1. **Problem Statement:** We have a frequency distribution of cylinder diameters and need to find the mean, median, and mode of the diameters. 2. **Given Data:** Length intervals (mm): 10-12, 13-15, 16-18, 19-21, 22-24, 25-27 Frequencies: 2, 7, 13, 15, 5, 3 3. **Step i: Calculate the Mean Diameter** - Use the midpoint of each class interval as the representative diameter. - Midpoints: $\frac{10+12}{2}=11$, $\frac{13+15}{2}=14$, $\frac{16+18}{2}=17$, $\frac{19+21}{2}=20$, $\frac{22+24}{2}=23$, $\frac{25+27}{2}=26$ - Multiply each midpoint by its frequency and sum: $$\sum f x = 2\times11 + 7\times14 + 13\times17 + 15\times20 + 5\times23 + 3\times26$$ $$= 22 + 98 + 221 + 300 + 115 + 78 = 834$$ - Total frequency $N = 2 + 7 + 13 + 15 + 5 + 3 = 45$ - Mean diameter: $$\bar{x} = \frac{\sum f x}{N} = \frac{834}{45} = 18.53$$ mm 4. **Step ii: Calculate the Median Diameter** - Median class is where cumulative frequency reaches $\frac{N}{2} = 22.5$ - Cumulative frequencies: 2, 9, 22, 37, 42, 45 - Median class is 19-21 (since cumulative frequency just exceeds 22.5 here) - Use median formula: $$L = 18.5 \text{ (lower boundary of median class)}$$ $$f_m = 15 \text{ (frequency of median class)}$$ $$F = 22 \text{ (cumulative frequency before median class)}$$ $$w = 3 \text{ (class width)}$$ $$\text{Median} = L + \frac{\frac{N}{2} - F}{f_m} \times w = 18.5 + \frac{22.5 - 22}{15} \times 3 = 18.5 + 0.1 = 18.6$$ mm 5. **Step iii: Calculate the Modal Diameter** - Mode is the midpoint of the modal class (class with highest frequency) - Highest frequency is 15 in class 19-21 - Use mode formula: $$L = 18.5$$ $$f_1 = 15$$ $$f_0 = 13 \text{ (frequency before modal class)}$$ $$f_2 = 5 \text{ (frequency after modal class)}$$ $$w = 3$$ $$\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times w = 18.5 + \frac{15 - 13}{2\times15 - 13 - 5} \times 3 = 18.5 + \frac{2}{12} \times 3 = 18.5 + 0.5 = 19.0$$ mm 6. **Step b: Probability the problem is solved by P, Q, or R** - Probabilities: $$P = \frac{2}{4} = 0.5, \quad Q = \frac{4}{7}, \quad R = \frac{7}{9}$$ - Probability problem not solved by each: $$P' = 1 - 0.5 = 0.5, \quad Q' = 1 - \frac{4}{7} = \frac{3}{7}, \quad R' = 1 - \frac{7}{9} = \frac{2}{9}$$ - Probability problem not solved by all three: $$P' \times Q' \times R' = 0.5 \times \frac{3}{7} \times \frac{2}{9} = \frac{3}{63} = \frac{1}{21}$$ - Probability problem solved by at least one: $$1 - \frac{1}{21} = \frac{20}{21} \approx 0.9524$$ **Final answers:** - Mean diameter = 18.53 mm - Median diameter = 18.6 mm - Modal diameter = 19.0 mm - Probability problem solved = $\frac{20}{21}$ or approximately 0.9524