A particular fruit's weights are normally distributed, with a mean of 551 grams and a standard deviation of 40 grams. The heaviest $13 \%$ of fruits weigh more than how many grams? Give your answer to the nearest gram.

Step 1 ：We are given a normal distribution with a mean of \(\mu = 551\) grams and a standard deviation of \(\sigma = 40\) grams. We are asked to find the weight above which the heaviest 13% of fruits lie. This is a question of finding a percentile in a normal distribution.

Step 2 ：In a normal distribution, percentiles can be found using the Z-score formula, which is \((X - \mu) / \sigma\), where X is the value we are looking for, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：However, in this case, we are given the percentile (87th percentile, because the heaviest 13% means the lightest 87%) and we need to find the corresponding value. This can be done by rearranging the Z-score formula to \(X = Z * \sigma + \mu\), where Z is the Z-score corresponding to the given percentile.

Step 4 ：The Z-score can be found using a Z-table or a function in a programming language that gives the Z-score for a given percentile. For the 87th percentile, the Z-score is approximately 1.126.

Step 5 ：Substituting the given values into the formula, we get \(X = 1.126 * 40 + 551\), which simplifies to \(X = 596.04\).

Step 6 ：Rounding to the nearest gram, we get \(X = 596\) grams.

Step 7 ：\(\boxed{596}\) grams is the weight above which the heaviest 13% of fruits lie.