1. **Problem Statement:** We have 10 battery drain times in minutes: 195, 203, 177, 186, 191, 225, 216, 202, 197, 21. We need to find: a. Range b. Mean c. Variance d. Standard deviation 2. **Formulas and Rules:** - Range = Maximum value - Minimum value - Mean ($\bar{x}$) = $\frac{\sum x_i}{n}$ - Variance ($s^2$) = $\frac{\sum (x_i - \bar{x})^2}{n-1}$ for sample variance - Standard deviation ($s$) = $\sqrt{s^2}$ 3. **Calculations:** **a. Range:** Maximum = 225, Minimum = 21 $$\text{Range} = 225 - 21 = 204$$ **b. Mean:** $$\bar{x} = \frac{195 + 203 + 177 + 186 + 191 + 225 + 216 + 202 + 197 + 21}{10} = \frac{1813}{10} = 181.30$$ **c. Variance:** Calculate each squared deviation: $$ (195 - 181.3)^2 = 13.7^2 = 187.69 $$ $$ (203 - 181.3)^2 = 21.7^2 = 470.89 $$ $$ (177 - 181.3)^2 = (-4.3)^2 = 18.49 $$ $$ (186 - 181.3)^2 = 4.7^2 = 22.09 $$ $$ (191 - 181.3)^2 = 9.7^2 = 94.09 $$ $$ (225 - 181.3)^2 = 43.7^2 = 1909.69 $$ $$ (216 - 181.3)^2 = 34.7^2 = 1204.09 $$ $$ (202 - 181.3)^2 = 20.7^2 = 428.49 $$ $$ (197 - 181.3)^2 = 15.7^2 = 246.49 $$ $$ (21 - 181.3)^2 = (-160.3)^2 = 25696.09 $$ Sum of squared deviations: $$ 187.69 + 470.89 + 18.49 + 22.09 + 94.09 + 1909.69 + 1204.09 + 428.49 + 246.49 + 25696.09 = 30878.10 $$ Sample variance: $$ s^2 = \frac{30878.10}{10 - 1} = \frac{30878.10}{9} = 3430.90 $$ **d. Standard deviation:** $$ s = \sqrt{3430.90} = 58.58 $$ 4. **Final answers:** - Range = 204.00 - Mean = 181.30 - Variance = 3430.90 - Standard deviation = 58.58