To determine the weight above which 3% of the sample means will fall when picking 20 fruits, we can use the properties of the normal distribution.
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Identify the parameters of the distribution:
- Population mean (\(\mu\)) = 380 grams
- Population standard deviation (\(\sigma\)) = 8 grams
- Sample size (\(n\)) = 20 fruits
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Calculate the standard deviation of the sample mean (\(\sigma_{\bar{x}}\)): \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{20}} \approx 1.7889 \text{ grams} \]
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Find the z-score corresponding to the top 3% tail of the distribution: Since we want the mean weight greater than a certain value, we need to find the z-score that corresponds to 97% of the distribution (as the top 3% is in the right tail).
You can look up the z-value for 0.97 in the standard normal (z) table or use a calculator: \[ z \approx 1.88079 \]
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Convert the z-score back to the corresponding weight: Using the formula for converting a z-score to an x-value: \[ x = \mu + z \cdot \sigma_{\bar{x}} \] Plugging in the values: \[ x = 380 + 1.88079 \cdot 1.7889 \approx 380 + 3.359 \approx 383.359 \]
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Round to the nearest gram: The value we calculated is approximately 383.359 grams, which rounds to 383 grams.
Therefore, 3% of the time, the mean weight of the 20 fruits will be greater than 383 grams.