# Simple question, difficult problem

#### bennyorit

##### New Member
Hi,
First I have to admit that my knowledge in statistics is very basic.
I'm a surgeon (that explains the reason).
I'm trying to conduct a meta-analysis for which the reported data in the articles I include must be in means and SD. I forgot to mention that I'm using RevMan 4.2 (from the Cochrane corporation) to do the calculations.

The problem is that most of the articles I included in my analysis reported their data as mean and range (not SD); example:
Duration of the operation was 165 (90-450) minutes. No SD Is there a way to calculate the SD if I know the mean, the range and the number of datapoints?

If there is a way to do that, please let me know as soon as possible.

Thanks #### quark

Is there a way to calculate the SD if I know the mean, the range and the number of datapoints?
It can't be done. Sorry.

#### bennyorit

##### New Member
I kind of figured that out myself already.
Thanks anyway.

#### PeterVincent

##### New Member
Bennyorit,

There is a way that you could get an approximation to the SD by using the empirical rule or Chebyshev's rule.

Chebyshev's rule: The proportion of observations that are within k standard deviations of the mean is at least 1 - 1/(k^2). This conservative rule applies to any distribution. If you were to estimate that at least 95% of the population values lie between 90-450, the range, the k value would be 4.472 so: 450 - 90 = 360, 2 * 4.472 = 8.944 standard deviations wide (4.472 each side of the mean). 360/8.944 = 40.25 ~~ 40. An estimate of SD is 40 minutes.

Empirical Rule: This only applies to Normal distributions. If you believe that the times are normally distributes this is less conservative and more precise.

Roughly 68% of the observations are within 1 standard deviation of the mean, roughly 95% of the observations are within 2 standard deviations of the mean, roughly 99.7% of the observations are within 1 standard deviation of the mean. If you were to estimate that 95% of the population values lie between 90-450, the corresponding value is 2, so: 450 - 90 = 360 is 2 * 2 = 4 standard deviations wide. 360/4 = 90. An estimate of SD is 90 minutes.

Peter