Page 1 1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age. Age (in years) 0 15 15 30 30 45 45 60 60 - 75 No. of patients 5 20 40 50 25 Sol: We prepare the cumulative frequency table, as shown below: Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0 15 5 5 15 30 20 25 30 45 40 65 45 60 50 115 60 75 25 140 Total N = = 140 Now, N = 140 = 70. The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 60. Thus, the median class is 45 60. l = 45, h = 15, f = 50, N = 140 and cf = 65. Now, Median = l + × h = 45 + × 15 = 45 + × 15 = 45 + 1.5 = 46.5 Page 2 1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age. Age (in years) 0 15 15 30 30 45 45 60 60 - 75 No. of patients 5 20 40 50 25 Sol: We prepare the cumulative frequency table, as shown below: Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0 15 5 5 15 30 20 25 30 45 40 65 45 60 50 115 60 75 25 140 Total N = = 140 Now, N = 140 = 70. The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 60. Thus, the median class is 45 60. l = 45, h = 15, f = 50, N = 140 and cf = 65. Now, Median = l + × h = 45 + × 15 = 45 + × 15 = 45 + 1.5 = 46.5 Hence, the median age is 46.5 years. 2. Compute mean from the following data: Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49 Number of Students 3 4 7 11 0 16 9 Sol: Class Frequency (f) Cumulative Frequency (cf) 0 7 3 3 7 14 4 7 14 21 7 14 21 - 28 11 25 28 35 0 25 35 42 16 41 42 49 9 50 N = = 50 Now, N = 50 = 25. The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42. Thus, the median class is 35 42. l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25. Now, Median = l + × h = 35 + 7 × = 35 + 0 = 35 Hence, the median age is 46.5 years. 3. The following table shows the daily wages of workers in a factory: 0 100 100 200 200 300 300 400 400 500 Number of workers 40 32 48 22 8 Find the median daily wage income of the workers. Sol: Class Frequency (f) Cumulative Frequency (cf) 0 100 40 40 100 200 32 72 200 300 48 120 300 400 22 142 400 500 8 150 N = = 150 Page 3 1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age. Age (in years) 0 15 15 30 30 45 45 60 60 - 75 No. of patients 5 20 40 50 25 Sol: We prepare the cumulative frequency table, as shown below: Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0 15 5 5 15 30 20 25 30 45 40 65 45 60 50 115 60 75 25 140 Total N = = 140 Now, N = 140 = 70. The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 60. Thus, the median class is 45 60. l = 45, h = 15, f = 50, N = 140 and cf = 65. Now, Median = l + × h = 45 + × 15 = 45 + × 15 = 45 + 1.5 = 46.5 Hence, the median age is 46.5 years. 2. Compute mean from the following data: Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49 Number of Students 3 4 7 11 0 16 9 Sol: Class Frequency (f) Cumulative Frequency (cf) 0 7 3 3 7 14 4 7 14 21 7 14 21 - 28 11 25 28 35 0 25 35 42 16 41 42 49 9 50 N = = 50 Now, N = 50 = 25. The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42. Thus, the median class is 35 42. l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25. Now, Median = l + × h = 35 + 7 × = 35 + 0 = 35 Hence, the median age is 46.5 years. 3. The following table shows the daily wages of workers in a factory: 0 100 100 200 200 300 300 400 400 500 Number of workers 40 32 48 22 8 Find the median daily wage income of the workers. Sol: Class Frequency (f) Cumulative Frequency (cf) 0 100 40 40 100 200 32 72 200 300 48 120 300 400 22 142 400 500 8 150 N = = 150 Now, N = 150 = 75. The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 300. Thus, the median class is 200 300. l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75. Now, Median, M = l + = 200 + = 200 + 6.25 = 206.25 Hence, the median daily wage income of the workers is Rs 206.25. 4. Calculate the median from the following frequency distribution table: Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45 Frequency 5 6 15 10 5 4 2 2 Sol: Class Frequency (f) Cumulative Frequency (cf) 5 10 5 5 10 15 6 11 15 20 15 26 20 25 10 36 25 30 5 41 30 35 4 45 35 40 2 47 40 45 2 49 N = = 49 Now, N = 49 = 24.5. The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 - 20. Thus, the median class is 15 20. l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5. Now, Median, M = l + = 15 + Page 4 1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age. Age (in years) 0 15 15 30 30 45 45 60 60 - 75 No. of patients 5 20 40 50 25 Sol: We prepare the cumulative frequency table, as shown below: Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0 15 5 5 15 30 20 25 30 45 40 65 45 60 50 115 60 75 25 140 Total N = = 140 Now, N = 140 = 70. The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 60. Thus, the median class is 45 60. l = 45, h = 15, f = 50, N = 140 and cf = 65. Now, Median = l + × h = 45 + × 15 = 45 + × 15 = 45 + 1.5 = 46.5 Hence, the median age is 46.5 years. 2. Compute mean from the following data: Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49 Number of Students 3 4 7 11 0 16 9 Sol: Class Frequency (f) Cumulative Frequency (cf) 0 7 3 3 7 14 4 7 14 21 7 14 21 - 28 11 25 28 35 0 25 35 42 16 41 42 49 9 50 N = = 50 Now, N = 50 = 25. The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42. Thus, the median class is 35 42. l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25. Now, Median = l + × h = 35 + 7 × = 35 + 0 = 35 Hence, the median age is 46.5 years. 3. The following table shows the daily wages of workers in a factory: 0 100 100 200 200 300 300 400 400 500 Number of workers 40 32 48 22 8 Find the median daily wage income of the workers. Sol: Class Frequency (f) Cumulative Frequency (cf) 0 100 40 40 100 200 32 72 200 300 48 120 300 400 22 142 400 500 8 150 N = = 150 Now, N = 150 = 75. The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 300. Thus, the median class is 200 300. l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75. Now, Median, M = l + = 200 + = 200 + 6.25 = 206.25 Hence, the median daily wage income of the workers is Rs 206.25. 4. Calculate the median from the following frequency distribution table: Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45 Frequency 5 6 15 10 5 4 2 2 Sol: Class Frequency (f) Cumulative Frequency (cf) 5 10 5 5 10 15 6 11 15 20 15 26 20 25 10 36 25 30 5 41 30 35 4 45 35 40 2 47 40 45 2 49 N = = 49 Now, N = 49 = 24.5. The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 - 20. Thus, the median class is 15 20. l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5. Now, Median, M = l + = 15 + = 15 + 4.5 = 19.5 Hence, the median = 19.5. 5. Given below is the number of units of electricity consumed in a week in a certain locality: Calculate the median. Sol: Class Frequency (f) Cumulative Frequency (cf) 65- 85 4 4 85 105 5 9 105 125 13 22 125 145 20 42 145 165 14 56 165 185 7 63 185 205 4 67 N = = 67 Now, N = 67 = 33.5. The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125 - 145. Thus, the median class is 125 145. l = 125, h = 20, f = 20, cf = c.f. of preceding class = 22 and = 33.5. Now, Median, M = l + = 125 + = 125 + 11.5 = 136.5 Hence, the median = 136.5. 6. Calculate the median from the following data: Class 65 85 85 105 105 125 125 145 145 165 165 185 185 200 Frequency 4 5 13 20 14 7 4 Height(in cm) 135 - 140 140 - 145 145 - 150 150 - 155 155 - 160 160 - 165 165 - 170 170 - 175 Frequency 6 10 18 22 20 15 6 3 Page 5 1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age. Age (in years) 0 15 15 30 30 45 45 60 60 - 75 No. of patients 5 20 40 50 25 Sol: We prepare the cumulative frequency table, as shown below: Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0 15 5 5 15 30 20 25 30 45 40 65 45 60 50 115 60 75 25 140 Total N = = 140 Now, N = 140 = 70. The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 60. Thus, the median class is 45 60. l = 45, h = 15, f = 50, N = 140 and cf = 65. Now, Median = l + × h = 45 + × 15 = 45 + × 15 = 45 + 1.5 = 46.5 Hence, the median age is 46.5 years. 2. Compute mean from the following data: Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49 Number of Students 3 4 7 11 0 16 9 Sol: Class Frequency (f) Cumulative Frequency (cf) 0 7 3 3 7 14 4 7 14 21 7 14 21 - 28 11 25 28 35 0 25 35 42 16 41 42 49 9 50 N = = 50 Now, N = 50 = 25. The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42. Thus, the median class is 35 42. l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25. Now, Median = l + × h = 35 + 7 × = 35 + 0 = 35 Hence, the median age is 46.5 years. 3. The following table shows the daily wages of workers in a factory: 0 100 100 200 200 300 300 400 400 500 Number of workers 40 32 48 22 8 Find the median daily wage income of the workers. Sol: Class Frequency (f) Cumulative Frequency (cf) 0 100 40 40 100 200 32 72 200 300 48 120 300 400 22 142 400 500 8 150 N = = 150 Now, N = 150 = 75. The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 300. Thus, the median class is 200 300. l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75. Now, Median, M = l + = 200 + = 200 + 6.25 = 206.25 Hence, the median daily wage income of the workers is Rs 206.25. 4. Calculate the median from the following frequency distribution table: Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45 Frequency 5 6 15 10 5 4 2 2 Sol: Class Frequency (f) Cumulative Frequency (cf) 5 10 5 5 10 15 6 11 15 20 15 26 20 25 10 36 25 30 5 41 30 35 4 45 35 40 2 47 40 45 2 49 N = = 49 Now, N = 49 = 24.5. The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 - 20. Thus, the median class is 15 20. l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5. Now, Median, M = l + = 15 + = 15 + 4.5 = 19.5 Hence, the median = 19.5. 5. Given below is the number of units of electricity consumed in a week in a certain locality: Calculate the median. Sol: Class Frequency (f) Cumulative Frequency (cf) 65- 85 4 4 85 105 5 9 105 125 13 22 125 145 20 42 145 165 14 56 165 185 7 63 185 205 4 67 N = = 67 Now, N = 67 = 33.5. The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125 - 145. Thus, the median class is 125 145. l = 125, h = 20, f = 20, cf = c.f. of preceding class = 22 and = 33.5. Now, Median, M = l + = 125 + = 125 + 11.5 = 136.5 Hence, the median = 136.5. 6. Calculate the median from the following data: Class 65 85 85 105 105 125 125 145 145 165 165 185 185 200 Frequency 4 5 13 20 14 7 4 Height(in cm) 135 - 140 140 - 145 145 - 150 150 - 155 155 - 160 160 - 165 165 - 170 170 - 175 Frequency 6 10 18 22 20 15 6 3 Sol: Class Frequency (f) Cumulative Frequency (cf) 135 140 6 6 140 145 10 16 145 150 18 34 150 155 22 56 155 160 20 76 160 165 15 91 165 170 6 97 170 175 3 100 N = = 100 Now, N = 100 = 50. The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 - 155. Thus, the median class is 150 155. l = 150, h = 5, f = 22, cf = c.f. of preceding class = 34 and = 50. Now, Median, M = l + = 150 + = 150 + 3.64 = 153.64 Hence, the median = 153.64. 7. Calculate the missing frequency from the following distribution, it being given that the median of distribution is 24. Sol: Class Frequency (fi) Cumulative Frequency (cf) 0 10 5 5 10 20 25 30 20 30 x x + 30 30 40 18 x + 48 40 50 7 x + 55 Median is 24 which lies in 20 30 Median class = 20 30 Let the unknown frequency be x. Class 0 10 10 20 20 30 30 40 40 - 50 Frequency 5 25 ? 18 7Read More

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