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Answer :
Final answer:
About 49.5% of people should have IQ scores between 90 and 110 using the 68-95-99.7 rule.
Explanation:
The IQ scores of a population are normally distributed with a mean of 100 and a standard deviation of 15. Using the 68-95-99.7 rule, we can determine the percentage of people who have IQ scores between two values. In this case, the values are (90, 110).
To calculate the percentage, we need to find the z-scores corresponding to the IQ values. The z-score formula is z = (x - μ) / σ, where x is the IQ score, μ is the mean, and σ is the standard deviation. For x = 90, the z-score is (90 - 100) / 15 = -2/3, and for x = 110, the z-score is (110 - 100) / 15 = 2/3.
The percentage of people with IQ scores between (90, 110) can be found using the standard normal distribution table or a calculator that provides probabilities for normal distributions. Using the table, we can find the probability of being below -2/3 and subtract it from the probability of being below 2/3. P(z < -2/3) ≈ 0.2525 and P(z < 2/3) ≈ 0.7475. So, the probability of z-scores between -2/3 and 2/3 is approximately 0.7475 - 0.2525 = 0.495. This corresponds to 49.5%.
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