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I have been going in rounds with this problem... I may be thinking "complicated", any advice?

I have the mean and total sample size (=number of data points) and I need to know what is the standard deviation (SD).

I know I can calculate back the sum of individual scores from the formal formula for calculation of the mean, i.e.:

M = Σ(X) / N

where X=individual data points N=number of data points

However after this step I am stuck. To find the SD using the variance I need to know the individual data points and which I don't have.

I then end up with two "unknown" variables, S2 and X in this formula:

S2 = Σ(X-M) 2 / N - 1

Thanks!


Thank you André and Jonathan. I now got some extra information: I am given the N and mean(maximum), e.g.: N=596, mean(maximum): 5.86(39.1); any extra advice?

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You are stuck. If the data points are all equal (they might be) your sample variance would be $0$. If they wiggle all over the place, the sample variance would be high. –  André Nicolas Dec 1 '12 at 21:46
    
What do you mean by "mean(maximum)"? –  alex.jordan Dec 1 '12 at 23:20
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If all you know is the mean and sample size, then no. The standard deviation could be 6 or $1.5\times 10^{10^{10}}$.

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Thank you! Andre and Jonathan. I now got some extra information: I am given the N and mean(maximum), e.g.: N=596, mean(maximum): 5.86(39.1); any extra advice? –  user50505 Dec 1 '12 at 22:47
    
If you knew what distribution the data came from then you could estimate the variance using the distribution of the maximum order statistic, probably, but it won't be a very good estimator. If you don't know what distribution it comes from then you're still in a pretty bad place. What are you trying to do here, anyway? –  Jonathan Christensen Dec 1 '12 at 23:34
    
Thanks Jonathan. I am calculating back from reported data, see if I can pool a few studies together, once i have same data. –  user50505 Dec 2 '12 at 20:58
    
Ah, I see. I don't think you'll be able to get any estimate of the standard deviation reasonable enough to pool with, unfortunately. Maximum order statistics tend to have fairly disperse distributions. –  Jonathan Christensen Dec 3 '12 at 15:17
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