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#1




95% of observations
A large study of serum folate levels in women aged 16  45years reveals that this parameter is normally distributed, with a mean of 5.0 ng/ml and a standard deviation of 0.5 ng/ml. According to the study results, 95% of serum folate observations in these patients will lie between the following limits.
A 4.0 and 6.0 ng/ml B 4.0 and 5.5 ng/ml C 4.5 and 5.5 ng/ml D 3.5 and 6.0 ng/ml E 3.5 and 6.5 ng/ml 
The above post was thanked by:  
usmle100 (06242010) 
#2




A
The correct answer is A 4.0 and 6.0 ng/ml...
1SD is .5 ng/ml....so 2SD = 1 ng/ml.... 95% CI = mean +/ 2SD....so....5 +/ 1 = 4  6 ng/ml.... 
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achistikbenny (06232010) 
#3




Correct .. but
Quote:
mean +/ 2 Z scores X SD/N 
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achistikbenny (06232010), aungnaingmyo (06242010) 
#4




1 sd  68
2 sd  95 3sd  99.7 so here its 95..ie 2sd...so 1 on each side of 5 which will be 51= 4 and 5+1=6 
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achistikbenny (06242010), aungnaingmyo (06242010) 
#5




You are all correct the ans is A. thanks for the explanations. I was actually confused and didn't know how to go about it. my fear has always been biostatistics but i thank God for this forum. the forum is actually helping me get thru it .thanks again

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aungnaingmyo (06242010) 
#6




Quote:
mean + or  [Z score x sd/sqrt(N)]...and Z is almost 2 (95% confidence Z score = 1.96)...so, you dont have to multiple Z score by two...and dont forget the square root in the denominator. 
#7




Hi bebix..
Can you please explain me what value should we take for 'N'..?? Because here they are saying large sample??.. 
#8




Quote:

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BiostatisticsEpidemiology 
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