Instruction
As we already know from the previous exercise, skewness is given by the formula
Skewness=\dfrac{\frac{1}{n}((x_1-\overline{x})^3+(x_2 - \overline{x})^3 + \ldots + (x_n - \overline{x})^3)}{\sqrt{(\frac{1}{n-1}((x_1-\overline{x})^2+\ldots+(x_n-\overline{x})^2))^3}}
Consider the following set:
1, 2, 3, 3, 4, 4, 4, 5
The corresponding histogram is shown below. It looks like a left-tailed histogram, so we'd expect skewness to be negative.
We can calculate the skewness for this dataset using the above formula. First, we find the arithmetic mean \overline{x}=3.25. The rest of the calculations are given in the table below:
| x_i | x_i-\overline{x} | (x_i-\overline{x})^2 | (x_i-\overline{x})^3 |
|---|---|---|---|
| 1 | -2.25 | 5.0625 | -11.390625 |
| 2 | -1.25 | 1.5625 | -1.953125 |
| 3 | -0.25 | 0.0625 | -0.015625 |
| 3 | -0.25 | 0.0625 | -0.015625 |
| 4 | 0.75 | 0.5625 | 0.421875 |
| 4 | 0.75 | 0.5625 | 0.421875 |
| 4 | 0.75 | 0.5625 | 0.421875 |
| 5 | 1.75 | 3.0625 | 5.359375 |
| SUM | 0 | 11.5 | -6.75 |
Skewness = \dfrac{\frac{1}{8}(-6.75)}{\sqrt{(\frac{1}{8-1}11.5})^3}=\frac{-0.84375}{\sqrt{1.642857^3}}=-0.4006951
A negative value implies an asymmetric histogram with a longer left tail. This histogram's asymmetry, however, is rather small.
