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Statistics Question 22 (3 points) Suppose we know that 80% of the time Hay Fever(HF) causes red eyes (R) in those with the disease. At

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Statistics

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Question 22 (3 points) Suppose we know that 80% of the time Hay Fever(HF) causes red eyes (R) in those with the disease. At any point in time 10% of the general population has Hay Fever. But, among the people who don't have Hay Fever 15% still have red eyes. Note also that we have no prior information about the percentage of the population that might have red eyes. You have red eyes. What is the probability that you have Hay Fever? 1:) 0.372 C) 0.56 None of the above (:3. 0.08 O 0.012 Question 23 (3 points) Now, suppose that in addition to red eyes, you also have congestion (C). We know that Hay Fever results in congestion 60% of the time. Furthermore, based on a recent study, if you have red eyes, you are 30% likely to also have congestion (so P(C|R) = 0.3). Assuming that congestion and red eyes are independent given Hay Fever, what is the updated probability that you have Hay Fever (given that you have both red eyes and congestion)? (T :1. 0.112 '1: None of the above (T j) 0.180 (f j) 0.067 (f :1. 0.744 Question 35 (3 points) Consider the following training data indicating whether bank customers received loans based on their credit history, income level, and debt. Instance Credit Income Debt Loan? good high high yes 2 good high low yes good median low yes Aw good median high no 5 good low low yes 6 bad high high yes 7 bad high low yes 8 bad median high no 9 bad low high no 10 bad low low no We want to use the ID3 decision tree learning algorithm to determine what feature/attribute should be the root of the decision tree. What is the information gain for the attribute Credit (i.e. Gain(Credit)). Select the closest answer. [Note: when computing entropies, use Log base 2]. 0.845 0.12 0.97 ONone of the aboveQuestion 36 (3 points) Again consider the following training data: Instance Credit Income Debt Loan? 1 good high high yes good high low yes good median low yes good median high no CO VOUIAWON good low low yes bad high high yes bad high low yes bad median high no 9 bad low high no 10 bad low low no When using ID3 decision tree learning algorithm to determine what feature/attribute should be the root of the decision tree, what is the information gain for the attribute income. Select the closest answer. [Note: when computing entropies, use Log base 2]. O 0.42 Oo 0.55 0.275Question 37 (2 points) Again consider the following training data and ID3 algorithm: Instance Credit Income Debt Loan? 1 good high high yes IN good high low yes good median low yes good median high no good low low yes bad high high yes bad high low yes 8 bad median high no 9 bad low high no 10 bad low low no Which attribute will be the root of the decision tree? Credit Income Debt The data is not sufficient to construct a modelQuestion 38 (2 points) In this problem, we'll use the same bank data for Naive Bayes classification (with discrete variables). Credit good good good good low low high median median low good low low median high 5'5' 5 E l- O m 3 low As an example, the conditional probability table for Credit is given below:l Credit Yes No Good 0.667 0.25 Bad 0.333 0.75 E.g., P(Credit=Good | No) is 0.25. Compute the conditional probability P(lncome=High |Yes). (f) 1.0 ()0A (1' j) 0.167 (f :1 0.667 Question 39 (2 points) Continuing with the Naive Bayes classification problem: Instance Credit Income Debt Loan? 1 good high high yes IN good high low yes good median low yes good median high no good low low yes 6 bad high high yes bad high low yes 8 bad median high no 9 bad low high no 10 bad low low no Compute the conditional probability P(Debt=High | No). 0.25 0.75 0.6 0.333Question 40 (3 points) Continuing with the Naive Bayes classification problem: Instance Credit Income Debt Loan? 1 good high high yes good high low yes good median low yes UI A WON good median high no good low low yes 6 bad high high yes bad high low yes bad median high no CO bad low high no 10 bad low low no Suppose a new instance X is to be classified: X = The posterior probability P(X | No) is (approximately): 0.0375 O 0.37 0.63 0.72

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