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#1
06-23-2010
 USMLE Forums Veteran Steps History: 1 + CK Posts: 234 Threads: 63 Thanked 128 Times in 63 Posts Reputation: 158
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 (06-24-2010)

#2
06-23-2010
 USMLE Forums Guru Steps History: --- Posts: 487 Threads: 69 Thanked 296 Times in 133 Posts Reputation: 316
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....
 The above post was thanked by: achistikbenny (06-23-2010)
#3
06-23-2010
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 742 Threads: 137 Thanked 2,461 Times in 441 Posts Reputation: 2564
Correct .. but

Quote:
 Originally Posted by khushboo 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....
Your answer is correct. But they don't call it Confidence Interval. This is just 95% of a normally distributed population. Which is as you said mean +/- 2 standard deviations, 95% Confidence Interval of the mean on the other hand is calculated by this formula
mean +/- 2 Z scores X SD/N
 The above post was thanked by: achistikbenny (06-23-2010), aungnaingmyo (06-24-2010)

#4
06-23-2010
 USMLE Forums Scout Steps History: Not yet Posts: 18 Threads: 2 Thanked 4 Times in 2 Posts Reputation: 14

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 5-1= 4
and 5+1=6
 The above post was thanked by: achistikbenny (06-24-2010), aungnaingmyo (06-24-2010)
#5
06-24-2010
 USMLE Forums Veteran Steps History: 1 + CK Posts: 234 Threads: 63 Thanked 128 Times in 63 Posts Reputation: 158

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
 The above post was thanked by: aungnaingmyo (06-24-2010)
#6
08-22-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,268 Times in 881 Posts Reputation: 3278

Quote:
 Originally Posted by rasheed Your answer is correct. But they don't call it Confidence Interval. This is just 95% of a normally distributed population. Which is as you said mean +/- 2 standard deviations, 95% Confidence Interval of the mean on the other hand is calculated by this formula mean +/- 2 Z scores X SD/N
Actually, the formula for a 95% CI is

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
08-22-2011
 USMLE Forums Guru Steps History: Not yet Posts: 324 Threads: 70 Thanked 239 Times in 144 Posts Reputation: 249

Hi bebix..
Can you please explain me what value should we take for 'N'..??
Because here they are saying large sample??..
#8
08-22-2011
 USMLE Forums Guru Steps History: Not yet Posts: 324 Threads: 70 Thanked 239 Times in 144 Posts Reputation: 249

Quote:
 Originally Posted by struggle Hi bebix.. Can you please explain me what value should we take for 'N'..?? Because here they are saying large sample??..
Oh yeah i got it..I didnt looked the question properly..they are asking for 95% population..

 Tags Biostatistics-Epidemiology

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