1. **Stating the problem:** We have data on the status pekerjaan (job status) of parents of students in classes A, B, and C. We want to determine which class has the most variation (heterogeneous) or is more homogeneous in terms of parental job status. 2. **Understanding the concept:** To measure variation or homogeneity in categorical data, we can use the concept of variance or standard deviation of the distribution of counts or proportions. Another approach is to use the coefficient of variation or entropy, but here we will use variance of the counts. 3. **Data given:** | Status | Class A | Class B | Class C | |--------|---------|---------|---------| | ASN | 20 | 25 | 30 | | TNI | 1 | 4 | 1 | | POLRI | 1 | 1 | 1 | | WIRASWASTA | 11 | 3 | 1 | | Total | 33 | 33 | 33 | 4. **Calculate proportions for each class:** For each class, calculate the proportion of each job status: Class A: $$p_{ASN} = \frac{20}{33}, p_{TNI} = \frac{1}{33}, p_{POLRI} = \frac{1}{33}, p_{WIRASWASTA} = \frac{11}{33}$$ Class B: $$p_{ASN} = \frac{25}{33}, p_{TNI} = \frac{4}{33}, p_{POLRI} = \frac{1}{33}, p_{WIRASWASTA} = \frac{3}{33}$$ Class C: $$p_{ASN} = \frac{30}{33}, p_{TNI} = \frac{1}{33}, p_{POLRI} = \frac{1}{33}, p_{WIRASWASTA} = \frac{1}{33}$$ 5. **Calculate variance of proportions for each class:** Variance formula for proportions $p_i$ with mean $\bar{p}$: $$\text{Var} = \frac{1}{n} \sum_{i=1}^n (p_i - \bar{p})^2$$ where $n=4$ (number of job categories). Calculate mean proportion for each class: $$\bar{p} = \frac{1}{4} \sum p_i = \frac{1}{4}$$ (since total proportions sum to 1, mean is always 0.25) Calculate variance for Class A: $$p = \left[\frac{20}{33}, \frac{1}{33}, \frac{1}{33}, \frac{11}{33}\right] = [0.606, 0.030, 0.030, 0.333]$$ $$\bar{p} = 0.25$$ $$\text{Var}_A = \frac{1}{4} ((0.606-0.25)^2 + (0.030-0.25)^2 + (0.030-0.25)^2 + (0.333-0.25)^2)$$ $$= \frac{1}{4} (0.127 + 0.048 + 0.048 + 0.007) = \frac{1}{4} (0.230) = 0.0575$$ Calculate variance for Class B: $$p = \left[\frac{25}{33}, \frac{4}{33}, \frac{1}{33}, \frac{3}{33}\right] = [0.758, 0.121, 0.030, 0.091]$$ $$\text{Var}_B = \frac{1}{4} ((0.758-0.25)^2 + (0.121-0.25)^2 + (0.030-0.25)^2 + (0.091-0.25)^2)$$ $$= \frac{1}{4} (0.258 + 0.017 + 0.048 + 0.025) = \frac{1}{4} (0.348) = 0.087$$ Calculate variance for Class C: $$p = \left[\frac{30}{33}, \frac{1}{33}, \frac{1}{33}, \frac{1}{33}\right] = [0.909, 0.030, 0.030, 0.030]$$ $$\text{Var}_C = \frac{1}{4} ((0.909-0.25)^2 + (0.030-0.25)^2 + (0.030-0.25)^2 + (0.030-0.25)^2)$$ $$= \frac{1}{4} (0.434 + 0.048 + 0.048 + 0.048) = \frac{1}{4} (0.578) = 0.1445$$ 6. **Interpretation:** - Class A variance = 0.0575 (lowest) - Class B variance = 0.087 - Class C variance = 0.1445 (highest) Higher variance means more variation in job status distribution. 7. **Conclusion:** Class C has the most varied (heterogeneous) parental job status distribution. Class A is the most homogeneous (least varied). This is because Class C has a very dominant ASN proportion but very low others, causing high variance. Class A has a more balanced distribution among job statuses. Hence, **Class C is the most varied, Class A is the most homogeneous.**