Statistics | Class 10 Mathematics Notes

Summary

This chapter covered the fundamental concepts of statistics focusing on measures of central tendency: mean, median, and mode. We learned how to calculate the mean for both ungrouped and grouped data using various methods such as the direct method, assumed mean method, and step-deviation method. The median was introduced as the middle value dividing data into two equal halves, and its calculation for grouped data was explained using cumulative frequencies. The mode was defined as the most frequently occurring value, with a formula to estimate it from grouped data. These measures help summarize large data sets by identifying central or typical values, aiding in data analysis and decision-making. The chapter also included several examples and activities to practice these concepts with real-world data.

📊 Diagram: Figures and tables throughout the chapter illustrating data sets, frequency distributions, and calculations.

🧪 Activity: Review and practice exercises provided at the end of the chapter to reinforce concepts.

🔗 Connection: Prepares students for advanced statistical concepts and applications in higher classes.

Table on page 2 (4×5)

|  f 1 to n. i Let us apply this formula to find the mean in the following example. Example 1 : The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students. Marks obtained 10 20 36 40 50 56 60 70 72 80 88 92 95 (x) i Number of 1 1 3 4 3 2 4 4 1 1 2 3 1 students ( f) i Solution: Recall that to find the mean marks, we require the product of each x with i the corresponding frequency f. So, let us put them in a column as shown in Table 13.1. i Table 13.1 Marks obtained (x ) Number of students ( f ) fx i i i i 10 1 10 20 1 20 . 36 3 108 40 4 160 50 3 150 56 2 112 60 4 240 70 4 280 72 1 72 80 1 80 88 2 176 | | | | |

| | Marks obtained (x ) i | Number of students ( f ) i | fx i i | | | | 10 20 . 36 40 50 56 60 70 72 80 88 | 1 1 3 4 3 2 4 4 1 1 2 | 10 20 108 160 150 112 240 280 72 80 176 | | | | 92 95 | 3 1 | 276 95 | | | | Total | f = 30 i | fx = 1779 i i | |

Table on page 2 (1×14)

| Marks obtained (x) i | 10 | 20 | 36 | 40 | 50 | 56 | 60 | 70 | 72 | 80 | 88 | 92 | 95 |

| Number of students ( f) i | 1 | 1 | 3 | 4 | 3 | 2 | 4 | 4 | 1 | 1 | 2 | 3 | 1 |

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 | |

Table on page 5 (3×7)

| | Class interval | Number of | Class mark | d = x – 47.5 i i | fd i i | |

| students ( f ) (x) i i 10 - 25 2 17.5 –30 –60 25 - 40 3 32.5 –15 –45 40 - 55 7 47.5 0 0 55 - 70 6 62.5 15 90 70 - 85 6 77.5 30 180 85 - 100 6 92.5 45 270 Total Σf = 30 Σfd = 435 i i i Σf d So, from Table 13.4, the mean of the deviations, d = i i . Σf i Now, let us find the relation between d and x . Since in obtaining d, we subtracted ‘a’ from each x, so, in order to get the mean i i x , we need to add ‘a’ to d . This can be explained mathematically as: Σf d Mean of deviations, d = i i Σf i Σf (x − a) So, d = i i Σf i Σf x Σf a = i i − i Σf Σf i i Σf = x − a i Σf i = x − a | | students ( f ) i | (x) i | | | | | | 10 - 25 25 - 40 40 - 55 55 - 70 70 - 85 85 - 100 | 2 3 7 6 6 6 | 17.5 32.5 47.5 62.5 77.5 92.5 | –30 –15 0 15 30 45 | –60 –45 0 90 180 270 | | | | Total | Σf = 30 i | | | Σfd = 435 i i | |

Table on page 6 (2×6)

| Class interval | f i | x i | d = x – a i i | x – a u = i i h | fu i i |

| 10 - 25 25 - 40 40 - 55 55 - 70 70 - 85 85 - 100 | 2 3 7 6 6 6 | 17.5 32.5 47.5 62.5 77.5 92.5 | –30 –15 0 15 30 45 | –2 –1 0 1 2 3 | –4 –3 0 6 12 18 | | Total | Σf = 30 i | | | | Σfu = 29 i i |

Table on page 8 (2×10)

| female teachers Number of 6 11 7 4 4 2 1 States/U.T. Source : Seventh All India School Education Survey conducted by NCERT Solution : Let us find the class marks, x, of each class, and put them in a column i (see Table 13.6): Table 13.6 Percentage of female Number of x i teachers States /U.T. ( f ) i 15 - 25 6 20 25 - 35 11 30 35 - 45 7 40 45 - 55 4 50 55 - 65 4 60 65 - 75 2 70 75 - 85 1 80 x 50 u i | female teachers | | | | | | | | | | | Number of States/U.T. | 6 | 11 | 7 | 4 | 4 | 2 | 1 | |

Table on page 9 (3×10)

| | Percentage of female teachers | Number of states/U.T. ( f) i | x i | d = x – 50 i i | x −50 u = i i 10 | fx i i | fd i i | fu i i | |

| 15 - 25 6 20 –30 –3 120 –180 –18 25 - 35 11 30 –20 –2 330 –220 –22 35 - 45 7 40 –10 –1 280 –70 –7 45 - 55 4 50 0 0 200 0 0 55 - 65 4 60 10 1 240 40 4 65 - 75 2 70 20 2 140 40 4 75 - 85 1 80 30 3 80 30 3 Total 35 1390 –360 –36 From the table above, we obtain Σf = 35, Σfx = 1390, i i i Σf d = – 360, Σfu = –36. i i i i Σf x 1390 Using the direct method, x = i i = =39.71 Σf 35 i Using the assumed mean method, Σf d ( −360) x = a + i i = 50 + =39.71 Σf 35 i Using the step-deviation method, Σf u  – 36 x = a +  i i × h = 50 +  ×10 = 39.71  Σf   35  i Therefore, the mean percentage of female teachers in the primary schools of rural areas is 39.71. Remark : The result obtained by all the three methods is the same. So the choice of method to be used depends on the numerical values of x and f. If x and f are i i i i | 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 75 - 85 | 6 11 7 4 4 2 1 | 20 30 40 50 60 70 80 | –30 –20 –10 0 10 20 30 | –3 –2 –1 0 1 2 3 | 120 330 280 200 240 140 80 | –180 –220 –70 0 40 40 30 | –18 –22 –7 0 4 4 3 | |

Table on page 10 (2×9)

| | Number of wickets | 20 - 60 | 60 - 100 | 100 - 150 | 150 - 250 | 250 - 350 | 350 - 450 | |

| Number of 7 5 16 12 2 3 bowlers Solution : Here, the class size varies, and the x,s are large. Let us still apply the step- i deviation method with a = 200 and h = 20. Then, we obtain the data as in Table 13.8. Table 13.8 d Number of Number of x d = x – 200 u = i u f i i i i 20 i i wickets bowlers taken ( f) i 20 - 60 7 40 –160 –8 –56 60 - 100 5 80 –120 –6 –30 100 - 150 16 125 –75 –3.75 –60 150 - 250 12 200 0 0 0 250 - 350 2 300 100 5 10 350 - 450 3 400 200 10 30 Total 45 –106 −106 −106 So, u = ⋅ Therefore, x = 200 + 20  = 200 – 47.11 = 152.89. 45  45  This tells us that, on an average, the number of wickets taken by these 45 bowlers | | | | | | | | | | | Number of bowlers | 7 | 5 | 16 | 12 | 2 | 3 | |

Table on page 14 (1×6)

| Number of families | 7 | 8 | 2 | 2 | 1 |

Table on page 11 (3×10)

| of the data obtained. 2. Collect the daily maximum temperatures recorded for a period of 30 days in your city. Present this data as a grouped frequency table. 3. Measure the heights of all the students of your class (in cm) and form a grouped frequency distribution table of this data. After all the groups have collected the data and formed grouped frequency distribution tables, the groups should find the mean in each case by the method which they find appropriate. EXERCISE 13.1 1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. Number of plants 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12 12 - 14 Number of houses 1 2 1 5 6 2 3 Which method did you use for finding the mean, and why? 2. Consider the following distribution of daily wages of 50 workers of a factory. Daily wages (in `) 500 - 520 520 -540 540 - 560 560 - 580 580 -600 Number of workers 12 14 8 6 10 Find the mean daily wages of the workers of the factory by using an appropriate method. 3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f. Daily pocket 11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25 | | | | | | | | | |