USMLE Rx question on standard deviation

[Sir Gillies]

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Hi guys,

I came across this question:

The mean blood glucose concentration for a cohort of 100 infants is 95 mg/dL. The standard deviation for these measurements is 15 mg/dL. What percentage of glucose measurements falls between 65 and 125 mg/dL?

Answer: 95% and they saythat the 95% CI is +/- 1.96.

I just can't understand how they got to the result. I would be really grateful if someone could explain it to me.
Thanks a lot in advance.

 

[ZGX]

Given the SD is 15.
Remeber SD is sigma. Based on Gaussian distribution, +1/-1 Sigma is 68%. +2/-2 Sigma is 95%, this is what we are looking for.
They are asking for the % of measurements that is between 65 and 125.
That is 2SD from the mean. In other words it is 30.
65+30=95 the mean, and 125-30=95 the mean.

2SD is considered 95% Confidence Interval.

For the Z-score. To calculate that it is what you (observed standard - mean) / SD = (125 - 95) / 15 = 30 / 15 = 2.
Z = 2 roughly 1.96+/-

 

[Sir Gillies]

That is a great reply!!! Thank you so much.

Am I correct in saying that in FA's Gaussian distribution +1/-1 (etc) equals +1/-1 times sd?

The thing that confused was how to get to the Z-score and I truly thank you for explaining it to me. But why do we even need a Z-score? Why not just use the SD value?

 

[ZGX]

That is a great reply!!! Thank you so much.

Am I correct in saying that in FA's Gaussian distribution +1/-1 (etc) equals +1/-1 times sd?

The thing that confused was how to get to the Z-score and I truly thank you for explaining it to me. But why do we even need a Z-score? Why not just use the SD value?

The Z-score tells us how far we are deviating away from the mean.
So in your question the Z score is +/- 1.96. That indicates a we are +2/-2 Standard Deviations from the mean.
Since SD is 15 based on your question. Two times that is 30, thus the Z-Score of 2 gives us the range of 65 to 125.

 

[Old Style Nanny]

The Z-score tells us how far we are deviating away from the mean.
So in your question the Z score is +/- 1.96. That indicates a we are +2/-2 Standard Deviations from the mean.
Since SD is 15 based on your question. Two times that is 30, thus the Z-Score of 2 gives us the range of 65 to 125.

Isn't that technically not completely right? The CI deals with SEM rather than SD, i.e. 95% CI is (Mean - 1.96*SEM) to (Mean + 1.96*SEM).

And, Standard Errors of Means is used when you do a study with multiple samples and you are projecting a CI for the population for purposes of extrapolation from the study, whereas the question asks what percentage of the glucose measurements fell within two values for which we need the Standard Deviation.

In short, I believe the answer explanation in USMLERx is wrong (even though the answer is correct) and shows poor conceptualization.

 

[Phloston]

There is nothing remotely incorrect about USMLE Rx's answer here.

+/-1.96SDs = a range of 60 here. +/-1.96SDs = 95% of the values.

SirGillies, I still have my subscription open to USMLE Rx. If you post the question ID#, I'll go take a look at it myself.

 

[Old Style Nanny]

There is nothing remotely incorrect about USMLE Rx's answer here.

I believe there is, on the grounds that the term "Confidence Interval" has no place in the steps required to arrive at the answer of +/- 1.96 SD.

Confidence interval is used in studies to show the reliability when the data is extrapolated to the population whereas the question simply asks what percentage of glucose measurement falls (implying, out of the already measured values) within the specified range.

 

[valuat]

I'm with Old Stule Nanny on this one. ICs are calculated with the Standard Error, which is equal to the sample standard deviation divided by the square root of sample size (N).

If still in doubt, check it out: http://en.wikipedia.org/wiki/Confidence_interval

 

[Sir Gillies]

Thanks guys for your replies. Unfortunately I can't find the question, but thanks a lot Phloston for offering to have a look at it - I really appreciate it!