Question

Consider a diagnostic test for a hypothetical disease based on measuring the amount of a

certain biomarker present in blood. High levels of the biomarker are often found in individuals

with the disease, but a number of non-disease conditions can also cause high levels

of the biomarker. Individuals without the disease have biomarker levels that are normally

distributed with mean 1.6 ng/mL (nanograms per milliliter of blood), and standard deviation

0.50 ng/mL. Individuals with the disease have biomarker levels that are normally

distributed with mean 5 ng/mL and standard deviation 1.2 ng/mL. Values of 2.5 ng/mL

and higher constitute a positive test result. In a population where 6% of individuals are thought to have the hypothetical disease,

calculate the probability that an individual who tests positive has the disease.

Answer 1

We are given here that:
P( disease ) = 0.06, therefore P(no disease) = 1 - 0.06 = 0.94

Now for the person without disease, we have here:

Therefore the probability of a positive test without the disease is computed here as:
P(+ | W) = P(X > 2.5)

Converting it to a standard normal variable, we have:
P(+ | W) = P(Z > (2.5 - 1.6)/ 0.5 )

P(+ | W) = P(Z > 1.8)

Getting it from the standard normal tables, we have:
P(+ | W) = 0.0359

Similarly now for the person with the disease, we have:

The positive test probability here is computed as:
P(+ | D) = P(X > 2.5)

P(+ | D) = P(Z > (2.5 - 5) / 1.2)

P(+ | D) = P(Z > -2.0833)

P(+ | D) = 0.9814

Using law of total probability now, we have:
P(+) = P(+ | D)P(D) + P(+ | W)P(W) = 0.9814*0.06 + 0.0359*0.94 = 0.0926

Therefore using Bayes theorem now, we get here:
P(D | +) = P(+ | D)P(D) / P(+) = 0.9814*0.06 / 0.0926 = 0.6357

Therefore 0.6357 is the required probability here.