Measure of Dispersion

Variance measures the spread or scatter of numbers in a dataset. It quantifies how far each number is from the mean. Variance is denoted by $\sigma^2$ (sigma squared) or Var(x). The standard deviation ($\sigma$, sigma) is the positive square root of the variance. It measures the spread of data about the mean. A larger standard deviation indicates greater variability, while a small standard deviation indicates observations are close to the mean. A standard deviation of zero means all observations are equal to the mean.

If $x_1, x_2, x_3, ..., x_n$ represent N observations with mean $\bar{x}$, then the variance is the average of the squares of the mean deviations:

$\sigma^2 = Var(x) = \frac{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + ... + (x_n - \bar{x})^2}{N}$

The variance of an ungrouped dataset is given by:

$Var(x) = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{N} \quad (4.19)$

The standard deviation of an ungrouped dataset is:

$\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{N}} \quad (4.20)$

Equation (4.19) can be expanded to obtain a simplified form:

$Var(x) = \frac{\sum_{i=1}^{n}(x_i^2 - 2x_i\bar{x} + \bar{x}^2)}{N}$

$Var(x) = \frac{\sum_{i=1}^{n}x_i^2 - 2\bar{x}\sum_{i=1}^{n}x_i + \sum_{i=1}^{n}\bar{x}^2}{N}$

Since $\bar{x} = \frac{\sum_{i=1}^{n}x_i}{N}$, then $\sum_{i=1}^{n}x_i = N\bar{x}$. Also, $\sum_{i=1}^{n}\bar{x}^2 = N\bar{x}^2$. Substituting these into the variance equation:

$Var(x) = \frac{\sum_{i=1}^{n}x_i^2 - 2\bar{x}(N\bar{x}) + N\bar{x}^2}{N}$

$Var(x) = \frac{\sum_{i=1}^{n}x_i^2 - 2N\bar{x}^2 + N\bar{x}^2}{N}$

$Var(x) = \frac{\sum_{i=1}^{n}x_i^2 - N\bar{x}^2}{N}$

$Var(x) = \frac{\sum_{i=1}^{n}x_i^2}{N} - \bar{x}^2$

Thus, the simplified formula for variance of ungrouped data is:

$\sigma^2 = Var(x) = \frac{\sum_{i=1}^{n}x_i^2}{N} - \bar{x}^2 \quad (4.21)$

The simplified formula for standard deviation of ungrouped data is:

$\sigma = \sqrt{\frac{\sum_{i=1}^{n}x_i^2}{N} - \bar{x}^2} \quad (4.22)$

Example 1

Find the variance and standard deviation of 6, 7, 10, 11, 11, 13, 16, 18, 25 correct the answer into two decimal places.

Solution:

Mean $\bar{x} = \frac{6 + 7 + 10 + 11 + 11 + 13 + 16 + 18 + 25}{9} = \frac{117}{9} = 13$

Variance: $\sigma^2 = \frac{280}{9} \approx 31.11$

Standard deviation: $\sigma = \sqrt{31.11} \approx 5.58$

Example 2

Find the variance and standard deviation of the set of data 6, 7, 10, 12, 13, 4, 8, 12.

Solution:

Mean $\bar{x} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = 9$

Variance: $\sigma^2 = \frac{722}{8} - 9^2 = 90.25 - 81 = 9.25$

Standard deviation: $\sigma = \sqrt{9.25} \approx 3.04$

Example 3

Find the mean, variance, and standard deviation of the following data set:

Solution:

Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{(110\times12)+(130\times29)+(150\times22)+(170\times4)+(190\times2)+(210\times1)}{70} = \frac{9660}{70} = 138$

Variance: $\sigma^2 = \frac{\sum f(x - \bar{x})^2}{N} = \frac{29120}{70} = 416$

Standard deviation: $\sigma = \sqrt{416} \approx 20.40$

For grouped data, the variance is calculated using the formula:

$\sigma^2 = \frac{\sum_{i=1}^{n}f_i(x_i - \bar{x})^2}{N}$

where $N$ is the total number of observations, $f_i$ is the frequency of the $x_i$ variable, and $\bar{x}$ is the mean of the distribution.

This formula can be expanded as follows:

$\sigma^2 = \frac{\sum_{i=1}^{n}f_i(x_i^2 - 2x_i\bar{x} + \bar{x}^2)}{N}$

$\sigma^2 = \frac{\sum_{i=1}^{n}f_ix_i^2 - 2\bar{x}\sum_{i=1}^{n}f_ix_i + \bar{x}^2\sum_{i=1}^{n}f_i}{N}$

Since $\bar{x} = \frac{\sum_{i=1}^{n}f_ix_i}{N}$ and $\sum_{i=1}^{n}f_i = N$, we can substitute:

$\sigma^2 = \frac{\sum_{i=1}^{n}f_ix_i^2 - 2\bar{x}(N\bar{x}) + \bar{x}^2N}{N}$

$\sigma^2 = \frac{\sum_{i=1}^{n}f_ix_i^2 - 2N\bar{x}^2 + N\bar{x}^2}{N}$

$\sigma^2 = \frac{\sum_{i=1}^{n}f_ix_i^2 - N\bar{x}^2}{N}$

$\sigma^2 = \frac{\sum_{i=1}^{n}f_ix_i^2}{N} - \bar{x}^2 \quad (4.23)$

The standard deviation of grouped data is:

$\sigma = \sqrt{\frac{\sum_{i=1}^{n}f_ix_i^2}{N} - \bar{x}^2} \quad (4.24)$

Variance using the coding method (where $x_i = A + cu_i$, $A$ is an assumed mean, $c$ is the class width, and $u_i$ is a coded value):

$\sigma^2 = \frac{\sum_{i=1}^{n}f_i(A + cu_i)^2}{N} - \left(\frac{\sum_{i=1}^{n}f_i(A + cu_i)}{N}\right)^2$

$\sigma^2 = \frac{\sum f_i(A^2 + 2Acu_i + c^2u_i^2)}{N} - \left(A + c\frac{\sum f_iu_i}{N}\right)^2$

$\sigma^2 = \frac{A^2\sum f_i + 2Ac\sum f_iu_i + c^2\sum f_iu_i^2}{N} - \left(A^2 + 2Ac\frac{\sum f_iu_i}{N} + c^2\left(\frac{\sum f_iu_i}{N}\right)^2\right)$

$\sigma^2 = A^2 + 2Ac\frac{\sum f_iu_i}{N} + c^2\frac{\sum f_iu_i^2}{N} - A^2 - 2Ac\frac{\sum f_iu_i}{N} - c^2\left(\frac{\sum f_iu_i}{N}\right)^2$

$\sigma^2 = c^2\left(\frac{\sum f_iu_i^2}{N} - \left(\frac{\sum f_iu_i}{N}\right)^2\right) \quad (4.25)$

The standard deviation using the coding method is:

$\sigma = c\sqrt{\frac{\sum f_iu_i^2}{N} - \left(\frac{\sum f_iu_i}{N}\right)^2} \quad (4.26)$

Example 1

Given the following frequency distribution table.

Find the mean and standard deviation by using the coding method.

Solution:

Class size $c = 4$, assumed mean $A = 14.5$.

Mean: $\bar{x} = A + c\frac{\sum f_iu_i}{N} = 14.5 + 4\left(\frac{54}{107}\right) \approx 14.5 + 2.02 \approx 16.52$

Standard deviation: $\sigma = c\sqrt{\frac{\sum f_iu_i^2}{N} - \left(\frac{\sum f_iu_i}{N}\right)^2} = 4\sqrt{\frac{504}{107} - \left(\frac{54}{107}\right)^2} \approx 4\sqrt{4.71 - 0.25} \approx 4\sqrt{4.46} \approx 4 \times 2.11 \approx 8.44$

Example 2

The scores of 96 students in an Advanced Mathematics test were recorded as follows.

Use the coding method to calculate each of the following, correct to 2 decimal places:

(a) Mean mark of the scores

(b) Variance of the scores

(c) Standard deviation of the scores

Solution:

Class width $c = 10$, assumed mean $A = 55$.

(a) Mean: $\bar{x} = A + c\frac{\sum fu}{N} = 55 + 10\left(\frac{-47}{96}\right) \approx 55 - 4.90 \approx 50.10$

(b) Variance: $\sigma^2 = c^2\left[\frac{\sum fu^2}{N} - \left(\frac{\sum fu}{N}\right)^2\right] = 10^2\left[\frac{479}{96} - \left(\frac{-47}{96}\right)^2\right] \approx 100[4.99 - 0.24] \approx 100 \times 4.75 \approx 475$

(c) Standard deviation: $\sigma = \sqrt{\sigma^2} = \sqrt{475} \approx 21.79$

Example 3

The masses in kilograms of 200 students in a certain school were recorded as follows.

Calculate the mean mass and standard deviation by the coding method.

Solution:

Class width $c = 5$, assumed mean $A = 52$.

Mean: $\bar{x} = A + c\frac{\sum fu}{N} = 52 + 5\left(\frac{70}{200}\right) = 52 + 1.75 = 53.75$

Standard deviation: $\sigma = c\sqrt{\frac{\sum fu^2}{N} - \left(\frac{\sum fu}{N}\right)^2} = 5\sqrt{\frac{412}{200} - \left(\frac{70}{200}\right)^2} = 5\sqrt{2.06 - 0.1225} = 5\sqrt{1.9375} \approx 5 \times 1.39 \approx 6.96$