1. **State the problem:** We are given data about the number of snacks and their corresponding points, along with averages and standard deviations. We want to understand how to calculate the mean (average) and standard deviation for these values. 2. **Formulas:** - The mean (average) of a set of values $x_1, x_2, ..., x_n$ is given by: $$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$ - The standard deviation (SD) is a measure of spread and is calculated as: $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ 3. **Calculate the mean for the number of snacks:** Given values: 0.9, 2.11, 1.95, 1.8, 1.87, 0.47, 0.92 $$\bar{x} = \frac{0.9 + 2.11 + 1.95 + 1.8 + 1.87 + 0.47 + 0.92}{7} = \frac{9.02}{7} = 1.2886$$ 4. **Calculate the standard deviation for the number of snacks:** Calculate each squared difference: $$(0.9 - 1.2886)^2 = 0.1511$$ $$(2.11 - 1.2886)^2 = 0.6743$$ $$(1.95 - 1.2886)^2 = 0.4383$$ $$(1.8 - 1.2886)^2 = 0.2615$$ $$(1.87 - 1.2886)^2 = 0.3383$$ $$(0.47 - 1.2886)^2 = 0.6693$$ $$(0.92 - 1.2886)^2 = 0.1359$$ Sum of squared differences: $$0.1511 + 0.6743 + 0.4383 + 0.2615 + 0.3383 + 0.6693 + 0.1359 = 2.6687$$ Divide by $n-1 = 6$: $$\frac{2.6687}{6} = 0.4448$$ Take the square root: $$s = \sqrt{0.4448} = 0.6669$$ 5. **Interpretation:** The average number of snacks is approximately $1.29$ with a standard deviation of approximately $0.67$, indicating the typical deviation from the mean. 6. **Note:** The values $\sqrt{2.5}$ and $\sqrt{9.5}$ correspond to the square roots of the sums of squares or variances related to the data, which align with the standard deviation calculations. **Final answers:** - Mean number of snacks: $1.29$ - Standard deviation: $0.67$