Two Shaft Lines: Combined Standard Deviation, Consistency and Recalibration
Two production lines machine the same shaft, with a target diameter of 25.00 mm and a customer tolerance of 25.00 ± 0.20 mm. A random sample of 100 shafts from Line A gave a mean of 25.02 mm and a standard deviation of 0.08 mm. A random sample of 60 shafts from Line B gave a mean of 24.98 mm and a standard deviation of 0.12 mm. The quality manager must decide which line should run the large export order, which needs the greatest consistency, and whether Line A should first be recalibrated to bring its mean exactly to 25.00 mm. Recalibration takes two days and stops Line A completely; the export order is due in three weeks. The plant accountant compares the two standard deviations alone and concludes that Line A is better. The customer has also asked for a single combined figure for the mean and standard deviation of all 160 sampled shafts, to be quoted in the inspection report.
1. Compute the combined mean and combined standard deviation of the 160 shafts and comment on the result.
Read the worked answer
Combined mean: (100 × 25.02 + 60 × 24.98) ÷ 160 = (2,502 + 1,498.8) ÷ 160 = 4,000.8 ÷ 160 = 25.005 mm.
Combined variance uses within-group and between-group parts. Deviations of group means from the grand mean: dA = 25.02 − 25.005 = 0.015; dB = 24.98 − 25.005 = −0.025.
Within-group sum of squares: 100(0.08²) + 60(0.12²) = 100(0.0064) + 60(0.0144) = 0.64 + 0.864 = 1.504. Between-group sum of squares: 100(0.015²) + 60(0.025²) = 100(0.000225) + 60(0.000625) = 0.0225 + 0.0375 = 0.060. Total = 1.564; variance = 1.564 ÷ 160 = 0.009775; SD = 0.0989 mm.
The combined SD (0.099) is larger than either line's SD (0.08 and 0.12 is larger only for B). The combined figure exceeds A's SD because the two means differ, adding between-group variation. Quoting only the combined SD would conceal that the lines are individually more precise than the pooled figure suggests.
2. Using coefficients of variation and the tolerance limits, assess which line is more consistent and which is more likely to meet the ±0.20 mm tolerance.
Read the worked answer
CV_A = 0.08 ÷ 25.02 × 100 = 0.320%. CV_B = 0.12 ÷ 24.98 × 100 = 0.480%. On relative variation, A is clearly the more consistent line, supporting the accountant's conclusion but for a better-stated reason.
Tolerance is 24.80 to 25.20 mm. Using Chebyshev's inequality, 1 − 1/k², with k measured to the nearer limit:
- Line A: distance to upper limit = 25.20 − 25.02 = 0.18; k = 0.18 ÷ 0.08 = 2.25; guaranteed within tolerance ≥ 1 − 1/5.0625 = 80.2%.
- Line B: distance to lower limit = 24.98 − 24.80 = 0.18; k = 0.18 ÷ 0.12 = 1.5; guaranteed ≥ 1 − 1/2.25 = 55.6%.
A is superior on both counts, but its off-centre mean wastes margin on the upper side: it has 0.22 mm of room below but only 0.18 mm above. B is also off-centre, in the opposite direction. Under an approximate normal model, the gap would be even wider in A's favour.
3. Should Line A be recalibrated before running the export order? Give a recommendation that weighs statistical and operational factors.
Read the worked answer
Recalibrating A shifts its mean from 25.02 to 25.00 while leaving its SD at about 0.08 (assuming the adjustment is a pure offset). The nearer tolerance distance becomes 0.20 mm, k = 2.5, so Chebyshev guarantees at least 84% within tolerance, up from 80.2%; under approximate normality the gain is larger, roughly from 97.6% to 98.8%.
Operational cost: two days' stoppage out of a three-week (about 15 working day) window, roughly 13% of the available time on that line, plus scheduling risk. B cannot fully substitute, being both less consistent and off-centre in the other direction.
Recommendation: run the export order on Line A without delay if the order can be completed inside the window, because A's CV of 0.320% is far better than B's 0.480% and its current expected defect rate is already low. Schedule recalibration immediately after the shipment, or during a planned gap if one exists. Recalibrating first is justified only if the customer's sampling plan is strict enough that a shift of 0.02 mm materially changes acceptance, and if spare capacity exists elsewhere to protect the delivery date.