math distribution of averages and distribution of standard deviation for comparison of bytes

Let me change my question:

I have a sequence of 512 numbers (these 512 numbers can have a value between 0 and 255) (in computer mode these numbers can be between 00000000 and 11111111 (bits)). For these 512 numbers I calculate the average value, standard deviation value and kurtosis value. Now I want to calculate the values for the distribution of averages, distribution of standard deviations and distribution of Byte values. Finaly, I want to compare the distributions from other sequesces of 512 numbers to find out the similarity (as percentage) that two sequences have. I dont understand how can I calculate these distributions.

for example Distribution of Averages is : the probability that an average chosen from all the averages of a memory block is of value B in the range of 0-255.

Dxb=Pr((B+1))>Xj>=B ,Xj=Average.

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To check I understand your question: you have two sequences of 512 bytes, call them $X$ and $Y$, and you want a measure of how similar they are? – Chris Taylor May 19 '11 at 10:41
yes exactly! i want to have a percent of their similarity. for example that sequences X and Y are 78% similar. – nomercy May 19 '11 at 10:46
To say that 'sequences X and Y are 78%' similar, in this approach, makes no sense to me. Before asking "how to calculate" something, you must be clear what results you expect. Say X and Y have the same values in distinct order- are they 100% similar? Say $x_i = y_i+1$ : the have same variance and higher centered moments, and similar average (difference =1) How similar would you call them? – leonbloy Jun 24 '11 at 14:06

For the new question, it appears that you have no problem calculating the mean, standard deviation, and kurtosis of one series. Then if you have a bunch of series and want to calculated the distribution of means you do it the same way. You have data that comes as $(i,mean_i,sd_i,kurt_i)$ and ignore where it came from. So you can calculate the mean and standard deviation of all the means. As you have not defined "the similarity as percentage" I don't know what to do with that.