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