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Discussion Starter · #1 ·
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
 

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Correct .. but

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
 

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Discussion Starter · #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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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. :)
 
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