Final answer:
To find the weight such that the heaviest 6% of fruits exceed it, we calculate the z-score for the 94th percentile and use it along with the mean and standard deviation to find the required weight, which is approximately 711.6 grams.
Step-by-step explanation:
The student has asked about the weight of a particular fruit that is normally distributed, where the heaviest 6% of fruits weigh more than a certain amount. To solve this problem, we would use the concept of a z-score and the standard normal distribution.
Given that the mean weight is 693 grams and the standard deviation is 12 grams, we are looking for a weight value (let's call it X) such that P(X > x) is 0.06. This requires finding the z-score that corresponds to the top 6% of the distribution. Using a standard normal distribution table or calculator, we find that the z-score for the top 6% is approximately 1.55.
The z-score formula is:
z = (X - mean) / standard deviation.
Solving for X gives us:
X = z * standard deviation + mean.
Plugging in our numbers:
X = 1.55 * 12 + 693.
X ≈ 711.6 grams.
Therefore, the heaviest 6% of these fruits weigh more than approximately 711.6 grams.