Calculate the sum of frequencies (sum f) and the sum of (frequency x mark) (sum fx).

Step 1: Calculate the sum of frequencies ( f) and the sum of (frequency × mark) ( fx). The given table is: | Marks (x) | 5 | 6 | 7 | 8 | 9 | 10 | |---|---|---|---|---|---|---| | Frequency (f) | 3x | x+1 | 4 | 7 | 8 | x | Sum of frequencies: f = 3x + (x+1) + 4 + 7 + 8 + x f = 3x + x + x + 1 + 4 + 7 + 8 f = 5x + 20 Sum of (frequency × mark): fx = (5 × 3x) + (6 × (x+1)) + (7 × 4) + (8 × 7) + (9 × 8) + (10 × x) fx = 15x + 6x + 6 + 28 + 56 + 72 + 10x fx = 15x + 6x + 10x + 6 + 28 + 56 + 72 fx = 31x + 162 Step 2: Use the given mean to find the value of x. The mean of the distribution is given as 7(7)/(15). Convert this to an improper fraction: 7(7)/(15) = (7 × 15 + 7)/(15) = (105 + 7)/(15) = (112)/(15) The formula for the mean is x = ( fx)/( f). (31x + 162)/(5x + 20) = (112)/(15) Cross-multiply: 15(31x + 162) = 112(5x + 20) 465x + 2430 = 560x + 2240 Rearrange the terms to solve for x: 2430 - 2240 = 560x - 465x 190 = 95x x = (190)/(95) x = 2 a) The value of x is 2 Step 3: Calculate the total number of students in 3G2. The total number of students is the sum of frequencies, f = 5x + 20. Substitute the value of x=2: f = 5(2) + 20 f = 10 + 20 f = 30 b) The total number of students in 3G2 is 30 Step 4: Calculate the probability that a student chosen at random obtained a mark not more than 6. "Not more than 6" means marks of 5 or 6. Frequency for mark 5 is 3x. Substitute x=2: 3(2) = 6. Frequency for mark 6 is x+1. Substitute x=2: 2+1 = 3. Number of students with marks not more than 6 = 6 + 3 = 9. Total number of students = 30 (from part b). The probability is: P(mark 6) = Number of students with marks 6Total number of students P(mark 6) = (9)/(30) Simplify the fraction: P(mark 6) = (3)/(10) c) The probability that a student chosen at random obtained a mark not more than 6 is (3)/(10) Send me the next one 📸