Let's solve the problem step by step as you requested. --- ### **Step 1: Introduction to Entropy in Decision Trees** **Concept:** In decision trees, **entropy** is a measure of impurity or disorder used to quantify the uncertainty in a dataset. It tells us how mixed the classes are in a node. The formula for entropy for a binary classification is: \[ \text{Entropy} = - p_1 \log_2(p_1) - p_2 \log_2(p_2) \] where \(p_1\) and \(p_2\) are the proportions of the two classes. --- ### **Step 2: Identify the Values Needed** From the root node in the tree: - **No Churn:** 376 - **Churned:** 104 - **Total:** \(376 + 104 = 480\) So: - \(p_{\text{No Churn}} = \frac{376}{480}\) - \(p_{\text{Churned}} = \frac{104}{480}\) --- ### **Step 3: Calculate the Proportions** Calculate each probability: - \(p_{\text{No Churn}} = \frac{376}{480} = 0.7833\) - \(p_{\text{Churned}} = \frac{104}{480} = 0.2167\) --- ### **Step 4: Substitute into the Entropy Formula** \[ \text{Entropy} = - (0.7833) \log_2(0.7833) - (0.2167) \log_2(0.2167) \] Calculate each term: - \(0.7833 \log_2(0.7833) = 0.7833 \times (-0.3519) = -0.2755\) - \(0.2167 \log_2(0.2167) = 0.2167 \times (-2.2091) = -0.4789\) Sum the positive values: \[ \text{Entropy} = 0.2755 + 0.4789 = 0.7544 \] --- ### **Step 5: Final Answer in Few Sentences** **The entropy in the root node is approximately** ***0.7544*** **(rounded to 4 decimal places).** --- ### **Step 6: One-Sentence Explanation of Each Step** 1. Introduced entropy as a measure of impurity in decision trees. 2. Identified the counts of each class at the root node. 3. Calculated the class proportions. 4. Substituted the proportions into the entropy formula and computed the value. 5. Stated the final entropy at the root node. 6. Summarized each step in one sentence.