1. Limitations of Mean Deviation

While mean deviation provides a good measure of spread, it uses absolute values (e.g., ∣xi−xˉ∣|x_i - \bar{x}|). This makes it algebraically cumbersome for more advanced statistical analysis because functions involving absolute values are not easily differentiable. To overcome this, statisticians prefer a method that removes negative signs by squaring the deviations. This approach leads to the concepts of Variance and Standard Deviation. 💡

2. Variance

Variance, denoted by σ2\sigma^2 (sigma squared), is defined as the average of the squared deviations from the mean. It provides a measure of how far a set of numbers is spread out from their average value.

  • A small variance indicates that the data points are tightly clustered around the mean, implying high consistency.
  • A large variance indicates that the data points are spread out over a wider range, implying low consistency.

A. Variance for Ungrouped Data

For a set of n observations x1,x2,…,xnx_1, x_2, \dots, x_n with mean xˉ\bar{x}: Variance (σ2)=∑i=1n(xi−xˉ)2n\text{Variance } (\sigma^2) = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n}

Example: Find the variance of the data set {2, 4, 6, 8}.

  1. Find the Mean: xˉ=2+4+6+84=204=5\bar{x} = \frac{2+4+6+8}{4} = \frac{20}{4} = 5.
  2. Find the Squared Deviations:
    • (2−5)2=(−3)2=9(2-5)^2 = (-3)^2 = 9
    • (4−5)2=(−1)2=1(4-5)^2 = (-1)^2 = 1
    • (6−5)2=(1)2=1(6-5)^2 = (1)^2 = 1
    • (8−5)2=(3)2=9(8-5)^2 = (3)^2 = 9
  3. Calculate Variance: σ2=9+1+1+94=204=5\sigma^2 = \frac{9+1+1+9}{4} = \frac{20}{4} = \mathbf{5}.

B. Variance for Grouped Data

For a frequency distribution with N total observations: Variance (σ2)=∑i=1nfi(xi−xˉ)2N,where N=∑fi\text{Variance } (\sigma^2) = \frac{\sum_{i=1}^{n} f_i (x_i - \bar{x})^2}{N}, \quad \text{where } N = \sum f_i

3. Standard Deviation

The variance is measured in squared units (e.g., if your data is in meters, the variance is in meters squared). This makes it difficult to interpret in the context of the original data. To solve this, we take the square root of the variance.

Standard Deviation, denoted by σ\sigma, is the positive square root of the variance. It is the most common and important measure of dispersion because it is expressed in the same units as the data.

Standard Deviation (σ)=Variance\text{Standard Deviation } (\sigma) = \sqrt{\text{Variance}}

Example: For the data set {2, 4, 6, 8} from the previous example, the variance was σ2=5\sigma^2 = 5. Therefore, the standard deviation is σ=5≈2.236\sigma = \sqrt{5} \approx \mathbf{2.236}. This means that, on average, the data points are about 2.236 units away from the mean of 5.

4. Shortcut Formula for Variance

Calculating each deviation (xi−xˉ)(x_i - \bar{x}) can be tedious if the mean xˉ\bar{x} is not a whole number. A more convenient computational formula can be used instead:

σ2=∑fixi2N−(∑fixiN)2=∑fixi2N−(xˉ)2\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left( \frac{\sum f_i x_i}{N} \right)^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2

In simpler terms: Variance = (Mean of the squares) - (Square of the mean).

Example (using the same ungrouped data {2, 4, 6, 8}):

  1. Find the Square of the Mean: We know xˉ=5\bar{x} = 5, so (xˉ)2=25(\bar{x})^2 = 25.
  2. Find the Mean of the Squares:
    • Squares of data (xi2x_i^2): 22=42^2=4, 42=164^2=16, 62=366^2=36, 82=648^2=64.
    • Sum of squares: ∑xi2=4+16+36+64=120\sum x_i^2 = 4+16+36+64=120.
    • Mean of squares: ∑xi2n=1204=30\frac{\sum x_i^2}{n} = \frac{120}{4} = 30.
  3. Apply the Formula: σ2=30−25=5\sigma^2 = 30 - 25 = \mathbf{5}. This matches our previous result.

5. Step-Deviation Method for Variance

For grouped data where the mid-points (xix_i) are equally spaced, this method simplifies calculations significantly.

Let 'a' be an assumed mean (usually the middle xix_i), and 'h' be the class size. We define a new variable yi=xi−ahy_i = \frac{x_i - a}{h}. The formula for variance becomes:

σ2=h2[∑fiyi2N−(∑fiyiN)2]\sigma^2 = h^2 \left[ \frac{\sum f_i y_i^2}{N} - \left( \frac{\sum f_i y_i}{N} \right)^2 \right]

Example: Find the variance for the data:

xix_i 3 8 13 18 23
fif_i 7 10 15 10 6
  1. Setup: Here, the class size is h=5h=5. Let's pick the assumed mean a=13a=13. The total frequency is N=48N=48.
  2. Calculation Table:
xix_i fif_i yi=xi−135y_i = \frac{x_i-13}{5} fiyif_i y_i yi2y_i^2 fiyi2f_i y_i^2
3 7 -2 -14 4 28
8 10 -1 -10 1 10
13 15 0 0 0 0
18 10 1 10 1 10
23 6 2 12 4 24
N=48 ∑fiyi=−2\sum f_i y_i = -2 ∑fiyi2=72\sum f_i y_i^2 = 72

  1. Apply the Formula: σ2=52[7248−(−248)2]\sigma^2 = 5^2 \left[ \frac{72}{48} - \left( \frac{-2}{48} \right)^2 \right] σ2=25[1.5−(−124)2]\sigma^2 = 25 \left[ 1.5 - \left( \frac{-1}{24} \right)^2 \right] σ2=25[1.5−1576]≈25(1.498)≈37.45\sigma^2 = 25 \left[ 1.5 - \frac{1}{576} \right] \approx 25(1.498) \approx \mathbf{37.45}.

    The standard deviation would be σ=37.45≈6.12\sigma = \sqrt{37.45} \approx 6.12.