1. **Problem Statement:** Find the values at one, two, and three standard deviations from the mean for each given normal distribution. 2. **Formula:** For a normal distribution with mean $\mu$ and standard deviation $\sigma$, the value at $k$ standard deviations from the mean is: $$x = \mu \pm k\sigma$$ where $k = 1, 2, 3$. 3. **Step-by-step calculations:** **5. Mean = 95, Standard Deviation = 12** - $x_1 = 95 \pm 1 \times 12 = 95 \pm 12 = 83, 107$ - $x_2 = 95 \pm 2 \times 12 = 95 \pm 24 = 71, 119$ - $x_3 = 95 \pm 3 \times 12 = 95 \pm 36 = 59, 131$ **6. Mean = 100, Standard Deviation = 15** - $x_1 = 100 \pm 15 = 85, 115$ - $x_2 = 100 \pm 30 = 70, 130$ - $x_3 = 100 \pm 45 = 55, 145$ **7. Mean = 60, Standard Deviation = 6** - $x_1 = 60 \pm 6 = 54, 66$ - $x_2 = 60 \pm 12 = 48, 72$ - $x_3 = 60 \pm 18 = 42, 78$ **8. Mean = 23.8, Standard Deviation = 5.2** - $x_1 = 23.8 \pm 5.2 = 18.6, 29.0$ - $x_2 = 23.8 \pm 10.4 = 13.4, 34.2$ - $x_3 = 23.8 \pm 15.6 = 8.2, 39.4$ **9. Mean = 676, Standard Deviation = 60** - $x_1 = 676 \pm 60 = 616, 736$ - $x_2 = 676 \pm 120 = 556, 796$ - $x_3 = 676 \pm 180 = 496, 856$ **10. Mean = 54.2, Standard Deviation = 12.3** - $x_1 = 54.2 \pm 12.3 = 41.9, 66.5$ - $x_2 = 54.2 \pm 24.6 = 29.6, 78.8$ - $x_3 = 54.2 \pm 36.9 = 17.3, 91.1$ 4. **Explanation:** For each distribution, we add and subtract multiples of the standard deviation from the mean to find the values at one, two, and three standard deviations away. This shows how data is spread around the mean in a normal distribution. **Final answers:** 5. $59, 71, 83, 95, 107, 119, 131$ 6. $55, 70, 85, 100, 115, 130, 145$ 7. $42, 48, 54, 60, 66, 72, 78$ 8. $8.2, 13.4, 18.6, 23.8, 29.0, 34.2, 39.4$ 9. $496, 556, 616, 676, 736, 796, 856$ 10. $17.3, 29.6, 41.9, 54.2, 66.5, 78.8, 91.1$