We are asked to find the probability that the mean weight of a sample of 40 trout exceeds 405.5 grams. Given the population parameters, we can apply the Central Limit Theorem (CLT) to address this problem.

Given:

• Population mean ($\mu$) = 402.7 grams
• Population standard deviation ($\sigma$) = 8.8 grams
• Sample size ($n$) = 40
• Sample mean threshold = 405.5 grams

Step 1: Calculate the standard error of the mean (SE)

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

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

Substituting the values:

$SE = \frac{8.8}{\sqrt{40}} = \frac{8.8}{6.3246} \approx 1.391$

Step 2: Convert the problem into a Z-score

The Z-score is calculated using the formula:

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

Where:

• $\bar{x}$ is the sample mean (405.5 grams)
• $\mu$ is the population mean (402.7 grams)
• $SE$ is the standard error calculated above.

Substitute the values:

$Z = \frac{405.5 - 402.7}{1.391} \approx \frac{2.8}{1.391} \approx 2.01$

Step 3: Find the probability corresponding to the Z-score

Using a Z-table or a standard normal distribution calculator, the probability of getting a Z-score greater than 2.01 is approximately 0.0222. This represents the area under the standard normal curve to the right of Z = 2.01.

Conclusion:

The probability that the mean weight of a sample of 40 trout exceeds 405.5 grams is approximately 0.0222 or 2.22%.

Do you need more details or further clarification?

Here are 5 related questions to expand on this topic:

1. How would the probability change if the sample size were increased to 100?
2. What is the probability that the mean weight is less than 400 grams?
3. Can we find the probability that a single trout weighs more than 405.5 grams?
4. What role does the Central Limit Theorem play in this type of problem?
5. How would the answer change if the standard deviation of the population were different?

Tip: Remember, as the sample size increases, the standard error decreases, leading to a more accurate estimate of the population mean.