Statistics | Class 10 Mathematics Notes

Mean of Grouped Data

When data is grouped into class intervals, the exact values of individual observations are not known. Instead, the data is presented as frequency distribution tables with class intervals and corresponding frequencies. To calculate the mean for grouped data, we use the class mark (mid-point of the class interval) as a representative value for all observations in that class. The class mark is calculated as (Upper class limit + Lower class limit) / 2. The mean is then calculated using the formula: x̄ = (Σfi×xi) / (Σfi), where xi is the class mark. This method is called the Direct Method. However, the mean calculated using class marks is an approximation because the exact values are unknown. The mean helps summarize large data sets efficiently and is widely used in statistics.

📊 Diagram: Table 13.3 showing class intervals and frequencies; Figure 4 on page 3; Table 13.4 showing calculation of mean using class marks

🧪 Activity: Students collect data such as heights of classmates, group them into class intervals, and calculate the mean using the direct method.

🔗 Connection: Leads to methods that simplify mean calculation for grouped data, such as the Assumed Mean and Step-Deviation methods.

Table on page 3 (1×7)

Table on page 4 (3×6)

| | Class interval | Number of students ( f) i | Class mark (x) i | f x i i | |

| | 10 - 25 25 - 40 | 2 3 | 17.5 32.5 | 35.0 97.5 | | | 40 - 55 7 47.5 332.5 55 - 70 6 62.5 375.0 70 - 85 6 77.5 465.0 85 - 100 6 92.5 555.0 Total Σ f = 30 Σ f x = 1860.0 i i i The sum of the values in the last column gives us Σ f x. So, the mean x of the i i given data is given by Σf x 1860.0 x = i i = = 62 Σf 30 i This new method of finding the mean is known as the Direct Method. We observe that Tables 13.1 and 13.3 are using the same data and employing the same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the two values is because of the mid-point assumption in Table 13.3, 59.3 being the exact mean, while 62 an approximate mean. Sometimes when the numerical values of x and f are large, finding the product i i of x and f becomes tedious and time consuming. So, for such situations, let us think of i i a method of reducing these calculations. We can do nothing with the f’s, but we can change each x to a smaller number i i so that our calculations become easy. How do we do this? What about subtracting a fixed number from each of these x’s? Let us try this method. i The first step is to choose one among the x’s as the assumed mean, and denote i it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that x i which lies in the centre of x , x , . . ., x . So, we can choose a = 47.5 or a = 62.5. Let 1 2 n us choose a = 47.5. | 40 - 55 55 - 70 70 - 85 85 - 100 | 7 6 6 6 | 47.5 62.5 77.5 92.5 | 332.5 375.0 465.0 555.0 | | | | Total | Σ f = 30 i | | Σ f x = 1860.0 i i | |