| Content On This Page | ||
|---|---|---|
| Variance: Definition and Calculation (for Ungrouped and Grouped Data) | Standard Deviation: Definition and Calculation | Properties of Variance and Standard Deviation |
Measures of Dispersion: Variance and Standard Deviation
Variance: Definition and Calculation (for Ungrouped and Grouped Data)
Definition and Concept
The Variance is a widely used measure of dispersion that quantifies the average variability or spread of the data points around their mean. It provides a single numerical value that summarizes how much the individual observations tend to deviate from the central value (the arithmetic mean).
The calculation of variance involves squaring the deviations of each observation from the mean. This squaring achieves two important goals:
- It ensures that positive deviations (values above the mean) and negative deviations (values below the mean) do not cancel each other out when summed. If we simply summed the deviations $(x_i - \bar{x})$, the sum would always be zero by definition of the mean.
- It gives greater weight to larger deviations. Squaring a larger deviation results in a much larger value than squaring a smaller deviation, thus making the variance (and its square root, the standard deviation) particularly sensitive to outliers.
The variance is typically denoted by:
- $\sigma^2$ (read as "sigma squared") when referring to the variance of an entire **population**.
- $s^2$ when referring to the variance of a **sample**. (Often, the denominator is $n-1$ instead of $n$ for sample variance, to provide an unbiased estimator of the population variance. However, in many introductory contexts, especially when treating the given data as the complete set of interest, the denominator $n$ or $N$ is used. We will primarily use the denominator $n$ or $N$ here, corresponding to $\sigma^2$).
The units of variance are the square of the units of the original data. For example, if the data represents weights in kilograms (kg), the variance will be in square kilograms (kg$^2$). This squared unit can make direct interpretation of the variance value itself less intuitive compared to the standard deviation.
Calculation for Ungrouped Data
For a set of $n$ individual observations $x_1, x_2, \dots, x_n$, the variance is calculated based on the deviations from the mean.
Method 1: Using the Definition Formula
This method directly applies the definition of variance as the mean of the squared deviations.
- Calculate the arithmetic mean, $\bar{x} = \frac{\sum x_i}{n}$.
- For each observation $x_i$, calculate its deviation from the mean: $x_i - \bar{x}$.
- Square each of these deviations: $(x_i - \bar{x})^2$.
- Sum all the squared deviations: $\sum_{i=1}^{n} (x_i - \bar{x})^2$.
- Divide the sum of squared deviations by the total number of observations, $n$.
Formula:
$\sigma^2 = \frac{\sum\limits_{i=1}^{n} (x_i - \bar{x})^2}{n}$
... (1)
Method 2: Using the Computational Formula (Shortcut Formula)
This formula is algebraically equivalent to the definition formula but is often easier to use for manual calculations, as it avoids calculating each individual deviation. It involves the sum of the squares of the observations and the mean.
Formula:
$\sigma^2 = \frac{\sum\limits_{i=1}^{n} x_i^2}{n} - (\bar{x})^2$
... (2)
This can also be written as: $\sigma^2 = \frac{1}{n} \left( \sum\limits_{i=1}^{n} x_i^2 - \frac{(\sum\limits_{i=1}^{n} x_i)^2}{n} \right)$
The formula states that variance is equal to the "Mean of the squares of the observations minus the square of the mean of the observations".
Derivation of the Computational Formula:
Starting from the definition formula:
$\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$
Expand the squared term $(x_i - \bar{x})^2 = x_i^2 - 2x_i\bar{x} + \bar{x}^2$:
$\sigma^2 = \frac{1}{n} \sum (x_i^2 - 2x_i\bar{x} + \bar{x}^2)$
Using summation properties ($\sum (a_i + b_i) = \sum a_i + \sum b_i$ and $\sum c a_i = c \sum a_i$, $\sum c = nc$):
$\sigma^2 = \frac{1}{n} \left( \sum x_i^2 - \sum (2x_i\bar{x}) + \sum \bar{x}^2 \right)$
$\sigma^2 = \frac{1}{n} \left( \sum x_i^2 - 2\bar{x} \sum x_i + n\bar{x}^2 \right)$
From the definition of mean, $\bar{x} = \frac{\sum x_i}{n}$, so $\sum x_i = n\bar{x}$. Substitute this into the equation:
$\sigma^2 = \frac{1}{n} \left( \sum x_i^2 - 2\bar{x} (n\bar{x}) + n\bar{x}^2 \right)$
$\sigma^2 = \frac{1}{n} \left( \sum x_i^2 - 2n\bar{x}^2 + n\bar{x}^2 \right)$
$\sigma^2 = \frac{1}{n} \left( \sum x_i^2 - n\bar{x}^2 \right)$
Distribute $\frac{1}{n}$:
$\sigma^2 = \frac{\sum x_i^2}{n} - \frac{n\bar{x}^2}{n}$
$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$
(Computational Formula Derived)
Calculation for Frequency Distributions (Ungrouped or Grouped Data)
When data is presented in a frequency distribution table, where $x_1, x_2, \dots, x_k$ are the distinct values (for ungrouped frequency distribution) or class marks (for grouped frequency distribution) and $f_1, f_2, \dots, f_k$ are their corresponding frequencies, the variance is calculated by incorporating the frequencies.
Let $N = \sum\limits_{i=1}^{k} f_i$ be the total frequency (total number of observations).
Method 1: Using the Definition Formula
This is an adaptation of the definition formula for frequency data.
- Calculate the mean: $\bar{x} = \frac{\sum f_i x_i}{N}$.
- For each value/class mark $x_i$, calculate its deviation from the mean: $x_i - \bar{x}$.
- Square each deviation: $(x_i - \bar{x})^2$.
- Multiply each squared deviation by its corresponding frequency $f_i$: $f_i (x_i - \bar{x})^2$.
- Sum these products: $\sum_{i=1}^{k} f_i (x_i - \bar{x})^2$.
- Divide the sum by the total frequency $N$.
Formula:
$\sigma^2 = \frac{\sum\limits_{i=1}^{k} f_i (x_i - \bar{x})^2}{N}$
... (3)
Method 2: Using the Computational Formula (Shortcut Formula)
This formula is the frequency distribution equivalent of the shortcut formula for ungrouped data.
Formula:
$\sigma^2 = \frac{\sum\limits_{i=1}^{k} f_i x_i^2}{N} - \left( \frac{\sum\limits_{i=1}^{k} f_i x_i}{N} \right)^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$
... (4)
Where $\sum f_i x_i^2 = f_1 x_1^2 + f_2 x_2^2 + \dots + f_k x_k^2$. This means square the $x_i$ value *before* multiplying by its frequency.
Method 3: Using the Step-Deviation Method (for Grouped Data with equal class width $h$)
When dealing with grouped data that has a constant class width ($h$), the Step-Deviation Method can significantly simplify calculations, especially if class marks are large. It uses step-deviations $u_i = \frac{x_i - A}{h}$, where $A$ is the assumed mean.
Formula:
$\sigma^2 = h^2 \left[ \frac{\sum\limits_{i=1}^{k} f_i u_i^2}{N} - \left(\frac{\sum\limits_{i=1}^{k} f_i u_i}{N}\right)^2 \right]$
... (5)
Where:
- $h$ is the class width.
- $u_i = \frac{x_i - A}{h}$ is the step-deviation.
- $\sum f_i u_i^2 = f_1 u_1^2 + f_2 u_2^2 + \dots + f_k u_k^2$. (Square $u_i$ *before* multiplying by $f_i$).
- $\sum f_i u_i$ is the sum used in the Step-Deviation method for the mean.
- $N = \sum f_i$ is the total frequency.
Derivation of the Step-Deviation Formula:
From the definition $u_i = \frac{x_i - A}{h}$, we have $x_i = A + h u_i$. Substitute this into the computational formula for frequency distributions (Formula 4):
$\sigma^2 = \frac{\sum f_i (A + h u_i)^2}{N} - \left(\frac{\sum f_i (A + h u_i)}{N}\right)^2$
Expand $(A + h u_i)^2 = A^2 + 2Ah u_i + h^2 u_i^2$ and $\sum f_i (A + h u_i) = \sum f_i A + \sum f_i h u_i = A \sum f_i + h \sum f_i u_i = AN + h \sum f_i u_i$.
$\sigma^2 = \frac{\sum f_i (A^2 + 2Ah u_i + h^2 u_i^2)}{N} - \left(\frac{AN + h \sum f_i u_i}{N}\right)^2$
$\sigma^2 = \frac{\sum f_i A^2 + \sum f_i 2Ah u_i + \sum f_i h^2 u_i^2}{N} - \left(\frac{AN}{N} + \frac{h \sum f_i u_i}{N}\right)^2$
$\sigma^2 = \frac{A^2 \sum f_i + 2Ah \sum f_i u_i + h^2 \sum f_i u_i^2}{N} - \left(A + h \frac{\sum f_i u_i}{N}\right)^2$
$\sigma^2 = \frac{A^2 N + 2Ah \sum f_i u_i + h^2 \sum f_i u_i^2}{N} - \left(A^2 + 2Ah \frac{\sum f_i u_i}{N} + h^2 \left(\frac{\sum f_i u_i}{N}\right)^2\right)$
$\sigma^2 = \left(\frac{A^2 N}{N} + \frac{2Ah \sum f_i u_i}{N} + \frac{h^2 \sum f_i u_i^2}{N}\right) - \left(A^2 + 2Ah \frac{\sum f_i u_i}{N} + h^2 \left(\frac{\sum f_i u_i}{N}\right)^2\right)$
$\sigma^2 = \left(A^2 + 2Ah \frac{\sum f_i u_i}{N} + h^2 \frac{\sum f_i u_i^2}{N}\right) - \left(A^2 + 2Ah \frac{\sum f_i u_i}{N} + h^2 \left(\frac{\sum f_i u_i}{N}\right)^2\right)$
The terms $A^2$ and $2Ah \frac{\sum f_i u_i}{N}$ cancel out:
$\sigma^2 = h^2 \frac{\sum f_i u_i^2}{N} - h^2 \left(\frac{\sum f_i u_i}{N}\right)^2$
Factor out $h^2$:
$\sigma^2 = h^2 \left[ \frac{\sum f_i u_i^2}{N} - \left(\frac{\sum f_i u_i}{N}\right)^2 \right]$
(Step-Deviation Formula Derived)
Example
Example 1. Find the variance of the data: 6, 7, 10, 12, 13, 4, 8, 12.
Answer:
Given: Dataset: 6, 7, 10, 12, 13, 4, 8, 12.
To Find: The variance ($\sigma^2$).
Solution:
This is ungrouped data with $n=8$ observations.
First, calculate the mean $\bar{x}$. Sum of observations $\sum x_i = 6+7+10+12+13+4+8+12 = 72$.
$\bar{x} = \frac{\sum x_i}{n} = \frac{72}{8} = 9$
... (i)
Method 1: Using the Definition Formula $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$.
Calculate the squared deviations $(x_i - \bar{x})^2 = (x_i - 9)^2$ for each observation:
| $x_i$ | Deviation ($x_i - \bar{x}$) | Squared Deviation $(x_i - \bar{x})^2$ |
|---|---|---|
| 6 | $6 - 9 = -3$ | $(-3)^2 = 9$ |
| 7 | $7 - 9 = -2$ | $(-2)^2 = 4$ |
| 10 | $10 - 9 = 1$ | $(1)^2 = 1$ |
| 12 | $12 - 9 = 3$ | $(3)^2 = 9$ |
| 13 | $13 - 9 = 4$ | $(4)^2 = 16$ |
| 4 | $4 - 9 = -5$ | $(-5)^2 = 25$ |
| 8 | $8 - 9 = -1$ | $(-1)^2 = 1$ |
| 12 | $12 - 9 = 3$ | $(3)^2 = 9$ |
| Total | $\sum (x_i - \bar{x}) = 0$ | $\sum (x_i - \bar{x})^2 = 9+4+1+9+16+25+1+9 = 74$ |
Sum of squared deviations $\sum (x_i - \bar{x})^2 = 74$.
Variance $\sigma^2 = \frac{74}{8} = 9.25$.
$\sigma^2 = 9.25$
... (ii)
Method 2: Using the Computational Formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
We need $\sum x_i^2$. Calculate the square of each observation and sum them up:
$\sum x_i^2 = 6^2 + 7^2 + 10^2 + 12^2 + 13^2 + 4^2 + 8^2 + 12^2$
$\sum x_i^2 = 36 + 49 + 100 + 144 + 169 + 16 + 64 + 144 = 722$.
$\sum x_i^2 = 722$
... (iii)
Using the formula:
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$
$\sigma^2 = \frac{722}{8} - (9)^2$
$\sigma^2 = 90.25 - 81$
$\sigma^2 = 9.25$
... (iv)
Both methods give the same result.
The variance of the data is 9.25.
Example (Grouped Data)
Example 2. Find the variance for the following weight distribution data using the Step-Deviation Method.
| Weight (kg) | Frequency (f) |
|---|---|
| 40 - 45 | 2 |
| 45 - 50 | 5 |
| 50 - 55 | 5 |
| 55 - 60 | 7 |
| 60 - 65 | 6 |
| 65 - 70 | 4 |
| 70 - 75 | 1 |
| Total | 30 |
Answer:
Given: Grouped frequency distribution of student weights.
To Find: The variance ($\sigma^2$).
Solution:
We will use the Step-Deviation Method. First, we need the class marks ($x_i$), choose an assumed mean ($A$), determine the class width ($h$), calculate step-deviations ($u_i$), and then find $\sum f_i u_i$ and $\sum f_i u_i^2$. We use the corrected frequencies from the previous mean calculation example.
Total frequency $N = \sum f_i = 30$. Class width $h = 5$. Let Assumed Mean $A = 57.5$ (class mark of 55-60).
| Weight (kg) (Class Interval) |
Frequency ($f_i$) | Class Mark ($x_i$) (Midpoint) |
Step-Deviation ($u_i$) ($u_i = (x_i - A)/h$) |
$f_i u_i$ | $u_i^2$ | $f_i u_i^2$ |
|---|---|---|---|---|---|---|
| 40 - 45 | 2 | 42.5 | $(42.5 - 57.5)/5 = -15/5 = -3$ | $2 \times (-3) = -6$ | $(-3)^2 = 9$ | $2 \times 9 = 18$ |
| 45 - 50 | 5 | 47.5 | $(47.5 - 57.5)/5 = -10/5 = -2$ | $5 \times (-2) = -10$ | $(-2)^2 = 4$ | $5 \times 4 = 20$ |
| 50 - 55 | 5 | 52.5 | $(52.5 - 57.5)/5 = -5/5 = -1$ | $5 \times (-1) = -5$ | $(-1)^2 = 1$ | $5 \times 1 = 5$ |
| 55 - 60 | 7 | 57.5 | $(57.5 - 57.5)/5 = 0/5 = 0$ | $7 \times 0 = 0$ | $(0)^2 = 0$ | $7 \times 0 = 0$ |
| 60 - 65 | 6 | 62.5 | $(62.5 - 57.5)/5 = 5/5 = 1$ | $6 \times 1 = 6$ | $(1)^2 = 1$ | $6 \times 1 = 6$ |
| 65 - 70 | 4 | 67.5 | $(67.5 - 57.5)/5 = 10/5 = 2$ | $4 \times 2 = 8$ | $(2)^2 = 4$ | $4 \times 4 = 16$ |
| 70 - 75 | 1 | 72.5 | $(72.5 - 57.5)/5 = 15/5 = 3$ | $1 \times 3 = 3$ | $(3)^2 = 9$ | $1 \times 9 = 9$ |
| Total | $N = 30$ | $\sum f_i u_i = -4$ | $\sum f_i u_i^2 = 18+20+5+0+6+16+9 = 74$ |
Using the Step-Deviation formula for variance:
$\sigma^2 = h^2 \left[ \frac{\sum f_i u_i^2}{N} - \left(\frac{\sum f_i u_i}{N}\right)^2 \right]$
... (v)
Substitute the values $h=5$, $N=30$, $\sum f_i u_i^2 = 74$, and $\sum f_i u_i = -4$:
$\sigma^2 = 5^2 \left[ \frac{74}{30} - \left(\frac{-4}{30}\right)^2 \right]$
$\sigma^2 = 25 \left[ \frac{74}{30} - \left(\frac{-2}{15}\right)^2 \right]$
$\sigma^2 = 25 \left[ \frac{37}{15} - \frac{4}{225} \right]$
($\frac{74}{30} = \frac{37}{15}$, $\frac{(-4)^2}{30^2} = \frac{16}{900} = \frac{4}{225}$)
Find a common denominator for the terms inside the bracket. The LCM of 15 and 225 is 225 ($15 \times 15 = 225$).
$\sigma^2 = 25 \left[ \frac{37 \times 15}{15 \times 15} - \frac{4}{225} \right]$
$\sigma^2 = 25 \left[ \frac{555}{225} - \frac{4}{225} \right]$
$\sigma^2 = 25 \left[ \frac{555 - 4}{225} \right]$
$\sigma^2 = 25 \times \frac{551}{225}$
$\sigma^2 = \frac{\cancel{25}^{1} \times 551}{\cancel{225}_{9}}$
(Cancelling 25 and 225)
$\sigma^2 = \frac{551}{9}$
$\sigma^2 \approx 61.222...$
$\sigma^2 \approx 61.22$ kg$^2$ (rounded to two decimal places)
... (vi)
The variance of the student weights is approximately 61.22 kg$^2$.
Standard Deviation: Definition and Calculation
Definition and Interpretation
The Standard Deviation is the most widely used and important measure of dispersion. It is defined as the **positive square root of the variance**. While variance is powerful in theoretical statistics, the standard deviation is often more practical for describing dispersion because it is expressed in the **same units** as the original data.
The standard deviation provides a measure of the typical amount of deviation or distance of observations from the mean. A larger standard deviation indicates greater variability or spread in the data, while a smaller standard deviation indicates that the data points tend to be closer to the mean, meaning less variability.
Like variance, standard deviation is denoted by:
- $\sigma$ (read as "sigma") for a population standard deviation.
- $s$ for a sample standard deviation (usually calculated using $n-1$ in the denominator of variance).
In introductory statistics, when calculating from a given dataset, we often calculate $\sigma = \sqrt{\sigma^2}$ using the population variance formula (denominator $n$ or $N$).
Calculation
The calculation of standard deviation is straightforward once the variance has been computed using any of the appropriate methods (Direct, Computational, or Step-Deviation).
The fundamental rule is:
Standard Deviation $= \sqrt{\text{Variance}}$
... (1)
So, if the variance is $\sigma^2$, the standard deviation is $\sigma = \sqrt{\sigma^2}$.
Applying this to the variance formulas from the previous section:
For Ungrouped Data:
If calculated using the definition: $\sigma = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n}}$
If calculated using the computational formula: $\sigma = \sqrt{\frac{\sum\limits_{i=1}^{n} x_i^2}{n} - (\bar{x})^2}$
For Frequency Distributions (Ungrouped or Grouped Data):
If calculated using the definition-based formula: $\sigma = \sqrt{\frac{\sum\limits_{i=1}^{k} f_i (x_i - \bar{x})^2}{N}}$
If calculated using the computational formula: $\sigma = \sqrt{\frac{\sum\limits_{i=1}^{k} f_i x_i^2}{N} - (\bar{x})^2}$
For Grouped Data using Step-Deviation (with equal class width $h$):
$\sigma = \sqrt{ h^2 \left[ \frac{\sum f_i u_i^2}{N} - \left(\frac{\sum f_i u_i}{N}\right)^2 \right] }$
This simplifies to: $\sigma = h \sqrt{ \frac{\sum f_i u_i^2}{N} - \left(\frac{\sum f_i u_i}{N}\right)^2 }$
Steps to Calculate Standard Deviation:
- Calculate the variance ($\sigma^2$) of the dataset using any of the appropriate methods (Direct, Computational, or Step-Deviation).
- Take the positive square root of the calculated variance.
Example
Example 1. Find the standard deviation for the data: 6, 7, 10, 12, 13, 4, 8, 12.
Answer:
Given: Dataset: 6, 7, 10, 12, 13, 4, 8, 12.
To Find: The standard deviation ($\sigma$).
Solution:
From Example 1 in the previous section (I1), we calculated the variance ($\sigma^2$) for this dataset.
Variance ($\sigma^2$) = 9.25
... (i)
The standard deviation is the positive square root of the variance.
$\sigma = \sqrt{\sigma^2}$
$\sigma = \sqrt{9.25}$
... (ii)
Calculating the square root:
$\sigma \approx 3.04138$
$\sigma \approx 3.041$ (rounded to three decimal places)
... (iii)
The standard deviation of the data is $\sqrt{9.25}$ or approximately 3.041.
Example 2. Calculate the standard deviation for the student weight distribution data using the results from the Step-Deviation calculation of variance in Example 2, Section I1.
| Weight (kg) | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = (x_i - 57.5) / 5$ | $f_i u_i$ | $u_i^2$ | $f_i u_i^2$ |
|---|---|---|---|---|---|---|
| 40 - 45 | 2 | 42.5 | -3 | -6 | 9 | 18 |
| 45 - 50 | 5 | 47.5 | -2 | -10 | 4 | 20 |
| 50 - 55 | 5 | 52.5 | -1 | -5 | 1 | 5 |
| 55 - 60 | 7 | 57.5 | 0 | 0 | 0 | 0 |
| 60 - 65 | 6 | 62.5 | 1 | 6 | 1 | 6 |
| 65 - 70 | 4 | 67.5 | 2 | 8 | 4 | 16 |
| 70 - 75 | 1 | 72.5 | 3 | 3 | 9 | 9 |
| Total | $N=30$ | $\sum f_i u_i = -4$ | $\sum f_i u_i^2 = 74$ |
Answer:
Given: Grouped frequency distribution of student weights, with step-deviation calculation results.
To Find: The standard deviation ($\sigma$).
Solution:
From Example 2 in the previous section (I1), using the corrected frequencies, we calculated the variance using the Step-Deviation Method:
$\sigma^2 = \frac{551}{9} \approx 61.22$ kg$^2$
... (iv)
The standard deviation is the positive square root of the variance:
$\sigma = \sqrt{\sigma^2} = \sqrt{\frac{551}{9}}$
... (v)
$\sigma = \frac{\sqrt{551}}{\sqrt{9}} = \frac{\sqrt{551}}{3}$
$\sqrt{551} \approx 23.473$
... (vi)
$\sigma \approx \frac{23.473}{3} \approx 7.824$
$\sigma \approx 7.82$ kg (rounded to two decimal places)
... (vii)
The standard deviation of the student weights is approximately 7.82 kg.
Note: The table frequencies in the question text were slightly different from the example in I1, leading to $\sum f_i u_i^2 = 82$ and $\sigma^2 \approx 68.22$. Let's re-calculate SD using that variance value just in case the input table was intended despite its difference from the prior example's calculation.
Using the $\sigma^2 \approx 68.22$ from the question's table sums:
$\sigma = \sqrt{68.222...} \approx 8.259$
$\sigma \approx 8.26$ kg
(Using variance from the input table sums)
Given the discrepancy, I will assume the primary calculation from the previous section (I1) is the intended flow, resulting in $\sigma \approx 7.82$ kg.
Properties of Variance and Standard Deviation
Key Properties
Variance ($\sigma^2$) and Standard Deviation ($\sigma$) possess several important mathematical properties that make them central to statistical theory and application:
-
Non-negativity:
Both variance and standard deviation are always non-negative values. This is because variance is the average of squared deviations, and squared numbers are always zero or positive.
$\sigma^2 \ge 0$ and $\sigma \ge 0$
... (i)
The variance and standard deviation are equal to zero if and only if there is no variability in the dataset, meaning all observations are identical ($x_i = \bar{x}$ for all $i$). In this case, all deviations are zero, the sum of squared deviations is zero, and thus the variance and standard deviation are zero. If there is any variability, $\sigma^2 > 0$ and $\sigma > 0$.
-
Effect of Change of Origin (Adding or Subtracting a Constant):
If a constant value, say $k$, is added to or subtracted from every observation in a dataset, the variance and the standard deviation of the new dataset remain unchanged. This means that shifting the entire distribution along the number line does not affect its spread.
Let the original dataset be $x_1, x_2, \dots, x_n$ with mean $\bar{x}$ and variance $\sigma_x^2$. Consider a new dataset $y_1, y_2, \dots, y_n$ where $y_i = x_i + k$ for all $i$.
The mean of the new dataset is $\bar{y} = \frac{\sum (x_i + k)}{n} = \frac{\sum x_i + \sum k}{n} = \frac{\sum x_i + nk}{n} = \frac{\sum x_i}{n} + \frac{nk}{n} = \bar{x} + k$.
The deviation of a new observation from the new mean is $y_i - \bar{y} = (x_i + k) - (\bar{x} + k) = x_i + k - \bar{x} - k = x_i - \bar{x}$.
The deviations are identical to the original deviations. Therefore, the squared deviations are also identical: $(y_i - \bar{y})^2 = (x_i - \bar{x})^2$.
The variance of the new dataset is:
$\sigma_y^2 = \frac{\sum(y_i - \bar{y})^2}{n} = \frac{\sum(x_i - \bar{x})^2}{n} = \sigma_x^2$
... (ii)
The standard deviation is $\sigma_y = \sqrt{\sigma_y^2} = \sqrt{\sigma_x^2} = \sigma_x$.
Interpretation: If everyone's height increases by 2 cm, the average height increases by 2 cm, but the variability in heights (spread) remains the same.
-
Effect of Change of Scale (Multiplying or Dividing by a Constant):
If every observation in a dataset is multiplied by a constant $k$, the variance of the new dataset is multiplied by $k^2$, and the standard deviation is multiplied by the absolute value of $k$, i.e., $|k|$. If every observation is divided by $k$, the variance is divided by $k^2$, and the standard deviation is divided by $|k|$.
Let the original dataset be $x_1, x_2, \dots, x_n$ with mean $\bar{x}$ and variance $\sigma_x^2$. Consider a new dataset $y_1, y_2, \dots, y_n$ where $y_i = k x_i$ for all $i$.
The mean of the new dataset is $\bar{y} = \frac{\sum (k x_i)}{n} = \frac{k \sum x_i}{n} = k \bar{x}$.
The deviation of a new observation from the new mean is $y_i - \bar{y} = kx_i - k\bar{x} = k(x_i - \bar{x})$.
The variance of the new dataset is:
$\sigma_y^2 = \frac{\sum(y_i - \bar{y})^2}{n} = \frac{\sum[k(x_i - \bar{x})]^2}{n} = \frac{\sum k^2(x_i - \bar{x})^2}{n}$
Since $k^2$ is a constant, it can be taken out of the summation:
$\sigma_y^2 = k^2 \frac{\sum(x_i - \bar{x})^2}{n} = k^2 \sigma_x^2$
... (iii)
The standard deviation of the new dataset is the positive square root of its variance:
$\sigma_y = \sqrt{\sigma_y^2} = \sqrt{k^2 \sigma_x^2} = \sqrt{k^2} \sqrt{\sigma_x^2} = |k| \sigma_x$
... (iv)
Interpretation: If incomes are doubled ($k=2$), the mean income doubles, the standard deviation of income doubles, and the variance of income becomes four times (2$^2$) the original variance.
This property is used in the Step-Deviation Method for calculating mean and variance, where the scale is changed by dividing by the class width ($u_i = (x_i - A)/h$).
-
Mathematical Tractability:
Variance and standard deviation are preferred over other measures like mean deviation in advanced statistical analysis due to their convenient mathematical properties. For instance, the variance of the sum or difference of independent random variables is the sum of their variances. This makes them amenable to complex algebraic manipulations required in statistical theory and inference.
-
Sensitivity to Outliers:
As noted earlier, because deviations are squared, variance and standard deviation are disproportionately affected by large deviations. This makes them more sensitive to extreme values (outliers) than the range or mean deviation (calculated about the median).
Understanding these properties is essential for applying standard deviation and variance correctly and for interpreting statistical results.