Average, Variance and Standard Deviation

Pocket

1. Average (Mean)

Average is the most frequently used metric to compute statistical features of populations or datasets.

Given n observations of, for example, body height $x_i~(i=1,2,3,\cdots,n)$, average $\bar{x}$ is defined as follows: $\displaystyle \bar{x} = \frac{1}{n}\times \sum_{i=1}^n x_i = \frac{1}{n}\times(x_1 + x_2 + x_3 + \cdots + x_n)$

Example: Let’s compute the average height of the following 5 individuals.

The height data of these 5 individuals is：178,165,171,163,173 (cm) $x_1$ 178cm $x_2$ 165cm $x_3$ 171cm $x_4$ 163cm $x_5$ 173cm $\displaystyle \bar{x} = \frac{1}{5}\times \sum_{i=1}^n x_i = \frac{1}{5}\times (x_1 + x_2 + x_3 + x_4 + x_5)\\~~~~=\frac{1}{5}\times (178 + 165 + 171 + 163 + 173) = 170 (cm)$

2. Variance

Variance estimates the deviation between the average and the data points (observations). The variance $V[x]$ is defined as follows: $\displaystyle V[x] = \frac{1}{n}\times \sum_{i=1}^n (x_i-\bar{x})^2\\ ~~~~~~= \frac{1}{n}\times \Bigl\{(x_1-\bar{x})^2 + (x_2-\bar{x})^2 + \cdots + (x_n-\bar{x})^2 \Bigr\}$

Example: Let’s calculate the Variance using the data of the previous example. $\displaystyle V[x] = \frac{1}{5}\times \sum_{i=1}^5 (x_i-\bar{x})^2\\ ~~~~~~ = \frac{1}{5}\times \Bigl\{(x_1-\bar{x})^2 + (x_2-\bar{x})^2 + (x_3-\bar{x})^2 + (x_4-\bar{x})^2 + (x_5-\bar{x})^2 \Bigr\}\\ ~~~~~ = \frac{1}{5}\times \Bigl\{(178-170)^2 + (165-170)^2 + (171-170)^2 + (163-170)^2 + (173-170)^2 \Bigr\}\\ ~~~~~ = \frac{1}{5}\times \Bigl\{8^2 + 5^2 + 1^2 + 7^2 + 3^2 \Bigr\}\\ ~~~~~ = 29.6 ~~(\mbox{cm}^2)$

It is worth pointing out that the unit of Variance (for example, using the above example the Variance unit is ( $\mbox{cm}^2$)) is not the original unit of the observation, which is (cm). Because the unit is different, and it is difficult to understand, we often use Standard Deviation as defined in the following section.

3. Standard Deviation

The Standard Deviation $S[x]$ is defined as the squared root of Variance $V[x]$. $S[x]=\sqrt{V[x]}$

Example: Let’s compute the Standard Deviation using data from the previous example. $S[x]=\sqrt{V[x]}=\sqrt{29.6}=5.4~~(\mbox{cm})$