Statistics | Class 11 Mathematics Notes

Variance and Standard Deviation

Variance and standard deviation are fundamental measures of dispersion that quantify the spread of data around the mean. Variance is defined as the average of the squared deviations from the mean, which emphasizes larger deviations due to squaring. Standard deviation is the positive square root of the variance and is preferred because it is expressed in the same units as the data, making interpretation more intuitive. The chapter explains the formulas for variance and standard deviation for ungrouped data and extends these concepts to discrete frequency distributions and grouped data. It introduces the step-deviation method to simplify calculations, especially for large data sets. The chapter provides detailed examples demonstrating the calculation of variance and standard deviation, including tabular methods that organize data, deviations, squared deviations, and their products with frequencies. Understanding these measures is crucial for assessing data reliability, comparing variability across data sets, and conducting further statistical analysis.

📊 Diagram: Figure 9 on page 6; Figure 10 on page 7; Table 14 on page 18; Table 15 on page 19

🧪 Activity: Students calculate variance and standard deviation for their data sets to understand data dispersion.

🔗 Connection: The next sections focus on variance and standard deviation calculations specifically for grouped data.

Table on page 18 (4×6)

| since variance involves the sum of squares of (x– x ). For this reason, the proper i measure of dispersion about the mean of a set of observations is expressed as positive square-root of the variance and is called standard deviation. Therefore, the standard deviation, usually denoted by σ , is given by 1 n σ= ∑(x −x)2 ... (1) n i i=1 Let us take the following example to illustrate the calculation of variance and hence, standard deviation of ungrouped data. Example 8 Find the variance of the following data: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 Solution From the given data we can form the following Table 13.7. The mean is calculated by step-deviation method taking 14 as assumed mean. The number of observations is n = 10 Table 13.7 x −14 Deviations from mean x d = i (x i– x ) i i 2 (x– x ) i 6 –4 –9 81 8 –3 –7 49 10 –2 –5 25 12 –1 –3 9 14 0 –1 1 16 1 1 1 18 2 3 9 20 3 5 25 | | | | | |

| | x i | x −14 d = i i 2 | Deviations from mean (x– x ) i | (x– x ) i | | | | 6 8 10 12 14 16 18 20 | –4 –3 –2 –1 0 1 2 3 | –9 –7 –5 –3 –1 1 3 5 | 81 49 25 9 1 1 9 25 | | | | 22 24 | 4 5 | 7 9 | 49 81 | |

Table on page 19 (4×8)

| and Variance (σ ) = ∑( x i x) = ×330 = 33 n 10 i=1 Thus Standard deviation (σ ) = 33=5. 74 13.5.2 Standard deviation of a discrete frequency distribution Let the given discrete frequency distribution be x : x , x , x ,. . . , x 1 2 3 n f : f , f , f ,. . . , f 1 2 3 n 1 n In this case standard deviation ( σ) = ∑ f (x − x)2 ... (2) i i N i=1 n where N=∑ f . i i=1 Let us take up following example. Example 9 Find the variance and standard deviation for the following data: x 4 8 11 17 20 24 32 i f 3 5 9 5 4 3 1 i Solution Presenting the data in tabular form (Table 13.8), we get Table 13.8 x f f x x – x (x −x)2 f i(x −x)2 i i i i i i i 4 3 12 –10 100 300 8 5 40 –6 36 180 11 9 99 –3 9 81 17 5 85 3 9 45 20 4 80 6 36 144 | | | | | | | |

| | x i | f i | f x i i | x – x i | (x −x)2 i | f (x −x)2 i i | | | | 4 8 11 17 20 | 3 5 9 5 4 | 12 40 99 85 80 | –10 –6 –3 3 6 | 100 36 9 9 36 | 300 180 81 45 144 | | | | 24 32 | 3 1 | 72 32 | 10 18 | 100 324 | 300 324 | |