1. **Problem Statement:** Calculate the standard deviation of the truck capacities given the frequency distribution: | Weight (000kg) | Number of Trucks | |------------------------|------------------| | 10 and less than 20 | 4 | | 20 and less than 30 | 6 | | 30 and less than 40 | 7 | | 40 and less than 50 | 3 | 2. **Step 1: Find the midpoints of each class interval.** - For 10 to less than 20: midpoint $= \frac{10 + 20}{2} = 15$ - For 20 to less than 30: midpoint $= \frac{20 + 30}{2} = 25$ - For 30 to less than 40: midpoint $= \frac{30 + 40}{2} = 35$ - For 40 to less than 50: midpoint $= \frac{40 + 50}{2} = 45$ 3. **Step 2: Calculate the mean ($\bar{x}$).** - Total trucks $n = 4 + 6 + 7 + 3 = 20$ - Sum of midpoints times frequency: $$\sum f x = 4 \times 15 + 6 \times 25 + 7 \times 35 + 3 \times 45 = 60 + 150 + 245 + 135 = 590$$ - Mean: $$\bar{x} = \frac{\sum f x}{n} = \frac{590}{20} = 29.5$$ 4. **Step 3: Calculate the variance ($\sigma^2$).** - Use formula: $$\sigma^2 = \frac{\sum f (x - \bar{x})^2}{n}$$ - Calculate each squared deviation times frequency: - For 15: $(15 - 29.5)^2 = 210.25$, contribution $= 4 \times 210.25 = 841$ - For 25: $(25 - 29.5)^2 = 20.25$, contribution $= 6 \times 20.25 = 121.5$ - For 35: $(35 - 29.5)^2 = 30.25$, contribution $= 7 \times 30.25 = 211.75$ - For 45: $(45 - 29.5)^2 = 240.25$, contribution $= 3 \times 240.25 = 720.75$ - Sum of contributions: $$841 + 121.5 + 211.75 + 720.75 = 1895$$ - Variance: $$\sigma^2 = \frac{1895}{20} = 94.75$$ 5. **Step 4: Calculate the standard deviation ($\sigma$).** - Standard deviation is the square root of variance: $$\sigma = \sqrt{94.75} \approx 9.74$$ **Final answer:** The standard deviation of the truck capacities is approximately **9.74 (000 kg)**.