13.1 Introduction In Class IX, you have studied the classification of given data into ungrouped as well as grouped frequency distributions. You have also learnt to represent the data pictorially in the form of various graphs such as bar graphs, histograms (including those of varying widths) and frequency polygons. In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median and mode. In this chapter, we shall extend the study of these three measures, i.e., mean, median and mode from ungrouped data to that of grouped data. We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution and how to draw cumulative frequency curves, called ogives. 13.2 Mean of Grouped Data The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations. From Class IX, recall that if x1 , x2 ,. . ., xn are observations with respective frequencies f 1 , f2 , . . ., fn , then this means observation x1 occurs f 1 times, x2 occurs f 2 times, and so on. Now, the sum of the values of all the observations = f 1 x1

• f2 x2
• . . . + f n xn , and the number of observations = f 1
• f2
• . . . + f n . So, the mean x of the data is given by x = 11 2 2 1 2 n n n fx fx fx ff f      Recall that we can write this in short form by using the Greek letter  (capital sigma) which means summation. That is, STATISTICS Rationalised 2023-24 172 MATHEMATICS x = 1 1 n i i i n i i f x f     which, more briefly, is written as x =   i i i f x f , if it is understood that i varies from 1 to n. 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 (xi ) Number of 1 1 3 4 3 2 4 4 1 1 2 31 students ( fi ) Solution: Recall that to find the mean marks, we require the product of each xi with the corresponding frequency f i . So, let us put them in a column as shown in Table 13.1. Table 13.1 Marks obtained (xi ) Number of students ( f i ) fi xi 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 92 3 276 95 1 95 Total f i = 30 f i xi = 1779 Rationalised 2023-24 STATISTICS 173 Now,    i i i f x x f = 1779 30 = 59.3 Therefore, the mean marks obtained is 59.3. In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean. Let us convert the ungrouped data of Example 1 into grouped data by forming class-intervals of width, say 15. Remember that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, e.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped frequency distribution table (see Table 13.2). Table 13.2 Class interval 10 - 25 25 - 40 40 - 55 55 - 70 70 - 85 85 - 100 Number of students 2 3 7 6 6 6 Now, for each class-interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. Recall that we find the mid-point of a class (or its class mark) by finding the average of its upper and lower limits. That is, Class mark = Upper class limit + Lower class limit 2 With reference to Table 13.2, for the class 10-25, the class mark is 10 25 2 + , i.e., 17.5. Similarly, we can find the class marks of the remaining class intervals. We put them in Table 13.3. These class marks serve as our xi ’s. Now, in general, for the ith class interval, we have the frequency f i corresponding to the class mark xi . We can now proceed to compute the mean in the same manner as in Example 1. Rationalised 2023-24 174 MATHEMATICS Table 13.3 Class interval Number of students ( f i ) Class mark (xi ) f i xi 10 - 25 2 17.5 35.0 25 - 40 3 32.5 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 i = 30 Σ f i x i = 1860.0 The sum of the values in the last column gives us Σ f i xi . So, the mean x of the given data is given by x = 1860.0 62 30 i i i f x f Σ = = Σ 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 i and f i are large, finding the product of xi and fi becomes tedious and time consuming. So, for such situations, let us think of a method of reducing these calculations. We can do nothing with the f i ’s, but we can change each x i to a smaller number so that our calculations become easy. How do we do this? What about subtracting a fixed number from each of these xi ’s? Let us try this method. The first step is to choose one among the xi ’s as the assumed mean, and denote it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that xi which lies in the centre of x1 , x2 , . . ., xn . So, we can choose a = 47.5 or a = 62.5. Let us choose a = 47.5. The next step is to find the difference di between a and each of the x i ’s, that is, the deviation of ‘a’ from each of the xi ’s. i.e., di = xi – a = xi – 47.5 Rationalised 2023-24 STATISTICS 175 The third step is to find the product of di with the corresponding f i , and take the sum of all the f i di ’s. The calculations are shown in Table 13.4. Table 13.4 Class interval Number of Class mark di = xi – 47.5 f i di students ( f i ) (xi ) 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 i = 30 Σf i di = 435 So, from Table 13.4, the mean of the deviations, d = i i i f d f Σ Σ . Now, let us find the relation between d and x . Since in obtaining di , we subtracted ‘a’ from each xi , so, in order to get the mean x , we need to add ‘a’ to d . This can be explained mathematically as: Mean of deviations, d = i i i f d f Σ Σ So, d = ( ) i i i f x a f Σ − Σ = i i i i i f x f a f f Σ Σ − Σ Σ = i i f x a f Σ − Σ = x a − So, x = a + d i.e., x = i i i f d a f Σ + Σ Rationalised 2023-24 176 MATHEMATICS Substituting the values of a, Σf i di and Σf i from Table 13.4, we get x = 435 47.5 47.5 14.5 62 30
• = + = . Therefore, the mean of the marks obtained by the students is 62. The method discussed above is called the Assumed Mean Method. Activity 1 : From the Table 13.3 find the mean by taking each of x i (i.e., 17.5, 32.5, and so on) as ‘a’. What do you observe? You will find that the mean determined in each case is the same, i.e., 62. (Why ?) So, we can say that the value of the mean obtained does not depend on the choice of ‘a’. Observe that in Table 13.4, the values in Column 4 are all multiples of 15. So, if we divide the values in the entire Column 4 by 15, we would get smaller numbers to multiply with f i . (Here, 15 is the class size of each class interval.) So, let ui = i x a h − , where a is the assumed mean and h is the class size. Now, we calculate ui in this way and continue as before (i.e., find f i ui and then Σ f i ui ). Taking h = 15, let us form Table 13.5. Table 13.5 Class interval f i xi di = xi – a ui = x – a i h f i ui 10 - 25 2 17.5 –30 –2 –4 25 - 40 3 32.5 –15 –1 –3 40 - 55 7 47.5 0 0 0 55 - 70 6 62.5 15 1 6 70 - 85 6 77.5 30 2 12 85 - 100 6 92.5 45 3 18 Total Σf i = 30 Σf i ui = 29 Let u = i i i f u f Σ Σ Here, again let us find the relation between u and x . Rationalised 2023-24 STATISTICS 177 We have, ui = i x a h  Therefore, u = ( ) 1 i i ii i i i x a f h fx a f fhf              = 1 ii i i i fx f a hf f            =   1 x a h  So, hu = x  a i.e., x = a + hu So, x = i i i f u a h f          Now, substituting the values of a, h, f i ui and f i from Table 14.5, we get x = 29 47.5 15 30         = 47.5 + 14.5 = 62 So, the mean marks obtained by a student is 62. The method discussed above is called the Step-deviation method. We note that :  the step-deviation method will be convenient to apply if all the di ’s have a common factor.  The mean obtained by all the three methods is the same.  The assumed mean method and step-deviation method are just simplified forms of the direct method.  The formula x = a + hu still holds if a and h are not as given above, but are any non-zero numbers such that ui = i x a h  . Let us apply these methods in another example. Rationalised 2023-24 178 MATHEMATICS Example 2 : The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section. Percentage of 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 75 - 85 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, xi , of each class, and put them in a column (see Table 13.6): Table 13.6 Percentage of female Number of xi teachers States /U.T. ( fi ) 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 Here we take a = 50, h = 10, then di = xi – 50 and 50 10 i i x u   . We now find di and ui and put them in Table 13.7. Rationalised 2023-24 STATISTICS 179 Table 13.7 Percentage of Number of xi di = xi – 50 −50 = 10 i i x u f i xi f i di f i ui female states/U.T. teachers ( f 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 i = 35, Σf i xi = 1390, Σf i di = – 360, Σf i ui = –36. Using the direct method, 1390 39.71 35 Σ = = = Σ i i i f x x f Using the assumed mean method, x = i i i f d a f Σ + Σ = ( 360) 50 39.71 35 −
• = Using the step-deviation method, x = – 36 50 10 39.71 35 i i i f u a h f   Σ  
• × = +   × =     Σ   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 xi and f i . If xi and f i are sufficiently small, then the direct method is an appropriate choice. If xi and f i are numerically large numbers, then we can go for the assumed mean method or step-deviation method. If the class sizes are unequal, and xi are large numerically, we can still apply the step-deviation method by taking h to be a suitable divisor of all the di ’s. Rationalised 2023-24 180 MATHEMATICS Example 3 : The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify? Number of 20 - 60 60 - 100 100 - 150 150 - 250 250 - 350 350 - 450 wickets Number of 7 5 16 12 2 3 bowlers Solution : Here, the class size varies, and the xi , s are large. Let us still apply the stepdeviation method with a = 200 and h = 20. Then, we obtain the data as in Table 13.8. Table 13.8 Number of Number of xi di = xi – 200 = 20 i i d u ui f 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 So, 106 45 − u = ⋅ Therefore, x = 200 + 106 20 45   −     = 200 – 47.11 = 152.89. This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89. Now, let us see how well you can apply the concepts discussed in this section! Rationalised 2023-24 STATISTICS 181 Activity 2 : Divide the students of your class into three groups and ask each group to do one of the following activities.

1. Collect the marks obtained by all the students of your class in Mathematics in the latest examination conducted by your school. Form a grouped frequency distribution 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
4. 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 121562 3 Which method did you use for finding the mean, and why?
5. 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.
6. 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 allowance (in `) Number of children 7 6 9 13 f 5 4 Rationalised 2023-24 182 MATHEMATICS
7. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. Number of heartbeats 65 - 68 68 - 71 71 - 74 74 - 77 77 - 80 80 - 83 83 - 86 per minute Number of women 2438742
8. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Number of mangoes 50 - 52 53 - 55 56 - 58 59 - 61 62 - 64 Number of boxes 15 110 135 115 25 Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
9. The table below shows the daily expenditure on food of 25 households in a locality. Daily expenditure 100 - 150 150 - 200 200 - 250 250 - 300 300 - 350 (in `) Number of 4 5 12 2 2 households Find the mean daily expenditure on food by a suitable method.
10. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: Concentration of SO2 (in ppm) Frequency 0.00 - 0.04 4 0.04 - 0.08 9 0.08 - 0.12 9 0.12 - 0.16 2 0.16 - 0.20 4 0.20 - 0.24 2 Find the mean concentration of SO2 in the air. Rationalised 2023-24 STATISTICS 183
11. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. Number of 0 - 6 6 - 10 10 - 14 14 - 20 20 - 28 28 - 38 38 - 40 days Number of 11 10 7 4 4 3 1 students
12. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. Literacy rate (in %) 45 - 55 55 - 65 65 - 75 75 - 85 85 - 95 Number of cities 3 10 11 8 3 13.3 Mode of Grouped Data Recall from Class IX, a mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency. Further, we discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining a mode of grouped data. It is possible that more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to problems having a single mode only. Let us first recall how we found the mode for ungrouped data through the following example. Example 4 : The wickets taken by a bowler in 10 cricket matches are as follows: 2645021323 Find the mode of the data. Solution : Let us form the frequency distribution table of the given data as follows: Number of 0123456 wickets Number of 1132111 matches Rationalised 2023-24 184 MATHEMATICS Clearly, 2 is the number of wickets taken by the bowler in the maximum number (i.e., 3) of matches. So, the mode of this data is 2. In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula: Mode = 1 0 10 2 2 f f l h ff f            where l = lower limit of the modal class, h = size of the class interval (assuming all class sizes to be equal), f 1 = frequency of the modal class, f 0 = frequency of the class preceding the modal class, f 2 = frequency of the class succeeding the modal class. Let us consider the following examples to illustrate the use of this formula. Example 5 : A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household: Family size 1 - 3 3 - 5 5 - 7 7 - 9 9 - 11 Number of 78221 families Find the mode of this data. Solution : Here the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5. So, the modal class is 3 – 5. Now modal class = 3 – 5, lower limit (l) of modal class = 3, class size (h) = 2 frequency ( f 1 ) of the modal class = 8, frequency ( f 0 ) of class preceding the modal class = 7, frequency ( f 2 ) of class succeeding the modal class = 2. Now, let us substitute these values in the formula : Rationalised 2023-24 STATISTICS 185 Mode = 1 0 10 2 2 f f l h ff f            = 8 7 2 3 2 3 3.286 287 2 7              Therefore, the mode of the data above is 3.286. Example 6 : The marks distribution of 30 students in a mathematics examination are given in Table 13.3 of Example 1. Find the mode of this data. Also compare and interpret the mode and the mean. Solution : Refer to Table 13.3 of Example 1. Since the maximum number of students (i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55. Therefore, the lower limit (l) of the modal class = 40, the class size ( h) = 15, the frequency ( f 1 ) of modal class = 7, the frequency ( f 0 ) of the class preceding the modal class = 3, the frequency ( f 2 ) of the class succeeding the modal class = 6. Now, using the formula: Mode = 1 0 102 2 f f l h ff f            , we get Mode = 7 3 40 15 14 6 3            = 52 So, the mode marks is 52. Now, from Example 1, you know that the mean marks is 62. So, the maximum number of students obtained 52 marks, while on an average a student obtained 62 marks. Remarks :
13. In Example 6, the mode is less than the mean. But for some other problems it may be equal or more than the mean also.
14. It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the average of the marks obtained by most Rationalised 2023-24 186 MATHEMATICS of the students. In the first situation, the mean is required and in the second situation, the mode is required. Activity 3 : Continuing with the same groups as formed in Activity 2 and the situations assigned to the groups. Ask each group to find the mode of the data. They should also compare this with the mean, and interpret the meaning of both. Remark : The mode can also be calculated for grouped data with unequal class sizes. However, we shall not be discussing it. EXERCISE 13.2
15. The following table shows the ages of the patients admitted in a hospital during a year: Age (in years) 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 Number of patients 6 11 21 23 14 5 Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
16. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Lifetimes (in hours) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 Frequency 10 35 52 61 38 29 Determine the modal lifetimes of the components.
17. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure : Expenditure (in `) Number of families 1000 - 1500 24 1500 - 2000 40 2000 - 2500 33 2500 - 3000 28 3000 - 3500 30 3500 - 4000 22 4000 - 4500 16 4500 - 5000 7 Rationalised 2023-24 STATISTICS 187
18. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher Number of states / U.T. 15 - 20 3 20 - 25 8 25 - 30 9 30 - 35 10 35 - 40 3 40 - 45 0 45 - 50 0 50 - 55 2
19. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Runs scored Number of batsmen 3000 - 4000 4 4000 - 5000 18 5000 - 6000 9 6000 - 7000 7 7000 - 8000 6 8000 - 9000 3 9000 - 10000 1 10000 - 11000 1 Find the mode of the data.
20. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data : Number of cars 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency 7 14 13 12 20 11 15 8 Rationalised 2023-24 188 MATHEMATICS 13.4 Median of Grouped Data As you have studied in Class IX, the median is a measure of central tendency which gives the value of the middle-most observation in the data. Recall that for finding the median of ungrouped data, we first arrange the data values of the observations in ascending order. Then, if n is odd, the median is the 1 2   n      th observation. And, if n is even, then the median will be the average of the th 2 n and the 1 th 2   n      observations. Suppose, we have to find the median of the following data, which gives the marks, out of 50, obtained by 100 students in a test : Marks obtained 20 29 28 33 42 38 43 25 Number of students 6 28 24 15 2 4 1 20 First, we arrange the marks in ascending order and prepare a frequency table as follows : Table 13.9 Marks obtained Number of students (Frequency) 20 6 25 20 28 24 29 28 33 15 38 4 42 2 43 1 Total 100 Rationalised 2023-24 STATISTICS 189 Here n = 100, which is even. The median will be the average of the 2 n th and the 1 2  n       th observations, i.e., the 50th and 51st observations. To find these observations, we proceed as follows: Table 13.10 Marks obtained Number of students 20 6 upto 25 6 + 20 = 26 upto 28 26 + 24 = 50 upto 29 50 + 28 = 78 upto 33 78 + 15 = 93 upto 38 93 + 4 = 97 upto 42 97 + 2 = 99 upto 43 99 + 1 = 100 Now we add another column depicting this information to the frequency table above and name it as cumulative frequency column. Table 13.11 Marks obtained Number of students Cumulative frequency 20 6 6 25 20 26 28 24 50 29 28 78 33 15 93 38 4 97 42 2 99 43 1 100 Rationalised 2023-24 190 MATHEMATICS From the table above, we see that: 50th observaton is 28 (Why?) 51st observation is 29 So, Median = 28 29 28.5 2 + = Remark : The part of Table 13.11 consisting Column 1 and Column 3 is known as Cumulative Frequency Table. The median marks 28.5 conveys the information that about 50% students obtained marks less than 28.5 and another 50% students obtained marks more than 28.5. Now, let us see how to obtain the median of grouped data, through the following situation. Consider a grouped frequency distribution of marks obtained, out of 100, by 53 students, in a certain examination, as follows: Table 13.12 Marks Number of students 0 - 10 5 10 - 20 3 20 - 30 4 30 - 40 3 40 - 50 3 50 - 60 4 60 - 70 7 70 - 80 9 80 - 90 7 90 - 100 8 From the table above, try to answer the following questions: How many students have scored marks less than 10? The answer is clearly 5. Rationalised 2023-24 STATISTICS 191 How many students have scored less than 20 marks? Observe that the number of students who have scored less than 20 include the number of students who have scored marks from 0 - 10 as well as the number of students who have scored marks from 10 - 20. So, the total number of students with marks less than 20 is 5 + 3, i.e., 8. We say that the cumulative frequency of the class 10 -20 is 8. Similarly, we can compute the cumulative frequencies of the other classes, i.e., the number of students with marks less than 30, less than 40, . . ., less than 100. We give them in Table 13.13 given below: Table 13.13 Marks obtained Number of students (Cumulative frequency) Less than 10 5 Less than 20 5 + 3 = 8 Less than 30 8 + 4 = 12 Less than 40 12 + 3 = 15 Less than 50 15 + 3 = 18 Less than 60 18 + 4 = 22 Less than 70 22 + 7 = 29 Less than 80 29 + 9 = 38 Less than 90 38 + 7 = 45 Less than 100 45 + 8 = 53 The distribution given above is called the cumulative frequency distribution of the less than type. Here 10, 20, 30, . . . 100, are the upper limits of the respective class intervals. We can similarly make the table for the number of students with scores, more than or equal to 0, more than or equal to 10, more than or equal to 20, and so on. From Table 13.12, we observe that all 53 students have scored marks more than or equal to
21. Since there are 5 students scoring marks in the interval 0 - 10, this means that there are 53 – 5 = 48 students getting more than or equal to 10 marks. Continuing in the same manner, we get the number of students scoring 20 or above as 48 – 3 = 45, 30 or above as 45 – 4 = 41, and so on, as shown in Table 13.14. Rationalised 2023-24 192 MATHEMATICS Table 13.14 Marks obtained Number of students (Cumulative frequency) More than or equal to 0 53 More than or equal to 10 53 – 5 = 48 More than or equal to 20 48 – 3 = 45 More than or equal to 30 45 – 4 = 41 More than or equal to 40 41 – 3 = 38 More than or equal to 50 38 – 3 = 35 More than or equal to 60 35 – 4 = 31 More than or equal to 70 31 – 7 = 24 More than or equal to 80 24 – 9 = 15 More than or equal to 90 15 – 7 = 8 The table above is called a cumulative frequency distribution of the more than type. Here 0, 10, 20, . . ., 90 give the lower limits of the respective class intervals. Now, to find the median of grouped data, we can make use of any of these cumulative frequency distributions. Let us combine Tables 13.12 and 13.13 to get Table 13.15 given below: Table 13.15 Marks Number of students ( f ) Cumulative frequency (cf ) 0 - 10 5 5 10 - 20 3 8 20 - 30 4 12 30 - 40 3 15 40 - 50 3 18 50 - 60 4 22 60 - 70 7 29 70 - 80 9 38 80 - 90 7 45 90 - 100 8 53 Now in a grouped data, we may not be able to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in Rationalised 2023-24 STATISTICS 193 a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves. But which class should this be? To find this class, we find the cumulative frequencies of all the classes and 2 n . We now locate the class whose cumulative frequency is greater than (and nearest to) 2 n  This is called the median class. In the distribution above, n = 53. So, 2 n = 26.5. Now 60 – 70 is the class whose cumulative frequency 29 is greater than (and nearest to) 2 n , i.e., 26.5. Therefore, 60 – 70 is the median class. After finding the median class, we use the following formula for calculating the median. Median = cf 2 + , n l h f             where l = lower limit of median class, n = number of observations, cf = cumulative frequency of class preceding the median class, f = frequency of median class, h = class size (assuming class size to be equal). Substituting the values 26.5, 2 n  l = 60, cf = 22, f = 7, h = 10 in the formula above, we get Median = 26.5 22 60 10 7          = 60 + 45 7 = 66.4 So, about half the students have scored marks less than 66.4, and the other half have scored marks more than 66.4. Rationalised 2023-24 194 MATHEMATICS Example 7 : A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained: Height (in cm) Number of girls Less than 140 4 Less than 145 11 Less than 150 29 Less than 155 40 Less than 160 46 Less than 165 51 Find the median height. Solution : To calculate the median height, we need to find the class intervals and their corresponding frequencies. The given distribution being of the less than type, 140, 145, 150, . . ., 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140 - 145, 145 - 150, . . ., 160 - 165. Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e., the frequency of class interval below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140 - 145 is 11 – 4 = 7. Similarly, the frequency of 145 - 150 is 29 – 11 = 18, for 150 - 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with the given cumulative frequencies becomes: Table 13.16 Class intervals Frequency Cumulative frequency Below 140 4 4 140 - 145 7 11 145 - 150 18 29 150 - 155 11 40 155 - 160 6 46 160 - 165 5 51 Rationalised 2023-24 STATISTICS 195 Now n = 51. So, 51 25.5 2 2 n = = . This observation lies in the class 145 - 150. Then, l (the lower limit) = 145, cf (the cumulative frequency of the class preceding 145 - 150) = 11, f (the frequency of the median class 145 - 150) = 18, h (the class size) = 5. Using the formula, Median = l + cf 2 n h f   −     ×     , we have Median = 25.5 11 145 5 18   −

• ×     = 145 + 72.5 18 = 149.03. So, the median height of the girls is 149.03 cm. This means that the height of about 50% of the girls is less than this height, and 50% are taller than this height. Example 8 : The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Class intervals Frequency 0 - 100 2 100 - 200 5 200 - 300 x 300 - 400 12 400 - 500 17 500 - 600 20 600 - 700 y 700 - 800 9 800 - 900 7 900 - 1000 4 Rationalised 2023-24 196 MATHEMATICS Solution : Class intervals Frequency Cumulative frequency 0 - 100 2 2 100 - 200 5 7 200 - 300 x 7 + x 300 - 400 12 19 + x 400 - 500 17 36 + x 500 - 600 20 56 + x 600 - 700 y 56 + x + y 700 - 800 9 65 + x + y 800 - 900 7 72 + x + y 900 - 1000 4 76 + x + y It is given that n = 100 So, 76 + x + y = 100, i.e., x + y = 24 (1) The median is 525, which lies in the class 500 – 600 So, l = 500, f = 20, cf = 36 + x, h = 100 Using the formula : Median = cf 2 , n l h f             we get 525 = 50 36 500 100 20     x       i.e., 525 – 500 = (14 – x) × 5 i.e., 25 = 70 – 5x i.e., 5x = 70 – 25 = 45 So, x =9 Therefore, from (1), we get 9 + y = 24 i.e., y = 15 Rationalised 2023-24 STATISTICS 197 Now, that you have studied about all the three measures of central tendency, let us discuss which measure would be best suited for a particular requirement. The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes, i.e., the largest and the smallest observations of the entire data. It also enables us to compare two or more distributions. For example, by comparing the average (mean) results of students of different schools of a particular examination, we can conclude which school has a better performance. However, extreme values in the data affect the mean. For example, the mean of classes having frequencies more or less the same is a good representative of the data. But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20, 21, 18, then the mean will certainly not reflect the way the data behaves. So, in such cases, the mean is not a good representative of the data. In problems where individual observations are not important, and we wish to find out a ‘typical’ observation, the median is more appropriate, e.g., finding the typical productivity rate of workers, average wage in a country, etc. These are situations where extreme values may be there. So, rather than the mean, we take the median as a better measure of central tendency. In situations which require establishing the most frequent value or most popular item, the mode is the best choice, e.g., to find the most popular T.V. programme being watched, the consumer item in greatest demand, the colour of the vehicle used by most of the people, etc. Remarks :

1. There is a empirical relationship between the three measures of central tendency : 3 Median = Mode + 2 Mean
2. The median of grouped data with unequal class sizes can also be calculated. However, we shall not discuss it here. Rationalised 2023-24 198 MATHEMATICS EXERCISE 13.3
3. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption (in units) Number of consumers 65 - 85 4 85 - 105 5 105 - 125 13 125 - 145 20 145 - 165 14 165 - 185 8 185 - 205 4
4. If the median of the distribution given below is 28.5, find the values of x and y. Class interval Frequency 0 - 10 5 10 - 20 x 20 - 30 20 30 - 40 15 40 - 50 y 50 - 60 5 Total 60
5. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. Rationalised 2023-24 STATISTICS 199 Age (in years) Number of policy holders Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100
6. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table : Length (in mm) Number of leaves 118 - 126 3 127 - 135 5 136 - 144 9 145 - 153 12 154 - 162 5 163 - 171 4 172 - 180 2 Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.) Rationalised 2023-24 200 MATHEMATICS
7. The following table gives the distribution of the life time of 400 neon lamps : Life time (in hours) Number of lamps 1500 - 2000 14 2000 - 2500 56 2500 - 3000 60 3000 - 3500 86 3500 - 4000 74 4000 - 4500 62 4500 - 5000 48 Find the median life time of a lamp.
8. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters 1 - 4 4 - 7 7 - 10 10 - 13 13 - 16 16 - 19 Number of surnames 6 30 40 16 4 4 Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
9. The distribution below gives the weights of 30 students of a class. Find the median weight of the students. Weight (in kg) 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 Number of students 2 3 8 6 6 3 2 13.5 Summary In this chapter, you have studied the following points:
10. The mean for grouped data can be found by : (i) the direct method : i i i f x x f Σ = Σ (ii) the assumed mean method : i i i f d x a f Σ = + Σ Rationalised 2023-24 STATISTICS 201 (iii) the step deviation method : i i i f u x a h f           , with the assumption that the frequency of a class is centred at its mid-point, called its class mark.
11. The mode for grouped data can be found by using the formula: Mode = 1 0 102 2 f f l h ff f            where symbols have their usual meanings.
12. The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.
13. The median for grouped data is formed by using the formula: Median = cf 2 n l h f              , where symbols have their usual meanings. A NOTE TO THE READER For calculating mode and median for grouped data, it should be ensured that the class intervals are continuous before applying the formulae. Same condition also apply for construction of an ogive. Further, in case of ogives, the scale may not be the same on both the axes. Rationalised 2023-24