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Page 1 1. Find the mode of the following distribution: Marks 10 20 20 30 30 40 40 50 50 60 Frequency 12 35 45 25 13 Sol: Here, the maximum class frequency is 45, and the class corresponding to this frequency is 30 40. So, the modal class is 30- 40. Now, Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10, frequency (f1) of the modal class = 45, frequency (f0) of class preceding the modal class = 35, frequency (f2) of class succeeding the modal class = 25 Now, let us substitute these values in the formula: Mode = l + × h = 30 + × 10 = 30 + × 10 = 30 + 3.33 = 33.33 Hence, the mode is 33.33. 2. Compute the mode of the following data: Class 0 20 20 40 40 60 60 80 80 100 Frequency 25 16 28 20 5 Sol: Here, the maximum class frequency is 28, and the class corresponding to this frequency is 40 60. So, the modal class is 40 60. Page 2 1. Find the mode of the following distribution: Marks 10 20 20 30 30 40 40 50 50 60 Frequency 12 35 45 25 13 Sol: Here, the maximum class frequency is 45, and the class corresponding to this frequency is 30 40. So, the modal class is 30- 40. Now, Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10, frequency (f1) of the modal class = 45, frequency (f0) of class preceding the modal class = 35, frequency (f2) of class succeeding the modal class = 25 Now, let us substitute these values in the formula: Mode = l + × h = 30 + × 10 = 30 + × 10 = 30 + 3.33 = 33.33 Hence, the mode is 33.33. 2. Compute the mode of the following data: Class 0 20 20 40 40 60 60 80 80 100 Frequency 25 16 28 20 5 Sol: Here, the maximum class frequency is 28, and the class corresponding to this frequency is 40 60. So, the modal class is 40 60. Now, Modal class = 40 60, lower limit ( l) of modal class = 40, class size ( h) = 20, frequency (f1) of the modal class = 28, frequency (f0) of class preceding the modal class = 16, frequency (f2) of class succeeding the modal class = 20 Now, let us substitute these values in the formula: Mode = l + × h = 40 + × 20 = 40 + × 20 = 40 + 12 = 52 Hence, the mode is 52. 3. Heights of students of class X are givee in the flowing frequency distribution Height (in cm) 150 155 155 160 160 165 165 170 170 - 175 Number of students 15 8 20 12 5 Find the modal height. Also, find the mean height. Compared and interpret the two measures of central tendency. Sol: Here, the maximum class frequency is 20, and the class corresponding to this frequency is 160 165. So, the modal class is 160 165. Now, Modal class = 160 165, lower limit ( l) of modal class = 160, class size ( h) = 5, frequency (f1) of the modal class = 20, frequency (f0) of class preceding the modal class = 8, frequency (f2) of class succeeding the modal class = 12 Now, let us substitute these values in the formula: Mode = l + × h = 160 + × 5 = 160 + × 5 = 160 + 3 = 163 Hence, the mode is 163. It represents that the height of maximum number of students is 163cm. Now, to find the mean let us put the data in the table given below: Page 3 1. Find the mode of the following distribution: Marks 10 20 20 30 30 40 40 50 50 60 Frequency 12 35 45 25 13 Sol: Here, the maximum class frequency is 45, and the class corresponding to this frequency is 30 40. So, the modal class is 30- 40. Now, Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10, frequency (f1) of the modal class = 45, frequency (f0) of class preceding the modal class = 35, frequency (f2) of class succeeding the modal class = 25 Now, let us substitute these values in the formula: Mode = l + × h = 30 + × 10 = 30 + × 10 = 30 + 3.33 = 33.33 Hence, the mode is 33.33. 2. Compute the mode of the following data: Class 0 20 20 40 40 60 60 80 80 100 Frequency 25 16 28 20 5 Sol: Here, the maximum class frequency is 28, and the class corresponding to this frequency is 40 60. So, the modal class is 40 60. Now, Modal class = 40 60, lower limit ( l) of modal class = 40, class size ( h) = 20, frequency (f1) of the modal class = 28, frequency (f0) of class preceding the modal class = 16, frequency (f2) of class succeeding the modal class = 20 Now, let us substitute these values in the formula: Mode = l + × h = 40 + × 20 = 40 + × 20 = 40 + 12 = 52 Hence, the mode is 52. 3. Heights of students of class X are givee in the flowing frequency distribution Height (in cm) 150 155 155 160 160 165 165 170 170 - 175 Number of students 15 8 20 12 5 Find the modal height. Also, find the mean height. Compared and interpret the two measures of central tendency. Sol: Here, the maximum class frequency is 20, and the class corresponding to this frequency is 160 165. So, the modal class is 160 165. Now, Modal class = 160 165, lower limit ( l) of modal class = 160, class size ( h) = 5, frequency (f1) of the modal class = 20, frequency (f0) of class preceding the modal class = 8, frequency (f2) of class succeeding the modal class = 12 Now, let us substitute these values in the formula: Mode = l + × h = 160 + × 5 = 160 + × 5 = 160 + 3 = 163 Hence, the mode is 163. It represents that the height of maximum number of students is 163cm. Now, to find the mean let us put the data in the table given below: Height (in cm) Number of students (fi) Class mark (xi) fi xi 150 155 15 152.5 2287.5 155 160 8 157.5 1260 160 165 20 162.5 3250 165 170 12 167.5 2010 170 175 5 172.5 862.5 Total fi = 60 fi xi = 9670 Mean = i i i i i f x f = = 161.17 Thus, mean of the given data is 161.17. It represents that on an average, the height of a student is 161.17cm. 4. Find the mode of the following distribution: Class interval 10 14 14 18 18 22 22 26 26 30 30 34 34 38 38 42 Frequency 8 6 11 20 25 22 10 4 Sol: As the class 26 30 has the maximum frequency, it is the modal class. Now, xk = 26, h = 4, fk = 25, fk-1 = 20, fk+1 = 22 Mode, M0 = xk + = 26 + = 26 + = (26 + 2.5) = 28.5 5. Given below is the distribution of total household expenditure of 200 manual workers in a city: Expenditure (in Rs) Number of manual workers 1000 1500 24 1500 2000 40 2000 2500 31 2500 3000 28 3000 3500 32 3500 4000 23 4000 4500 17 4500 5000 5 Page 4 1. Find the mode of the following distribution: Marks 10 20 20 30 30 40 40 50 50 60 Frequency 12 35 45 25 13 Sol: Here, the maximum class frequency is 45, and the class corresponding to this frequency is 30 40. So, the modal class is 30- 40. Now, Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10, frequency (f1) of the modal class = 45, frequency (f0) of class preceding the modal class = 35, frequency (f2) of class succeeding the modal class = 25 Now, let us substitute these values in the formula: Mode = l + × h = 30 + × 10 = 30 + × 10 = 30 + 3.33 = 33.33 Hence, the mode is 33.33. 2. Compute the mode of the following data: Class 0 20 20 40 40 60 60 80 80 100 Frequency 25 16 28 20 5 Sol: Here, the maximum class frequency is 28, and the class corresponding to this frequency is 40 60. So, the modal class is 40 60. Now, Modal class = 40 60, lower limit ( l) of modal class = 40, class size ( h) = 20, frequency (f1) of the modal class = 28, frequency (f0) of class preceding the modal class = 16, frequency (f2) of class succeeding the modal class = 20 Now, let us substitute these values in the formula: Mode = l + × h = 40 + × 20 = 40 + × 20 = 40 + 12 = 52 Hence, the mode is 52. 3. Heights of students of class X are givee in the flowing frequency distribution Height (in cm) 150 155 155 160 160 165 165 170 170 - 175 Number of students 15 8 20 12 5 Find the modal height. Also, find the mean height. Compared and interpret the two measures of central tendency. Sol: Here, the maximum class frequency is 20, and the class corresponding to this frequency is 160 165. So, the modal class is 160 165. Now, Modal class = 160 165, lower limit ( l) of modal class = 160, class size ( h) = 5, frequency (f1) of the modal class = 20, frequency (f0) of class preceding the modal class = 8, frequency (f2) of class succeeding the modal class = 12 Now, let us substitute these values in the formula: Mode = l + × h = 160 + × 5 = 160 + × 5 = 160 + 3 = 163 Hence, the mode is 163. It represents that the height of maximum number of students is 163cm. Now, to find the mean let us put the data in the table given below: Height (in cm) Number of students (fi) Class mark (xi) fi xi 150 155 15 152.5 2287.5 155 160 8 157.5 1260 160 165 20 162.5 3250 165 170 12 167.5 2010 170 175 5 172.5 862.5 Total fi = 60 fi xi = 9670 Mean = i i i i i f x f = = 161.17 Thus, mean of the given data is 161.17. It represents that on an average, the height of a student is 161.17cm. 4. Find the mode of the following distribution: Class interval 10 14 14 18 18 22 22 26 26 30 30 34 34 38 38 42 Frequency 8 6 11 20 25 22 10 4 Sol: As the class 26 30 has the maximum frequency, it is the modal class. Now, xk = 26, h = 4, fk = 25, fk-1 = 20, fk+1 = 22 Mode, M0 = xk + = 26 + = 26 + = (26 + 2.5) = 28.5 5. Given below is the distribution of total household expenditure of 200 manual workers in a city: Expenditure (in Rs) Number of manual workers 1000 1500 24 1500 2000 40 2000 2500 31 2500 3000 28 3000 3500 32 3500 4000 23 4000 4500 17 4500 5000 5 Find the average expenditure done by maximum number of manual workers. Sol: As the class 1500-2000 has the maximum frequency, it is the modal class. Now, xk = 1500, h = 500, fk = 40, fk-1 = 24, fk+1 = 31 Mode, M0 = xk + = 1500 + = 1500 + = (1500 + 320) = 1820 Hence, mode = Rs 1820 6. Calculate the mode from the following data: Monthly salary (in Rs) No of employees 0 5000 90 5000 10000 150 10000 15000 100 15000 20000 80 20000 25000 70 25000 30000 10 Sol: As the class 5000-10000 has the maximum frequency, it is the modal class. Now, xk = 5000, h = 5000, fk = 150, fk-1 = 90, fk+1 = 100 Mode, M0 = xk + = 5000 + = 5000 + = (5000 + 2727.27) = 7727.27 Hence, mode = Rs 7727.27 7. Compute the mode from the following data: Age (in years) 0 5 5 10 10 15 15 20 20 25 25 30 30 - 35 No of patients 6 11 18 24 17 13 5 Sol: As the class 15 20 has the maximum frequency, it is the modal class. Now, xk = 15, h = 5, fk = 24, fk-1 = 18, fk+1 = 17 Page 5 1. Find the mode of the following distribution: Marks 10 20 20 30 30 40 40 50 50 60 Frequency 12 35 45 25 13 Sol: Here, the maximum class frequency is 45, and the class corresponding to this frequency is 30 40. So, the modal class is 30- 40. Now, Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10, frequency (f1) of the modal class = 45, frequency (f0) of class preceding the modal class = 35, frequency (f2) of class succeeding the modal class = 25 Now, let us substitute these values in the formula: Mode = l + × h = 30 + × 10 = 30 + × 10 = 30 + 3.33 = 33.33 Hence, the mode is 33.33. 2. Compute the mode of the following data: Class 0 20 20 40 40 60 60 80 80 100 Frequency 25 16 28 20 5 Sol: Here, the maximum class frequency is 28, and the class corresponding to this frequency is 40 60. So, the modal class is 40 60. Now, Modal class = 40 60, lower limit ( l) of modal class = 40, class size ( h) = 20, frequency (f1) of the modal class = 28, frequency (f0) of class preceding the modal class = 16, frequency (f2) of class succeeding the modal class = 20 Now, let us substitute these values in the formula: Mode = l + × h = 40 + × 20 = 40 + × 20 = 40 + 12 = 52 Hence, the mode is 52. 3. Heights of students of class X are givee in the flowing frequency distribution Height (in cm) 150 155 155 160 160 165 165 170 170 - 175 Number of students 15 8 20 12 5 Find the modal height. Also, find the mean height. Compared and interpret the two measures of central tendency. Sol: Here, the maximum class frequency is 20, and the class corresponding to this frequency is 160 165. So, the modal class is 160 165. Now, Modal class = 160 165, lower limit ( l) of modal class = 160, class size ( h) = 5, frequency (f1) of the modal class = 20, frequency (f0) of class preceding the modal class = 8, frequency (f2) of class succeeding the modal class = 12 Now, let us substitute these values in the formula: Mode = l + × h = 160 + × 5 = 160 + × 5 = 160 + 3 = 163 Hence, the mode is 163. It represents that the height of maximum number of students is 163cm. Now, to find the mean let us put the data in the table given below: Height (in cm) Number of students (fi) Class mark (xi) fi xi 150 155 15 152.5 2287.5 155 160 8 157.5 1260 160 165 20 162.5 3250 165 170 12 167.5 2010 170 175 5 172.5 862.5 Total fi = 60 fi xi = 9670 Mean = i i i i i f x f = = 161.17 Thus, mean of the given data is 161.17. It represents that on an average, the height of a student is 161.17cm. 4. Find the mode of the following distribution: Class interval 10 14 14 18 18 22 22 26 26 30 30 34 34 38 38 42 Frequency 8 6 11 20 25 22 10 4 Sol: As the class 26 30 has the maximum frequency, it is the modal class. Now, xk = 26, h = 4, fk = 25, fk-1 = 20, fk+1 = 22 Mode, M0 = xk + = 26 + = 26 + = (26 + 2.5) = 28.5 5. Given below is the distribution of total household expenditure of 200 manual workers in a city: Expenditure (in Rs) Number of manual workers 1000 1500 24 1500 2000 40 2000 2500 31 2500 3000 28 3000 3500 32 3500 4000 23 4000 4500 17 4500 5000 5 Find the average expenditure done by maximum number of manual workers. Sol: As the class 1500-2000 has the maximum frequency, it is the modal class. Now, xk = 1500, h = 500, fk = 40, fk-1 = 24, fk+1 = 31 Mode, M0 = xk + = 1500 + = 1500 + = (1500 + 320) = 1820 Hence, mode = Rs 1820 6. Calculate the mode from the following data: Monthly salary (in Rs) No of employees 0 5000 90 5000 10000 150 10000 15000 100 15000 20000 80 20000 25000 70 25000 30000 10 Sol: As the class 5000-10000 has the maximum frequency, it is the modal class. Now, xk = 5000, h = 5000, fk = 150, fk-1 = 90, fk+1 = 100 Mode, M0 = xk + = 5000 + = 5000 + = (5000 + 2727.27) = 7727.27 Hence, mode = Rs 7727.27 7. Compute the mode from the following data: Age (in years) 0 5 5 10 10 15 15 20 20 25 25 30 30 - 35 No of patients 6 11 18 24 17 13 5 Sol: As the class 15 20 has the maximum frequency, it is the modal class. Now, xk = 15, h = 5, fk = 24, fk-1 = 18, fk+1 = 17 Mode, M0 = xk + = 15 + = 15 + = (15 + 2.3) = 17.3 Hence, mode = 17.3 years 8. Compute the mode from the following series: Sol: As the class 85 95 has the maximum frequency, it is the modal class. Now, xk = 85, h = 10, fk = 32, fk-1 = 30, fk+1 = 6 Mode, M0 = xk + = 85 + = 85 + = (85 + 0.71) = 85.71 Hence, mode = 85.71 9. Compute the mode from the following data: Class interval 1 5 6 10 11 15 16 20 21 25 26 30 31 35 36 40 41 45 46 50 Frequency 3 8 13 18 28 20 13 8 6 4 Sol: Clearly, we have to find the mode of the data. The given data is an inclusive series. So, we will convert it to an exclusive form as given below: Class interval 0.5 5.5 5.5 10.5 10.5 15.5 15.5 20.5 20.5 25.5 25.5 30.5 30.5 35.5 35.5 40.5 40.5 45.5 45.5 50.5 Frequency 3 8 13 18 28 20 13 8 6 4 As the class 20.5 25.5 has the maximum frequency, it is the modal class. Now, xk = 20.5, h = 5, fk = 28, fk-1 = 18, fk+1 = 20 Mode, M0 = xk + = 20.5 + = 20.5 + = (20.5 + 2.78) Size 45 55 55 65 65 75 75 85 85 95 95 105 105 - 115 Frequency 7 12 17 30 32 6 10Read More
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