To solve this problem, we need to find the probability that the sample mean is 29.7 grams or less in a random sample of 36 farm-raised trout. This involves using the Central Limit Theorem and the properties of the normal distribution.

Here's the approach:

1. Given data:

   • Population mean $\mu = 32$ grams
   • Population standard deviation $\sigma = 7.5$ grams
   • Sample size $n = 36$
   • Sample mean $\bar{x} = 29.7$ grams

2. Standard error of the mean: The standard error (SE) is calculated as: $SE = \frac{\sigma}{\sqrt{n}} = \frac{7.5}{\sqrt{36}} = \frac{7.5}{6} = 1.25$

3. Z-score calculation: We now calculate the Z-score, which is the number of standard errors the sample mean is away from the population mean: $Z = \frac{\bar{x} - \mu}{SE} = \frac{29.7 - 32}{1.25} = \frac{-2.3}{1.25} = -1.84$

4. Finding the probability: Using the Z-table or a calculator, find the probability corresponding to a Z-score of $-1.84$.

   The probability for $Z \leq -1.84$ is approximately 0.0329 (from the Z-table).

Thus, the probability that the sample mean is 29.7 grams or less is 0.0329 or 3.29%.

Would you like further clarification or more details on any step?

Here are 5 related questions to expand your understanding:

1. What is the Central Limit Theorem and how does it apply to this problem?
2. How would the probability change if the sample size increased?
3. What is the significance of the Z-score in normal distribution problems?
4. Why do we use the standard error instead of the population standard deviation when working with sample means?
5. How would the result differ if the population had a non-normal distribution?

Tip: When solving problems involving the Central Limit Theorem, always check if the sample size is large enough to assume normality for the sample mean distribution.