1. **Problem Statement:** Find the variance of the midexam results for classes 5A and 5B. Given data: - 5A scores: 90, 92, 89, 95, 79, 88, 94, 96 - 5B scores: 93, 89, 90, 94, 93, 80, 89, 95 2. **Formula for Variance:** Variance $V$ of a data set is given by: $$V = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$$ where $n$ is the number of data points, $x_i$ are the data points, and $\bar{x}$ is the mean. 3. **Calculate Mean for 5A:** $$\bar{x}_{5A} = \frac{90 + 92 + 89 + 95 + 79 + 88 + 94 + 96}{8} = \frac{723}{8} = 90.375$$ 4. **Calculate Variance for 5A:** Calculate each squared deviation: - $(90 - 90.375)^2 = 0.1406$ - $(92 - 90.375)^2 = 2.6406$ - $(89 - 90.375)^2 = 1.8906$ - $(95 - 90.375)^2 = 21.3906$ - $(79 - 90.375)^2 = 129.8906$ - $(88 - 90.375)^2 = 5.6406$ - $(94 - 90.375)^2 = 13.1406$ - $(96 - 90.375)^2 = 31.6406$ Sum of squared deviations: $$0.1406 + 2.6406 + 1.8906 + 21.3906 + 129.8906 + 5.6406 + 13.1406 + 31.6406 = 206.375$$ Variance: $$V(5A) = \frac{206.375}{8} = 25.7969$$ 5. **Calculate Mean for 5B:** $$\bar{x}_{5B} = \frac{93 + 89 + 90 + 94 + 93 + 80 + 89 + 95}{8} = \frac{723}{8} = 90.375$$ 6. **Calculate Variance for 5B:** Calculate each squared deviation: - $(93 - 90.375)^2 = 6.8906$ - $(89 - 90.375)^2 = 1.8906$ - $(90 - 90.375)^2 = 0.1406$ - $(94 - 90.375)^2 = 13.1406$ - $(93 - 90.375)^2 = 6.8906$ - $(80 - 90.375)^2 = 107.6406$ - $(89 - 90.375)^2 = 1.8906$ - $(95 - 90.375)^2 = 21.3906$ Sum of squared deviations: $$6.8906 + 1.8906 + 0.1406 + 13.1406 + 6.8906 + 107.6406 + 1.8906 + 21.3906 = 159.875$$ Variance: $$V(5B) = \frac{159.875}{8} = 19.9844$$ **Final answers:** - Variance of 5A: $25.80$ (rounded to two decimals) - Variance of 5B: $19.98$ (rounded to two decimals)