1. Problem 27: Given mode = 55, find $x$ in frequency distribution: | Class | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 | | Frequency | 10 | 7 | x | 15 | 10 | 12 | Mode formula for grouped data: $$\text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$$ Where: - $L$ = lower class boundary of modal class - $f_1$ = frequency of modal class - $f_0$ = frequency of class before modal class - $f_2$ = frequency of class after modal class - $h$ = class width Since mode is 55, modal class is 45-60 because 55 lies in this range. Given: $L = 45$ $f_1 = 15$ $f_0 = x$ (frequency of 30-45 class) $f_2 = 10$ (frequency of 60-75 class) $h = 15$ Plugging values: $$55 = 45 + \left( \frac{15 - x}{2 imes15 - x - 10} \right) \times 15$$ Simplify: $$10 = 15 \times \frac{15 - x}{30 - x - 10} = 15 \times \frac{15 - x}{20 - x}$$ Divide both sides by 15: $$\frac{10}{15} = \frac{15 - x}{20 - x}$$ Simplify left side: $$\frac{2}{3} = \frac{15 - x}{20 - x}$$ Cross multiply: $$2(20 - x) = 3(15 - x)$$ $$40 - 2x = 45 - 3x$$ Bring variables to one side: $$-2x + 3x = 45 - 40$$ $$x = 5$$ Answer for problem 27: $x = 5$. --- 2. Problem 28: Find mode of: | Class | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | | Frequency | 45 | 30 | 75 | 20 | 35 | 15 | Identify modal class (highest frequency): 75 in 20-25. Using mode formula again: $L = 20$, $f_1=75$, $f_0=30$, $f_2=20$, $h=5$ $$\text{Mode} = 20 + \left( \frac{75 - 30}{2\times75 - 30 - 20} \right) \times 5 = 20 + \left( \frac{45}{150 - 50} \right) \times 5 = 20 + \left( \frac{45}{100} \right) \times 5 = 20 + 2.25 = 22.25$$ Answer for problem 28: Mode = 22.25. --- 3. Problem 29: Find mode of: | Marks | 0-10 | 10-20 | 20-0 | 30-40 | 40-50 | 50-60 | | Number of Students | 4 | 6 | 7 | 12 | 5 | 6 | Note: The third class is likely 20-30 instead of 20-0 (typo). Modal class frequency: 12 at 30-40 class. Parameters: $L=30$, $f_1=12$, $f_0=7$, $f_2=5$, $h=10$ Mode: $$30 + \left( \frac{12 - 7}{2\times12 - 7 - 5} \right) \times 10 = 30 + \left( \frac{5}{24 - 12} \right) \times 10 = 30 + \left( \frac{5}{12} \right) \times 10 = 30 + 4.17 = 34.17$$ Answer for problem 29: Mode = 34.17. --- 4. Problem 30: Find mode of: | Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 | 120-140 | | Frequency | 6 | 8 | 10 | 12 | 6 | 5 | 3 | Modal class frequency: 12 at 60-80. Parameters: $L=60$, $f_1=12$, $f_0=10$, $f_2=6$, $h=20$ Mode: $$60 + \left( \frac{12 - 10}{2\times12 - 10 - 6} \right) \times 20 = 60 + \left( \frac{2}{24 - 16} \right) \times 20 = 60 + \left( \frac{2}{8} \right) \times 20 = 60 + 5 = 65$$ Answer for problem 30: Mode = 65.