2. Given the following data set, we apply a naïve Bayesian (NB) classifier to predict the Health status. Each patient is described by Temp={H,N}, BP={H,N}, Test1=(P,N} and Test2=(P,N}. The class is Health={D,H). ID Temp BP Testl Test2 Health 1234 H P N D 6754 H P. D. 2345 N H P. D. 2457 H. N D 3567 N H P D 7532 1456 N N 1468 N 8989 H N H. 9876 H N N N H. For a new patient, ID 9999, with Temp=H, BP=H, Test1=N, Test2=P, is he/she Healthy (H) or Diseased (D)? Show (a) the value of each term in your NB classifier, and (b) your prediction. Answer: P_temp(H/D)=3/5, P_temp(N/D)=2/5 P_temp(H/H)=3/5, P_temp(N/H)=2/5 P_BP(H/D)=4/5, P_BP(N/D)=1/5 P_BP(H/H)=1/5, P_BP(N/H)=4/5 P_Test1(P/D)=4/5, P_Test1(N/D)=1/5 P_Test1(P/H)=0/5, P_Test1(N/H)=5/5 P_Test2(P/D)=3/5, P_Test2(N/D)=2/5 P_Test2(P/H)=1/5, P_Test2(N/H)=4/5 new patient A: Temp=H, BP=H, Test1=N, Test2=P S((H,H,N, P),D) = 3/5 * 4/5 * 1/5 * 3/5 = 0.0576 S((H,H,N,P),H) = 3/5 * 1/5 * 5/5 * 1/5 = 0.024 S((H,H,N,P),D) > S((H,H,N,P),H) So predict A as class D. %D
2. Given the following data set, we apply a naïve Bayesian (NB) classifier to predict the Health status. Each patient is described by Temp={H,N}, BP={H,N}, Test1=(P,N} and Test2=(P,N}. The class is Health={D,H). ID Temp BP Testl Test2 Health 1234 H P N D 6754 H P. D. 2345 N H P. D. 2457 H. N D 3567 N H P D 7532 1456 N N 1468 N 8989 H N H. 9876 H N N N H. For a new patient, ID 9999, with Temp=H, BP=H, Test1=N, Test2=P, is he/she Healthy (H) or Diseased (D)? Show (a) the value of each term in your NB classifier, and (b) your prediction. Answer: P_temp(H/D)=3/5, P_temp(N/D)=2/5 P_temp(H/H)=3/5, P_temp(N/H)=2/5 P_BP(H/D)=4/5, P_BP(N/D)=1/5 P_BP(H/H)=1/5, P_BP(N/H)=4/5 P_Test1(P/D)=4/5, P_Test1(N/D)=1/5 P_Test1(P/H)=0/5, P_Test1(N/H)=5/5 P_Test2(P/D)=3/5, P_Test2(N/D)=2/5 P_Test2(P/H)=1/5, P_Test2(N/H)=4/5 new patient A: Temp=H, BP=H, Test1=N, Test2=P S((H,H,N, P),D) = 3/5 * 4/5 * 1/5 * 3/5 = 0.0576 S((H,H,N,P),H) = 3/5 * 1/5 * 5/5 * 1/5 = 0.024 S((H,H,N,P),D) > S((H,H,N,P),H) So predict A as class D. %D
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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
Transcribed Image Text:2.
Given the following data set, we apply a naïve Bayesian (NB) classifier to predict the Health status.
Each patient is described by Temp={H,N}, BP={H,N}, Test1={P,N} and Test2={P,N}. The class is Health={D,H}.
ID
Temp
ВР
Test1
Test2
Health
1234
H
H.
N
6754
H
N
N
P
D
2345
N
H
P
P
2457
H
H.
P
N
D
3567
N
H
P
7532
N
H.
N
N
H
1456
N
N
N
H
1468
H
N
N
N
H
8989
H
N
N
P
H
9876
H
N
N
N
H
For a new patient, ID 9999, with Temp=H, BP=H, Test1=N, Test2=P, is he/she Healthy (H) or Diseased (D)?
Show (a) the value of each term in your NB classifier, and (b) your prediction.
Answer:
P_temp(H/D)=3/5, P_temp(N/D)=2/5
P_temp(H/H)=3/5, P_temp(N/H)=2/5
P_BP(H/D)=4/5, P_BP(N/D)=1/5
P_BP(H/H)=1/5, P_BP(N/H)=4/5
P_Test1(P/D)=4/5, P_Test1(N/D)=1/5
P_Test1(P/H)=0/5, P_Test1(N/H)=5/5
P_Test2(P/D)=3/5, P_Test2(N/D)=2/5
P_Test2(P/H)=1/5, P_Test2(N/H)=4/5
new patient A: Temp=H, BP=H, Test1=N, Test2=P
S((H,H,N,P),D) = 3/5 * 4/5 * 1/5 * 3/5 = 0.0576
S((H,H,N,P),H) = 3/5 * 1/5 * 5/5 * 1/5 = 0.024
S((H,H,N,P),D) > S((H,H,N,P),H)
So predict A as class D.
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