A random sample of 36 trout from this pond was caught and weighed, and they showed a mean weight of 560g with standard deviation of 84g. What would the upper limit of the required confidence interval be?

To compute the upper limit of the confidence interval for the mean weight of trout in the pond, we will use the following steps:

1. **Determine the sample mean and standard deviation**:
- Sample Mean (\( \bar{x} \)) = 560g
- Sample Standard Deviation (s) = 84g
- Sample size (n) = 36

2. **Choose the confidence level**:
We need to specify a confidence level. Common levels are 90%, 95%, and 99%. For this example, let's assume a 95% confidence level.

3. **Find the critical value (z-score)**:
For a 95% confidence level, the critical value (z) can be found using a z-table or calculator:
- The critical value for a 95% confidence level is approximately 1.96.

4. **Calculate the standard error (SE)**:
\[
SE = \frac{s}{\sqrt{n}} = \frac{84}{\sqrt{36}} = \frac{84}{6} = 14
\]

5. **Calculate the margin of error (ME)**:
\[
ME = z \times SE = 1.96 \times 14 = 27.44
\]

6. **Calculate the upper limit of the confidence interval**:
\[
\text{Upper limit} = \bar{x} + ME = 560 + 27.44 = 587.44
\]

Thus, the upper limit of the 95% confidence interval for the mean weight of trout is approximately **587.44 grams**.