1. **State the problem:** We want to find the percentage of people with IQ scores between 81 and 127, given that IQ scores are normally distributed with mean $\mu = 100$ and standard deviation $\sigma = 15$. 2. **Formula and rules:** For a normal distribution, the probability that a value $X$ lies between $a$ and $b$ is given by: $$P(a < X < b) = P\left(\frac{a - \mu}{\sigma} < Z < \frac{b - \mu}{\sigma}\right)$$ where $Z$ is the standard normal variable with mean 0 and standard deviation 1. 3. **Calculate the z-scores:** $$z_1 = \frac{81 - 100}{15} = \frac{-19}{15} = -1.27$$ $$z_2 = \frac{127 - 100}{15} = \frac{27}{15} = 1.8$$ 4. **Find the cumulative probabilities:** Using standard normal distribution tables or a calculator, $$P(Z < -1.27) = 0.1020$$ $$P(Z < 1.8) = 0.9641$$ 5. **Calculate the area between the z-scores:** $$P(-1.27 < Z < 1.8) = P(Z < 1.8) - P(Z < -1.27) = 0.9641 - 0.1020 = 0.8621$$ 6. **Convert to percentage:** $$0.8621 \times 100 = 86.2\%$$ **Final answer:** Approximately **86.2%** of people have an IQ score between 81 and 127.