A person who has non-functional cone type won't be able to identify the_____________of
1 answer:
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The data set is missing in the question. The data set is given in the attachment.
Solution :
a). In the table, there are four positive examples and give number of negative examples.
Therefore,
and
The entropy of the training examples is given by :
= 0.9911
b). For the attribute all the associating increments and the probability are :
+ -
T 3 1
F 1 4
Th entropy for is given by :
= 0.7616
Therefore, the information gain for is
0.9911 - 0.7616 = 0.2294
Similarly for the attribute the associating counts and the probabilities are :
+ -
T 2 3
F 2 2
Th entropy for is given by :
= 0.9839
Therefore, the information gain for is
0.9911 - 0.9839 = 0.0072
Class label split point entropy Info gain
1.0 + 2.0 0.8484 0.1427
3.0 - 3.5 0.9885 0.0026
4.0 + 4.5 0.9183 0.0728
5.0 -
5.0 - 5.5 0.9839 0.0072
6.0 + 6.5 0.9728 0.0183
7.0 +
7.0 - 7.5 0.8889 0.1022
The best split for observed at split point which is equal to 2.
c). From the table mention in part (b) of the information gain, we can say that produces the best split.
