1. **Problem statement:** We want to find the percentile of a person who is 1.67 standard deviations (sigma) above a group, which itself is 1.33 sigma above another group, which is 1 sigma above the whole general population. 2. **Understanding the problem:** Each group is shifted by a certain number of standard deviations from the previous group. The total number of sigmas above the general population is the sum of these shifts: $$1 + 1.33 + 1.67 = 3.99$$ So the person is 3.99 sigma above the general population mean. 3. **Formula and concept:** The percentile corresponding to a z-score (number of sigmas) in a normal distribution is given by the cumulative distribution function (CDF) of the standard normal distribution: $$\text{Percentile} = \Phi(z) \times 100$$ where $\Phi(z)$ is the CDF of the standard normal at $z$. 4. **Calculate the percentile:** Using standard normal distribution tables or a calculator for $z=3.99$: $$\Phi(3.99) \approx 0.999968$$ 5. **Interpretation:** This means the person is at approximately the 99.9968th percentile of the whole general population. **Final answer:** The person is in the 99.997th percentile of the general population (rounded to three decimal places).