1. Given the data set $x = \{6, 11, 13, 17, 9, 12\}$ with mean $\mu = 11.33$. 2. To verify the population variance, we use the formula: $$\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}$$ where $N=6$ is the number of data points. 3. Calculate each squared deviation: $$(6-11.33)^2 = (-5.33)^2 = 28.41$$ $$(11-11.33)^2 = (-0.33)^2 = 0.11$$ $$(13-11.33)^2 = 1.67^2 = 2.79$$ $$(17-11.33)^2 = 5.67^2 = 32.11$$ $$(9-11.33)^2 = (-2.33)^2 = 5.43$$ $$(12-11.33)^2 = 0.67^2 = 0.45$$ 4. Sum the squared deviations: $$28.41 + 0.11 + 2.79 + 32.11 + 5.43 + 0.45 = 69.3$$ 5. Divide by the number of data points $N=6$: $$\sigma^2 = \frac{69.3}{6} = 11.55$$ 6. Rounded to two decimal places, the population variance is $11.55$, which approximately matches the given value $11.56$. Final answers: - Mean = $11.33$ - Population variance = $11.56$ (rounded as provided)