Answer:
The probability that a randomly selected person gets incorrect result is 2.2 × 10⁻⁴
Explanation:
The parameters given are;
The accuracy of the test for a person who has the respiratory synctial virus = 97%
The accuracy of the test for a person who does not have the respiratory synctial virus = 99%
We have;
a = TP =
b = FP
c = FN
d = TN
a/(a + c) = 0.97
d/(d + b) = 0.99
a/(a + b) = 0.97*0.0055/(0.97*0.0055 + (1 - 0.99)*(1-0.0055))
PPV = 0.349 = 34.9%
Therefore, we have;
a/(a + c) = 0.97 and
a/(a + b) = 0.349
0.97(a + c) =0.349(a + b)
(0.97 - 0.349)a = 0.349·b - 0.97·c
a = (0.349·b - 0.97·c)0.621
b × (1 - 0.0055) = (1 - 0.97)×(1 - 0.0055)
b = 1 - 0.97 = 0.03
Similarly,
c = 1 - 0.99 = 0.01
The proportion of the population that have false positive and false negative = 0.03 + 0.01 = 0.04 = 4%
The probability that a randomly selected person gets incorrect result = 0.04×0.0055 = 0.00022.