Suppose that we have a blood test that has an 80% chance of detecting the presen

Question

Suppose that we have a blood test that has an 80% chance of detecting the presence of the L-virus when it is present, and the chance of the test mistakenly indicating presence of the virus in a non-infected individual is only 1%. Furthermore, we will assume that only 0.01% of the population is infected with the L-virus. (a) Calculate the marginal probability that a person randomly chosen from the population will get a negative test. (b) Calculate the conditional probability of a randomly chosen subjects being infected with the virus given that they had a positive test. (c) Suppose that we perform 3500 tests on subjects drawn independently from the population. Calculate the probability that more than 33 will test positive. You can use an appropriate approximation if necessary.

Explanation / Answer

LET Probability of negative test and postive test are P(NT) and P(PT) respectively,

probabilty of infected with virus and not infected with Virus are P(D) and P(ND) respectively

here P(D) =0.0001

and P(ND) =1-0.0001 =0.9999

P(PT|D) =0.80

and P(PT|ND) =0.01

a) probability that test will be positive =P(PT) =P(D)*P(PT|D)+P(ND)*P(PT|ND) =0.0001*0.8+0.9999*0.01=0.010079

hence probability that a person will get a negaive test =P(NT)=1-P(PT) =1-0.010079=0.989921

b) conditional probability P(D|PT) =P(D)*P(PT|D)/P(PT) =0.0001*0.8/0.010079=0.007937

c)probability of posiive test p=0.010079

therefore mean number of positive test =np =3500*0.10079=35.2765

and std deviation =(np(1-p))1/2 =5.9094

from normal approximation probability that more then 33 will test positive=P(X>=33) =1-P(X<=32)

=1-P(Z<(32.5-35.2765)/5.9094) =1-P(Z<-0.46985)=1-0.3192=0.6808

please revert for any clarification required

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