Jeff you didn't answer my question only confused me more.
Because the whole top unsquared would be zero, squaring it would always result in zero. It is the difference of the individual terms that must be squared, then summed, then divided, and then the root taken. Because I don't understand what you mean. what did the others have me figuring out. I've been working on this question for a long time and still have not come any closer to understanding. Just when I think i'm on my way to understanding and completing the equation you stated this. Could you explain in simpler terms what you mean. I'm a women returning to school after 30years and I'm so fustrated.
I hate to jump in so far into the lifetime of this question, but perhaps a
completely different perspective will help.
STARTING OVER FROM THE VERY BEGINNING
An example of 5 rolls of a die:
1 6 3 4 4
First step. Take the mean of the five numbers = (1+6+3+4+4)/5 = 3.6
That is a measure of the "central tendency" of the experimental data.
Now we need a measure of the dispersion or the
width of the distribution. The statistic we choose is called the "standard deviation," which I take to be defined as "The square root of the mean of the squares minus the square of the mean." Operationally that can be found on the same pass through the data as finding the mean.
Second step. Take the mean of the squares = (1+36+9+16+16)/5 = 15.6
Then sigma = sqrt[15.6 - 3.6^2] = 1.62
Note that if all five numbers are the
same - whatever the value - this gives a standard deviation of zero, because every one of the five deviations is zero. That is clearly the smallest.
A different procedure to find sigma is first to find the mean, then the deviations from the mean, and then the squares of those deviations.
mean square deviation = [(1-3.6)^2 + (6-3.6)^2 + (3-3.6)^2 + (4-3.6)^2 + (4-3.6)^2]/5 = 2.64
taking the square root --> sigma = 1.62
When n is small (and n=5
is small), either of these procedures produces a "biased" estimate of the standard deviation. Many people multiply by sqrt[n/(n-1)] for an "unbiased" estimator. The rationale for that is that one of the n degrees of freedom has been used up in finding the mean of the sample. That is a dfine point that will not affect comparing sigmas for different sets of 5 digits.
To make sigma as large as possible, the deviations from the mean have to be large. If repeats are allowed, these two sets have max sigma:
1 1 1 6 6
1 1 6 6 6
If repeats are not allowed, then think about omitting one of the six numbers. Omitting either
1 or
6 results in a set that is as compact as possible, hence will have minimum standard deviation.
Omitting one of the numbers in the middle, either
3 or
4, will be the most spread out you can get, hence maximum sigma.
Hope this helps.