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nbme21/Block 3/Question#24 (52.9 difficulty score)
Diastolic blood pressures are obtained in two ...

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+9
submitted by amarousis(23),
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so orf hsit oen ouy aehv to oolk at het soicilatd lobdo rsupesre nad tha'st hte aeuvsl or'yeu ppsuesdo to era.d tno hte ubensrm in hte ml.sconu Leki rguop 'sX medo si 07 eeaubcs ti sha atth alevu 23 .emsit purog sy' dmeo si 08 bueaecs ti apasrpe 20 m.esti fro e,mdnia oyu wluod have ot iwert teh istoailcd mnureb 025-10 hwo eerv anmy msite ti aseprap dan tneh nifd teh ldi.emd iytrkc neoisq.ut

sahusema  Wow. I hate this. I only looked at the number of participants and completely ignored the Diastolic BP readings +5
ma_rad  Everyone commented how to get the mode right. But there is an easier way to realize that the median in Y is higher without all the calculations. If you see, the last Diastolic BP in group X is 110 (as there are ZERO people with 120). While group Y has 8 people with 120 DBP. This automatically shifts the median to the higher side. I got this wrong though at first I didn't pay attention to the "0" number at group X for 120 DBP +9
brasel  I think there's another easy way to find the median without writing out every value. There are 100 total people in each group, so that means the median (if the DBPs are written in ascending order, which they are) is the 50th person. Group X: 8 + 12 + 30 = 50, so median is 70 Group Y: 2 + 8 + 10 + 20 +10 = 50, so median is 90 +8
mangotango  I did it the way @brasel explained. The way @ma_rad did it could give the incorrect answer in some cases (e.g. Group X had 0 ppl with 120 BP but a ton of ppl for 110 BP etc. + Group Y had 10 ppl with 120 BP but basically none with 110 BP etc.). In this question that way worked but it's not always guaranteed since median doesn't sway with outliers, but mean does. // FA 2019 pg. 261 +

+5
submitted by yb_26(258),
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yrrso for teh ,iknapn hpoe that liwl hpel oesemon

fadila  median should be: 70 for group X ; 85 for group Y median here is the avarage of Diastolic BP measured betweeen the 50 & 51 patients (since the number of patients in each group in even; Median= (n+1)/2 -> (100+1)/2= 50.5) Group X = (70+70)/2 = 70 Group Y = (80+90)/2 = 85 +1

+2
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rnyEveeo dtocenmme woh ot teg hte mdoe .itgrh uBt rethe si na esiear awy to lezeira atth eth mdaien ni Y is rigehh howutti all teh nliaucls.acto fI you e,se teh lsta ocDlaiist BP in gopru X is 101 (sa hrete rea ERZO ploepe ihwt 0.1)2 hWlei prguo Y has 8 eeplop htwi 102 PD.B hiTs claalmtoaiytu isshtf eth nmdiea to eht ihgreh .ised

I ogt isht wongr hghuot ta tsrif I dt'din pay ennittaot to eth ""0 enmbur ta pgoru X rfo 021 DPB

+1
submitted by yb_26(258),
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syror rof hte ,nkapin hope ttha lilw ehpl eeomnos

yb_26  tried to attach photo, sorry for that( +
yb_26  +2

+0
submitted by hajj(0),
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acn oennya xaeinpl h?sti i konw niemda rfo y si igrhhe by iuactlalonc btu x sah wot dosme os who moec y sah hiegrh o?dem

lispectedwumbologist  The mode in X is 32 and the mode in Y is 80 +
lispectedwumbologist  The mode in X is 70 and the mode in Y is 80* +1
hajj  Thank you! +
hungrybox  Just checking in so I could feel smart about getting this right despite bombing the rest of the test lmao +4
usmleuser007  can someone please explain the median in this +
nala_ula  The median can be known by first assembling the numbers in order from least to greater. If it's an uneven number set, the number in the middle is the median (for example: 4, 10, 12, 20, 27 = median is 12 since this is the number in the middle); if the numbers are even then you have to take the two values in the middle, add them up and divide them by 2 [for example: 4, 10, 12, 12, 20, 27 = (12+12)/2 = 12]. Page 261 on FA 2019 explains it as well. Not sure if I explained it well... good luck on the test, people! +
dubin johnson  Can someone please explain how the mode for Y than X. Not sure how we got the values above. Thanks! +
dubin johnson  I mean how is the mode for Y greater than mode for x? +1
sgarzon15  Mode is the one that repeats the most once you list them in order +
usmile1  Median would be the BP value that the person in the 50th percentile of each group would have. So for group X, to find the 50th percent value, I added 8 + 12 + 32 = 52, which is right above 50, so the median would be 70 mmHg for group X. Doing the same thing for group Y, 2+8+10+20+ 18 = 58; the 50th percentile would fall in group that had a BP of 90 mmHg. which makes the median higher for group Y. hope that isn't wrong, and helps someone! +4
poisonivy  I did it the same way! not pretty sure if it is the right way to do it, but it gave me the right answer! +