The top node had 10 high taste items and 20 not high taste items. We determined the Gini score should be .5555. For the H2S > 7 split, the > 7 child had 7 of 10 high and 3 of 10 not high. We determined the Gini score to be .58. The This program will calculate purity measures for a leaf node and display the Gini score and other values calculated. Please enter the number of objects in this leaf node of type I 7 Please enter the number of objects in this leaf node of type II 3 The Gini score for this node is 0.580 Assume that P() is the proportion or probability of an object of one type occurring. The formula for the Gini score is : Gini = (P(Type I objects))2 + (P(Type II objects))2 This program will calculate purity measures for a leaf node and display the Gini score and other values calculated. Please enter the number of objects in this leaf node of type I 7 Please enter the number of objects in this leaf node of type II 3 The Gini score for this node is 0.580 Assume that P() is the proportion or probability of an object of one type occurring. The formula for the Gini score is : Gini = (P(Type I objects))2 + (P(Type II objects))2