Final answer:
The weight at which the heaviest 9% of the fruits weigh more is approximately 524.74 grams. This is calculated using the z-score that corresponds to the top 9% of a normal distribution with a mean of 510 grams and a standard deviation of 11 grams.
Step-by-step explanation:
To find the weight at which the heaviest 9% of the fruits weigh more, we need to determine the z-score that corresponds to the top 9% of a normal distribution. We would then use the mean and standard deviation to find the specific weight.
First, we find the z-score that corresponds to the cumulative area of 0.91 (since 100% - 9% = 91%). Consulting a standard normal distribution table or using a calculator with normal distribution functions, we find a z-score of approximately 1.34.
Next, we use the z-score formula:
Z = (X - μ) / σ
Where Z is the z-score (1.34 in this case), X is the weight we are looking for, μ is the mean (510 grams), and σ is the standard deviation (11 grams). Isolating X, we get:
X = Z * σ + μX = 1.34 * 11 + 510X ≈ 14.74 + 510X ≈ 524.74 grams
Therefore, the heaviest 9% of these fruits weigh more than approximately 524.74 grams.