Monday, 27 August 2018

A Graphic Method of Calculating Standard Deviation,

From a series of posts on the Standard Deviation in 2008:

I’ve always been interested in the methods of graphic solutions to equations. Fortunately, over the years Dave Renfro has been willing to keep me in mind when he runs across old journal articles and send me copies, dozens over the last few years on just this topic. So, when I found a patent for a mechanical machine that would do standard deviations, I thought there had to be a graphic method also. I set about trying to construct one, and what follows is one such method. I have confined myself to three points, but the method would generalize to any number of data values.
The first step is to plot the points on the y-axis of a coordinate grid, and draw a horizontal line through the mean of the values. The first image shows such a construction starting with the values a=7, b=2, and c=3, with the mean, M, at 4. [If images are not sharp, click on them to open a full sized image]

Clever observers will have noticed another point on the graph at (-1,4), one to the left of the mean point, M. We will use this point to square the length of the deviation of A (the distance from M to A). To Do this we make MA the mean proportional between 1 and MA2. This can be easily done by the method of similar triangles. We make a line from the point one left of M to point A to complete a right triangle. Then if we construct a perpendicular to this line at A, it will intersect the line through the mean at a new point I call AA. The distance M to AA is the square of the length MA.. we have found the square of the deviation of point A.

Now we just need to transpose this line to another graph for safekeeping where we will add it to the other squared deviations. We repeat the previous process with points B and C, and copy each of the lengths sequentially onto a line, thus finding the sum of the squares of the deviations.

There seems to be one image of the sum of the squares of the deviations added together that doesn't come up, so you can find it here

Now we employ similar triangles again to find the mean of the sum of the squares of the deviations. We need to divide the length of the sum (Origin to CC) by n=3. To do this we drop a perpendicular from one end of the line (0-CC) that is three units long. We connect the far end (CC) to the other end of the three unit line(see figure below) and then construct a parallel to O-CC one unit away from the end. By the properties of similar triangles, we know that P-MSS is the Mean of the sum of the squares (O-CC divided by three)… we have found the variance.

Now all that is left is to take the square root, and we do this by reversing the method we used to find the square of the deviations, we find the mean proportional between P-MSS and one. So we construct a point one to the left of the segment )-MSS and then bisect this new segment to find the center. Now we create a circle with a centered at this midpoint and using the MSS + 1 as a diameter. The positive distance from where this point crosses the x-axis to the point P is the standard deviation of the three points… Which I have calculated out with two decimal points accuracy…(although I used a negative sign …Oops).. .

The method would be the same, of course for any number of points.

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