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Answer:
Given that IQ scores are approximately normally distributed with a mean of 100 and a standard deviation of 15, we can determine the proportion of people with IQs above 130. First, we calculate the Z-score for an IQ of 130, which is \( \frac{130 - 100}{15} = 2 \). Referring to the standard normal distribution, the area to the right of a Z-score of 2 is about 0.0228. Therefore, approximately 2.28% of people have an IQ above 130.