NBME 23 200 (NBME Answers)

+20
submitted by colonelred_(107),

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albbtArteuti srik = ndiiencec in oxseepd – nciiencde ni oeeuxdpsn

= 0/0013,0 )eskor(ms - 3,00/003 (smo)onnekrs
= 30.0 - .001
= .020 s(o eth baitaurbltet ksir si utbao 2%)

gpnlAypi ti ot a itoanpluop fo 000,:10

= 2.00 * 100,00
= 002

charcot_bouchard What if i tell you this is a ques of Attributable risk % in exposed? AR= 0.02 / IR in exposed (30/1000) = 0.6667 30 case in 1000. So 300 case in 10,000 0.6667 x 300 = 200 or in another word 66% cases of 100 lung cancer cases in smokers is actually due to smoking. so in 300 cases of smokers 200 is actually due to smoking +4

charcot_bouchard This is a mind fuck. Lemme tell u guys if any consolation while doing the ques during test i did it with AR = 0.02; NNH = 1/0.02 = 50. 50 persons smoke to cause 1 cancer. 10K smoke to cause 200 cancer. +3

ls3076 Sorry if this is a stupid question. Why is it incorrect to simply apply the same proportion (30 cancer per 1000 smokers) to 10,000 smokers? +2

krewfoo99 @is3076 Thats exactly what is did. I still dont understand how that is wrong. But i guess they want us to think about it in terms of AR +

hhsuperhigh @Is3076 and @Krewfoo99, If a person doesn't smoke, the natural risk of getting lung cancer is 30/3000=1%. The smoker's risk is 30/1000=3%. This 3% is not purely contributed by smoking, but mixed with the natural risk. So for calculating the pure contribution made by smoking, you should use 3%-1% which is 2%. And this 2% is the pure contribution of smoking. Not all smokers get lung cancer, the same thing, not all lung cancer among smokers are attributed by smoking. They may get lung cancer anyway despite smoking or not. +12