41}} \approx \frac{3.6}{6.403} \approx 0.562$

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Step 3: Convert the sample mean to a z-score

The z-score is calculated as:

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

Substitute the values:

$z = \frac{15.6 - 14}{0.562} \approx \frac{1.6}{0.562} \approx 2.847$

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Step 4: Find the probability from the z-score

We now find the probability corresponding to a z-score of $z = 2.847$ using the standard normal distribution table or a calculator.

The cumulative probability to the left of $z = 2.847$ is approximately:

$P(Z \leq 2.847) \approx 0.9978$

Thus, the probability that the mean is greater than 15.6 inches is:

$P(\bar{x} > 15.6) = 1 - P(Z \leq 2.847) \approx 1 - 0.9978 = 0.0022$

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Final Answer:

The probability that the mean length is greater than 15.6 inches is 0.0022 (rounded to four decimal places).

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Let me know if you'd like further clarification or more examples! Here are 5 related questions you might find useful:

1. What is the significance of the Central Limit Theorem in probability calculations?
2. How does increasing the sample size affect the standard error of the sample mean?
3. How do you find probabilities when the data is not normally distributed?
4. What happens if the population standard deviation is unknown?
5. How do z-scores relate to percentiles in a normal distribution?

Tip: Always check if the sample size is sufficiently large (typically $n > 30$) for the Central Limit Theorem to apply when working with non-normal populations!