1. **Problem Statement:** We have lengths of rods from two machines, A and B. We need to calculate the interquartile range (IQR) and standard deviation for each machine, determine which machine is more consistent, and check if either machine fails the industry standard of maximum deviation ±0.5 cm from the mean. 2. **Formulas and Important Rules:** - The interquartile range (IQR) is calculated as $$IQR = Q_3 - Q_1$$ where $Q_1$ is the first quartile (25th percentile) and $Q_3$ is the third quartile (75th percentile). - The standard deviation (SD) measures spread around the mean and is calculated as $$SD = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ where $\bar{x}$ is the sample mean. - Consistency means less variability, so the machine with smaller SD and IQR is more consistent. - The industry standard requires all rods to be within $\pm 0.5$ cm of the mean. 3. **Calculate for Machine A:** - Data sorted: 49.7, 49.8, 49.9, 49.9, 50.0, 50.1, 50.1, 50.2, 50.2, 50.3 - Median (middle value) between 5th and 6th values: $$\frac{50.0 + 50.1}{2} = 50.05$$ - First quartile $Q_1$ is median of first 5 values: 49.7, 49.8, 49.9, 49.9, 50.0 Median of these is 49.9 - Third quartile $Q_3$ is median of last 5 values: 50.1, 50.1, 50.2, 50.2, 50.3 Median of these is 50.2 - IQR for Machine A: $$IQR_A = 50.2 - 49.9 = 0.3$$ - Mean $\bar{x}_A = \frac{49.8 + 50.1 + 50.2 + 49.9 + 50.0 + 50.3 + 50.1 + 49.7 + 49.9 + 50.2}{10} = 50.02$$ - Calculate variance: $$\sum (x_i - \bar{x}_A)^2 = (49.8-50.02)^2 + (50.1-50.02)^2 + ... + (50.2-50.02)^2 = 0.268$$ - Standard deviation: $$SD_A = \sqrt{\frac{0.268}{10-1}} = \sqrt{0.02978} = 0.1725$$ 4. **Calculate for Machine B:** - Data sorted: 49.1, 49.3, 49.5, 49.6, 50.0, 50.2, 50.3, 50.5, 50.8, 50.9 - Median between 5th and 6th values: $$\frac{50.0 + 50.2}{2} = 50.1$$ - First quartile $Q_1$ median of first 5 values: 49.1, 49.3, 49.5, 49.6, 50.0 Median is 49.5 - Third quartile $Q_3$ median of last 5 values: 50.2, 50.3, 50.5, 50.8, 50.9 Median is 50.5 - IQR for Machine B: $$IQR_B = 50.5 - 49.5 = 1.0$$ - Mean $\bar{x}_B = \frac{49.5 + 50.5 + 50.8 + 49.3 + 50.2 + 49.6 + 50.9 + 49.1 + 50.3 + 50.0}{10} = 50.12$$ - Calculate variance: $$\sum (x_i - \bar{x}_B)^2 = (49.5-50.12)^2 + (50.5-50.12)^2 + ... + (50.0-50.12)^2 = 3.156$$ - Standard deviation: $$SD_B = \sqrt{\frac{3.156}{9}} = \sqrt{0.3518} = 0.5931$$ 5. **Compare Consistency:** - Machine A has $IQR = 0.3$ and $SD = 0.1725$ - Machine B has $IQR = 1.0$ and $SD = 0.5931$ - Since Machine A has smaller IQR and SD, it produces more consistent rods. 6. **Check Industry Standard ±0.5 cm:** - For Machine A: Mean = 50.02, range allowed = [49.52, 50.52] - Minimum rod length = 49.7, maximum = 50.3, all within range. - For Machine B: Mean = 50.12, range allowed = [49.62, 50.62] - Minimum rod length = 49.1 (below 49.62), maximum = 50.9 (above 50.62) - Machine B fails the industry standard. **Final answers:** - Machine A: $IQR = 0.3$, $SD = 0.1725$ - Machine B: $IQR = 1.0$, $SD = 0.5931$ - More consistent machine: Machine A - Machine B fails the industry standard requirement.