1. **Problem:** Find the mean and standard deviation of the annual salaries: 85, 75, 77, 79, 80, 78, 82, 78, 73, 64 (in thousands). 2. **Formula:** - Mean: $$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$ - Standard deviation: $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ 3. **Calculate the mean:** $$\bar{x} = \frac{85 + 75 + 77 + 79 + 80 + 78 + 82 + 78 + 73 + 64}{10} = \frac{771}{10} = 77.1$$ 4. **Calculate each squared deviation:** $$(85 - 77.1)^2 = 62.41$$ $$(75 - 77.1)^2 = 4.41$$ $$(77 - 77.1)^2 = 0.01$$ $$(79 - 77.1)^2 = 3.61$$ $$(80 - 77.1)^2 = 8.41$$ $$(78 - 77.1)^2 = 0.81$$ $$(82 - 77.1)^2 = 24.01$$ $$(78 - 77.1)^2 = 0.81$$ $$(73 - 77.1)^2 = 16.81$$ $$(64 - 77.1)^2 = 171.61$$ 5. **Sum of squared deviations:** $$62.41 + 4.41 + 0.01 + 3.61 + 8.41 + 0.81 + 24.01 + 0.81 + 16.81 + 171.61 = 292.9$$ 6. **Calculate standard deviation:** $$s = \sqrt{\frac{292.9}{10 - 1}} = \sqrt{\frac{292.9}{9}} = \sqrt{32.544} = 5.705$$ **Final answer:** Mean = 77.1, Standard deviation = 5.705 This corresponds to option (B).