we will consider the town of Springfield, where there are 100,000 adults. Consid
ID: 3122710 • Letter: W
Question
we will consider the town of Springfield, where there are 100,000 adults. Consider a different genetic disease, which affects only 0.1% of the U.S. population. The test for this disease is also 98% accurate. All the adults in Springfield were also tested for this disease.
How many of the residents of Springfield are likely to have the disease?
How many of the people who actually have the disease get a positive test result?
How many of the people who do not have the disease get a positive test result?
Of the people who get a positive test result, how many of them have the disease? Convert this to a percentage: What percent of people who get a positive result actually have the disease?
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Explanation / Answer
Solution:
total number of adults of springfield = N = 100000
Assume that Adults having disease = D
therefore P(D)= 0.1% = 0.001
Therefore
(A) Number of the residents of Springfield are likely to have the disease = N x P(D) = 100000 x 0.001 = 100
(B) since 100 people are likely to have the disease and probablitlity of test is 98% accurate
the people who actually have the disease get a positive test result = 100 P(T) = 100 x 0.98 = 98
(where P(T) = probability of accurate result )
Number of people who actually have the get a positive thest result = 98
(C) Similerly (100000-1000) 99000 are the adults who don't have disease and probability of not getting accurate test result P(nT) = 100%-98% = 2% = 0.02
therefore
Number of the people who do not have the disease get a positive test result = 99000*0.02 = 1980
(D) total number of people who gets positive test result = 98 + 1980 = 2078
percent of people who get a positive result actually have the disease = (98/2078)x100 = 4.71 %
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