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Answer :
Final answer:
The percentile rank of a sample mean weight of 66 lbs from a normally distributed population with a mean of 65 lbs and standard deviation of 3 lbs is the (b) 90th percentile.
Explanation:
To find the percentile rank for a sample mean, we first use the Z-score formula, which is Z = (X - μ) / (σ / √n), where X is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.
Substituting the given values into the formula: Z = (66 - 65) / (3 / √15) = 1.29. The Z-score 1.29 corresponds to a percentile rank of approximately 90th percentile, or 0.9. Thus, the sample mean of 66 pounds falls in the 90th percentile.
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