Thank you so much for the replies - I really appreciate your time!
However I'm wondering if that is the whole picture.
Eg. consider two 10 minute periods:
The first has all the 1 minute means the same, but all the 1 minute SDs large.
The second has all the 1 minute means varying widely, but all the 1 minute SDs small.
Then the formulas above would say the 10 minute SD would be larger for the first 10 minute period than the second. Whereas actually the 10 minute SD should be larger for the second 10 minute period than the first.
I guess I'm asking if the answer should include a term for the 1 minutes means????
Sorry - not very good at expressing myself - hope that makes sense!
And thank you again - more thoughts appreciated!
Because I am looking for the standard deviation of the 1 second data - and the standard deviation of the 1 minute data will be smaller than that. Can I combine the standard deviation of the 1 minute data with the combined standard deviations of the 1 minutes periods?
I'm not looking for an accurate measurement (if you see what I mean) - it's just that the 10 minute standard deviation is "standard" measurement and can tell you something empirically about the turbulence of the wind (which is what I'm interested in).
I have set up an Excel spreadsheet with random example numbers and found that I can approximate the 10 minute standard deviation by a simple equation combining the standard deviations and the means. However the answers vary so I fear it's not something for which a precise answer is possible. I will use my approximation and try to make sure the datalogger is set up better next time!!
Thank you everyone!!
Sigma means 'The sum of'. So if for example I want to find the mean of a set of 20 numbers, I would write.....Sigma (the sum of) x/20
x bar (mean)=∑x/n