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#1




Percent chance of a test being abnormal
A medical student accidentally sticks his finger after drawing blood on a woman, who is having elective surgery for removal of a benign breast mass. As part of the normal routine in handling accidental needle sticks, a baseline liver profile is ordered. The profile consists of a total bilirubin, serum aspartate aminotransferase (AST), serum alanine aminotransferase (ALT), serum alkaline phosphatase, and serum γglutamyltransferase. What is the approximate percent chance of one of these five tests having a value outside the normal reference interval for the test?
A. 10% B. 13% C. 23% D. 35% E. 47% From usmle consult
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mayursn39 (09272011) 
#4




c  23 %
Normal referance range mean +/ 2sd  95 % distribution So Chance of having all 5 tests to be normal  0.95*0.95*0.95*0.95*0.95 = 0.87 And chance of having 1 test outside of normal range = 10.87 = 0.23 
#5




jaimin ,your explanation was more credible until you made the subtraction error. isnt 10.87=0.13?
Most of us suck in maths, that's why we are all here.Dont take it personally, even i could have done the same. i will stick with answer A. I honestly wouldnt have time to a math in the exam and examiners are not expecting us to do it too. So its usually 10%(lowest) or 47%(highest). moreover needlestick injury related HIV,HBV,HCV is <2%. Viral hepatitis should be elevating the liver enzymes in case of a transmission,but still the probablity is low, but definitely not 47%. If you have hepatitis you shall most likely have AST,ALT elevated. 
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mayursn39 (09272011) 
#6




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ya ,i know ..i am not gud at maths .. But my answer is correct. C 23 % corrected calculation 0.95*0.95*0.95*0.95*0.95 = 0.77 
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#8




i have a quite different outlook
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mayursn39 (09272011) 
#9




well 13 % numbers are wrong calculation.so ignore it.
The right ones are there in the following posts. i think its a routine post exposure evaluation of blood borne diseases in needlstick injuries and the question asked here is the probablity of at least single lab test turning outside the normal range among 5 of them. 
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mayursn39 (09272011) 
#10




Answer is C
Most normal ranges are established by adding and subtracting 2 standard deviations (SD) from the mean of the test, which encompasses 95% of the normal population. Therefore, 5% of the normal population will be outside out of the normal range (outlier). The likelihood of an outlier increases as the number of tests ordered increases. The likelihood of an outlier = 100 − (0.95n × 100), where n is the number of tests ordered. In this patient, five tests were ordered: 100 − (0.955 × 100) = 100 − (0.77 × 100) = 23% chance of an outlier in one of the five tests
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