This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to the questions based on the provided data: The given frequency distribution table is: | Marks Obtained | No. of Students (f) | |---|---| | 0-20 | 2 | | 20-40 | x | | 40-60 | 8 | | 60-80 | 5 | | 80-100 | 1 | The first quartile (Q_1) of the given data is 35. First, we construct the cumulative frequency (cf) table: | Marks Obtained | No. of Students (f) | Cumulative Frequency (cf) | |---|---|---| | 0-20 | 2 | 2 | | 20-40 | x | 2 + x | | 40-60 | 8 | 2 + x + 8 = 10 + x | | 60-80 | 5 | 10 + x + 5 = 15 + x | | 80-100 | 1 | 15 + x + 1 = 16 + x | The total number of students is N = 16 + x. a) Find the first quartile class. Step 1: Identify the class interval where the first quartile value (35) falls. Since 20 < 35 < 40, the first quartile class is 20-40. The first quartile class is 20-40. b) Find the value of x. Step 1: Use the formula for the first quartile (Q_1): Q_1 = L + (N)/(4) - cff × h From part (a), the first quartile class is 20-40. L = 20 (lower boundary of the quartile class) f = x (frequency of the quartile class) cf = 2 (cumulative frequency of the class preceding the quartile class) h = 40 - 20 = 20 (class width) Q_1 = 35 (given) N = 16 + x (total number of students) Step 2: Substitute the values into the formula: 35 = 20 + (16+x)/(4) - 2x × 20 Step 3: Solve for x: 35 - 20 = (16+x)/(4) - 2x × 20 15 = (16+x-8)/(4)x × 20 15 = (8+x)/(4)x × 20 15 = (8+x)/(4x) × 20 15 = (8+x)/(x) × 5 Divide both sides by 5: 3 = (8+x)/(x) 3x = 8+x 3x - x = 8 2x = 8 x = (8)/(2) x = 4 The value of x is 4. c) If modal class is 40-60 then what is the value of mode? Step 1: Update the frequency table with x=4: | Marks Obtained | No. of Students (f) | |---|---| | 0-20 | 2 | | 20-40 | 4 | | 40-60 | 8 | | 60-80 | 5 | | 80-100 | 1 | Step 2: Use the formula for the mode for grouped data: Mode = L + (f_m - f_1)/(2f_m - f_1 - f_2) × h The modal class is given as 40-60. L = 40 (lower boundary of the modal class) f_m = 8 (frequency of the modal class) f_1 = 4 (frequency of the class preceding the modal class, 20-40) f_2 = 5 (frequency of the class succeeding the modal class, 60-80) h = 60 - 40 = 20 (class width) Step 3: Substitute the values into the formula: Mode = 40 + (8 - 4)/(2(8) - 4 - 5) × 20 Mode = 40 + (4)/(16 - 4 - 5) × 20 Mode = 40 + (4)/(12 - 5) × 20 Mode = 40 + (4)/(7) × 20 Mode = 40 + (80)/(7) Mode = 40 + 11.42857... Mode ≈ 51.43 The value of the mode is 51.43. d) Find the ratio of students who are above and below the first quartile class. Step 1: Identify the first quartile class. From part (a), the first quartile class is 20-40. Step 2: Calculate the number of students below the first quartile class. Students below the 20-40 class are those in the 0-20 class. Number of students below the first quartile class = 2. Step 3: Calculate the number of students above the first quartile class. Students above the 20-40 class are those in the 40-60, 60-80, and 80-100 classes. Number of students above the first quartile class = 8 + 5 + 1 = 14. Step 4: Find the ratio of students above to students below the first quartile class. Ratio = (Number of students above) : (Number of students below) Ratio = 14 : 2 Step 5: Simplify the ratio. Divide both sides by 2: Ratio = 7 : 1 The ratio of students who are above and below the first quartile class is 7:1.