mitty
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In a large sample of randomly selected healthy adults, low-density lipoprotein (LDL) cholesterol values have a mean of 110 mg/dL, a standard deviation (SD) of 10 mg/dL, and a standard error of the mean (SEM) of 3 mg/dL. Assuming that the LDL cholesterol values are distributed normally, what is the best estimate of the normal limits for LDL cholesterol in this population?
A. 80 to 140 mg/dL
B. 90 to 130 mg/dL
C. 100 to 120 mg/dL
D. 104 to 116 mg/dL
E. 107 to 113 mg/dL
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| vanshita
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| study_ing
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nice question..helps pinpoint dif btw sd and sem plus their use in questions..
any explanations as to y choose one over the other?
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| Moctopod
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I'll go for B
The SEM quantifies how accurately you know the true mean of the population.
The SD quantifies the degree of scatter within the data (i.e. how much the data values differ from each other). We want an estimate for the normal limits in the population, which is usually given as sample mean+/- 2SD.
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| docarchana
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formula is mean +- SE. so 107 to 113 is answer.
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| usmle_mogul
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I agree with B, mean +/- 2 SD = 95% of pupulation. SEM is a distractor here I guess, to indicate how large n was for this study.
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| vallia
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guys please explain your calculation...I'm kinda lost here
mine was
confidence interval= mean+/_1,96x SEM= 110 +/_5,88= 104-116, answer D
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| fongch
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The question is asking for the estimate of the normal limits rather than confidence interval, thus the answer must be mean+/- sem. Personally I will go with D.
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| mitty
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The answer is B.See moctopod's and usmle_mogul 's explanations.
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| klimt
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Right. It's B
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| docarchana
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very good q.
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| drfax
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mctopod explain 2 SD plz, y ????
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| equinoxe
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should be B, it's about a distribution with known mean and SD and _not_ about a sample of a population with known SEM (quality of the sample) and Z scores that should approximate the real (in this case known) population. The q asks about where an individual would be expected to be found, not how good the sample was at aproximating the real population.
the sample says that the real population - the mean is in fact
mean and CI = Z x SEM => mean = 110 +/- 2 x 3 = 110 +/- 6 (z =2 for 95%) = 104-116
individual expected to be found in +/- 2 SD (95%) = 110 +/- 20 = 90-130
to combine the two results would yield - is this thing possible? any help? please?
104-20 to 116+20
84 - 136
I'd go for B since I understand they ask about the distribution/ population and not the sample quality (the "real" mean )
- 95% the real mean should be 110 +/- 6 (z x sem)
- 95% of population in 2SD from 110
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| jsknetsdude
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In a large sample of randomly selected healthy adults, low-density lipoprotein (LDL) cholesterol values have a mean of 110 mg/dL, a standard deviation (SD) of 10 mg/dL, and a standard error of the mean (SEM) of 3 mg/dL. Assuming that the LDL cholesterol values are distributed normally, what is the best estimate of the normal limits for LDL cholesterol in this population?
A. 80 to 140 mg/dL
B. 90 to 130 mg/dL
C. 100 to 120 mg/dL
D. 104 to 116 mg/dL
E. 107 to 113 mg/dL
Confidenc Interval = Mean +/- Z-Score X SEM Mean =110 Z-Score= for 95% CI = 2 and for 99% CI = 2.5 SEM=3 since "best estimate of the normal limits " there fore CI should be 99 % 110 +/- 2.5 X 3
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