Finally, the main character of this article, "standard deviation". In "mean deviation", the absolute value was used to change minus to plus, but in "standard deviation", it was squared and minus was taken. In other words, the idea is exactly the same, only the negative removal method is different.

As shown below, there is a similarity in that the origin is symmetrical with the origin as the folding point, and the negative value is changed to positive.

(code is here) (The blue line is the absolute value functiong(x)=|x|, The red line is the square functionf(x)=x^2is)

In the previous section, as shown in the figure below, the difference from the average value was expressed by the length of the line.

This time, the image of evaluating the difference by the square is to evaluate the degree of deviation from the average value with the area of the square as shown here.

Let's add up these areas and take the average.

Expressing this as a mathematical formula,

The sum of the squares of the deviations from the average and the average is called the "variance". At this point, the "standard deviation" is just one step away.

The formula shown at the beginning of this article is reprinted as follows.

\sigma = \sqrt{ {1 \over n} \sum_{i=1}^n(x_i - \bar{x})^2}

The difference from "dispersion" is whether or not to calculate the route. "Dispersion" is also an indicator of how much data is scattered, but the unit has changed because we are thinking about the area. So you can get the area back to length by taking the root $ \ sqrt {\ } $.

By defining that root $ \ sqrt {\ } $ is the value when squared, $ \ sqrt {9} = \ sqrt {3 ^ 2} = 3 $ or $ \ sqrt {25} = \ sqrt Represents the relationship {5 ^ 2} = 5 $. It's the reverse calculation of $ 3 ^ 2 = 9 $. Since $ 25 $ can be thought of as the area of a square of $ 5 \ times5 $,

If you write it in a mathematical formula, it will look like this. Calculate root $ \ sqrt {\ } $ for the entire variance. You can think of it as an "operation to extract the length of the side" with respect to the average value of the above area.

That is, the standard deviation is

standard deviation= \sqrt{Distributed}

It becomes.

It looks like this in Python code.

standard_deviation.py

x = [96, 63, 85, 66, 91, 89, 77]
ave = np.average(x)
total = 0
for i in range(len(x)):
    total += (x[i] - ave)**2
print np.sqrt(total/len(x))

Also, the reason why I changed it to "square" instead of "absolute value" is a little difficult, but since the processing of the absolute value used for mean deviation is mathematically difficult to handle, I adopted a square that is easy to handle. One point is that it was done. After that, as a concept of distance, it is rather natural to take a route by squared Three squares theorem. I think that is also one of the reasons.

Summary

• Mathematical handling becomes convenient
• Because the concept of distance can be defined as squared and routed in the first place

I think. (A little more detailed explanation will be given in another article: grin :)

</ i> 5. Example 1: Understanding with a graph

Deviation value= { (Score-average) \over standard deviation} \times 10 + 50