Statistics | Class 11 Mathematics Notes

Variance and Standard Deviation for Grouped Data

This section extends the concepts of variance and standard deviation to grouped data, where data is classified into class intervals with corresponding frequencies. The class mid-points are used as representative values for the classes. The chapter explains the step-by-step procedure to calculate the mean, variance, and standard deviation for grouped data using tabular methods. It introduces the formula for variance as σ² = (Σ f × (x – x̄)²) / N, where x̄ is the mean of grouped data, f is the frequency, and N is the total frequency. The standard deviation is the square root of variance. The chapter also demonstrates the use of the step-deviation method to simplify calculations by choosing an assumed mean and class width. Detailed examples illustrate the process, including the calculation of deviations, squared deviations, and their products with frequencies. This section emphasizes the importance of these measures in analyzing grouped data, which is common in real-world statistical problems.

📊 Diagram: Table 6 on page 10; Table 7 on page 10; Figure 15 on page 12; Table 8 on page 12

🧪 Activity: Students practice calculating variance and standard deviation for grouped data using the step-deviation method.

🔗 Connection: This section concludes the detailed study of measures of dispersion and prepares students for summary and application.

Table on page 10 (1×8)

Table on page 10 (2×6)

| Marks obtained | Number of students f i | Mid-points x i | fx i i | x −x i | f x −x i i |

| 10-20 20-30 30-40 40-50 50-60 60-70 70-80 | 2 3 8 14 8 3 2 | 15 25 35 45 55 65 75 | 30 75 280 630 440 195 150 | 30 20 10 0 10 20 30 | 60 60 80 0 80 60 60 |

Table on page 12 (3×9)

| | Marks obtained | Number of students | Mid-points | x −45 d = i i 10 | f d i i | x −x i | f x −x i i | |

| f x i i 10-20 2 15 – 3 – 6 30 60 20-30 3 25 – 2 – 6 20 60 30-40 8 35 – 1 – 8 10 80 40-50 14 45 0 0 0 0 50-60 8 55 1 8 10 80 60-70 3 65 2 6 20 60 70-80 2 75 3 6 30 60 40 0 400 7 ∑ f d i i Therefore x=a+ i=1 ×h N 0 = 45+ ×10=45 40 1 7 400 and M . D. (x ) = ∑ f x −x = = 10 i i N 40 i=1 A Note The step deviation method is applied to compute x . Rest of the procedure is same. (ii) Mean deviation about median The process of finding the mean deviation about median for a continuous frequency distribution is similar as we did for mean deviation about the mean. The only difference lies in the replacement of the mean by median while taking deviations. Let us recall the process of finding median for a continuous frequency distribution. | 10-20 20-30 30-40 40-50 50-60 60-70 70-80 | f i 2 3 8 14 8 3 2 | x i 15 25 35 45 55 65 75 | – 3 – 2 – 1 0 1 2 3 | – 6 – 6 – 8 0 8 6 6 | 30 20 10 0 10 20 30 | 60 60 80 0 80 60 60 | |