A biomedical marker Y is used to screen for a particular disease. Among people with the disease, the distribution of Y is Normal, with mean 1000 and standard deviation 200. Among people without the disease, the distribution is Normal with mean 300 and standard deviation 100.

Because of these distributions, people screened are sent for further investigation if the result of their screening test is too high.

If someone is deemed worthy of investigation after the screening test, the next steps are expensive and difficult for the person involved. Those administering the test decide that among people without the disease, they should only carry out these further tests for 0.1% of people. What threshold for Y should be set for this criterion?

Report the value to one decimal place.

Question

A biomedical marker Y is used to screen for a particular disease.  Among people with the disease, the distribution of Y is Normal, with mean 1000 and standard deviation 200.  Among people without the disease, the distribution is Normal with mean 300 and standard deviation 100.

Because of these distributions, people screened are sent for further investigation if the result of their screening test is too high.

If someone is deemed worthy of investigation after the screening test, the next steps are expensive and difficult for the person involved.  Those administering the test decide that among people without the disease, they should only carry out these further tests for 0.1% of people.  What threshold for Y should be set for this criterion?

Report the value to one decimal place.

BlackTom AI · Accepted Answer

The problem asks for a screening threshold Y such that only 0.1% of people without the disease would surpass it, given the non-diseased group follows a Normal distribution with mean 300 and standard deviation 100.
First, interpret the criterion: we want P(Y > t | no disease) = 0.001, where t is the threshold.
Since Y | no disease ~ Normal(mean = 300, sd = 100), standardize t to a Z-score: z = (t - 300) / 100.
We need P(Z > z) = 0.001 for Z ~ N(0,1). Looking up or computing the standard normal tail probability, z ≈ 3.0902 because P(Z > 3.0902) ≈ 0.001.
Now back-transform to t: t = 300 + z * 100 = 300 + 3.0902 * 100 ≈ 300 + 309.02 = 609.02.
Rounding to one decimal place, the threshold should be 609.0.
To verify the logic: if we set a threshold around 609, only about 0.1% of the non-diseased group, which has a mean of 300 and sd of 100, would exceed it, aligning with the given cutoff criterion.
If the threshold were lower, a larger fraction of non-diseased individuals would exceed it, violating the 0.1% constraint; if higher, even fewer would be tested, which also contradicts the specified rate.
This approach relies purely on properties of the normal distribution and standard normal tail probabilities, with no influence from the diseased distribution since the criterion explicitly concerns only those without the disease.

MAST90007_2025_WIN 2025 Quiz 1

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