10. (1.5-10) Suppose we want to investigate the percentage of abused children in

Question

10. (1.5-10) Suppose we want to investigate the percentage of abused children in a certain population. To do this, doctors examine some of these children taken at random from that population. However, doctors are not perfect: They sometimes classify an abused child (A+) as one not abused (D) or they classify a nonabused child (AT) as one that is abused (D"). Suppose these error rates are P(D-IA+-0.08 and P(D+IA") = 0.05, respectively, thus, P(D+IA+) = 0.92 and P(D-IA") = 0.95 are the probabilities of the correct decisions. Let us pretend that only 2% of all children are abused, that is, P(A+) = 0.02 ad P(A-) = 0.98. (a) Select a child at random. What is the probability that the doctor classifies this child as abused? That is, compute (b) Compute P(AD+)andP(AD+ (c) Compute P(A D andP(A ID (d) Are the probabilities in (b) and (c) alarming? This happens because the error rates of 0.08 and 0.05 are high relative to the fraction 0.02 of abused children in the population.

Explanation / Answer

Ans:

a)P(D+)=0.02*0.92+0.98*0.05

=0.0184+0.049

=0.0674

P(D-)=1-0.0674=0.9326

b)P(A-/D+)=P(D+/A-)*P(A-)/P(D+)=0.05*0.98/0.0674=0.727

P(A+/D+)=1-0.727=0.273

c)P(A-/D-)=P(D-/A-)*P(A-)/P(D-)=0.95*0.98/0.9326=0.9983

P(A+/D-)=1-0.9983=0.0017

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