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anonymous
 one year ago
A new medical screening test is used to detect a rare, nonlifethreatening condition. If a person has this condition, the test always detects it. Approximately 0.2% of the population has the condition. Over many trials, the test returns a positive result 6% of the time. Julio takes the test and gets a positive result. To the nearest tenth of a percent, what is the probability that Julio actually has the condition?
anonymous
 one year ago
A new medical screening test is used to detect a rare, nonlifethreatening condition. If a person has this condition, the test always detects it. Approximately 0.2% of the population has the condition. Over many trials, the test returns a positive result 6% of the time. Julio takes the test and gets a positive result. To the nearest tenth of a percent, what is the probability that Julio actually has the condition?

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marihelenh
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal Earlier when I was working on those percent problems, were you there and trying them also? I think I have an idea, but I will need you to check it.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437440977572:dw

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.0That is not what I got. That isn't even how I interpreted the question.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0sorry don't care for stats

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 stats

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@marihelenh percent ok probability part of stats need to be verified

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1D = person has disease T = test returns positive `If a person has this condition, the test always detects it` means P(TD) = 1 `Approximately 0.2% of the population has the condition` > P(D) = 0.002 `Over many trials, the test returns a positive result 6% of the time` > P(T) = 0.06 Use Bayes Theorem to get \[\Large P(DT) = \frac{P(TD)*P(D)}{P(T)}\] \[\Large P(DT) = \frac{1*0.002}{0.06}\] \[\Large P(DT) \approx 0.0333\] So if the test returns positive, then there is a 3.33% chance (roughly) that he has the condition.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.0Aahhh, I would have had it right golly gee! That was what I got but in a much simpler way.
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