Final answer:
To determine the mean weight that is exceeded 4% of the time, we calculate the standard error and use the z-score for the 96th percentile. For a sample size of 19 fruits with a mean weight of 223 grams and standard deviation of 14 grams, the mean weight is approximately 228.61 grams.
Step-by-step explanation:
We need to find the mean weight of the fruit such that it exceeds a certain value 4% of the time. Since we're dealing with the sampling distribution of the mean, we should use the standard error (SE) of the mean to calculate this, which is the standard deviation (SD) divided by the square root of the sample size (n).
The formula for SE is: SE = SD / √n
Then, we use the z-score associated with the 4% upper tail of a standard normal distribution, which is approximately 1.75 (z-score for 96% to the left). The formula to find the weight corresponding to this z-score is: X = μ + z × SE.
In this case, μ (mean) = 223 grams, SD = 14 grams, and n = 19. Thus, SE = 14 / √19 ≈ 3.21 grams.
The calculation for the cutoff weight for the top 4% will then be: X = 223 + 1.75 × 3.21, which yields X ≈ 228.61 grams.
Therefore, 4% of the time, the mean weight of the 19 randomly picked fruits will be greater than 228.61 grams.