(a) What do you understand by control chart for fraction defective? Explain its construction. Give the theoretical distribution…

(a) Control chart for fraction defective (p-chart)

A control chart for fraction defective plots, sample by sample, the proportion of defective items pᵢ = dᵢ/nᵢ, where dᵢ is the number defective in a sample of size nᵢ. It is an attribute chart used when each item is classified only as defective or non-defective. Construction: Step 1: Take m samples (preferably 20 to 25) and record dᵢ. Step 2: Estimate the centre line p̄ = Σdᵢ/Σnᵢ. Step 3: Compute the 3-sigma limits UCL = p̄ + 3√[p̄(1 − p̄)/n] and LCL = p̄ − 3√[p̄(1 − p̄)/n], taking LCL = 0 if it is negative. If n varies, compute the limits for each sample with its own nᵢ. Step 4: Plot the pᵢ against sample number with the centre line and limits. Step 5: A point outside the limits, or a run or trend, signals an assignable cause; it is investigated and removed, and the limits are recomputed from the remaining samples. Theoretical distribution: each item is a Bernoulli trial with defective probability p, so dᵢ ~ Binomial(n, p), E(pᵢ) = p and Var(pᵢ) = p(1 − p)/n. For large n the binomial is approximated by a normal distribution, so 3-sigma limits include about 99·73% of in-control points.

(b) p-chart for the 24 daily samples

Step 1: n = 50 each day and total defectives = 49 + 39 = 88, so p̄ = 88/1200 = 0·0733. Step 2: Standard error = √(0·0733 × 0·9267/50) = 0·0369. Step 3: UCL = 0·0733 + 3(0·0369) = 0·1839, and LCL = 0·0733 − 0·1106 < 0, so LCL = 0. Step 4: The daily proportions all lie within 0 and 0·1839 except day 11: p = 12/50 = 0·24 > UCL. The process is not in control; day 11 has an assignable cause. Step 5: Drop day 11. The remaining 23 samples have 76 defectives, so p̄ = 76/1150 = 0·0661 and SE = √(0·0661 × 0·9339/50) = 0·0351. Step 6: Revised limits are UCL = 0·0661 + 0·1054 = 0·1715, CL = 0·0661 and LCL = 0. All 23 remaining points fall within them, with no run or trend. Limits suggested for subsequent use: CL = 0·0661, UCL = 0·1715, LCL = 0 (for samples of 50).

(c) Optimum replacement policy for 1000 bulbs

The probabilities 0·10, 0·25, 0·50, 0·70, 1·00 are cumulative (they reach 1 at week 5), so the failure probabilities in weeks 1 to 5 are fᵢ = 0·10, 0·15, 0·25, 0·20, 0·30, with mean life 3·45 weeks. Step 1: Expected failures in week n including failures of replaced bulbs: Nₙ = 1000 fₙ + Σ N(n−i) fᵢ, giving N₁ = 100, N₂ = 160, N₃ = 281, N₄ = 277·1. Step 2: Policy: replace all bulbs at the end of week T at $1 each, and replace individually at $3 each the bulbs failing in weeks 1 to T − 1 (failures in week T are covered by the group replacement). Step 3: Average weekly cost = [1000 + 3 × (N₁ + … + N(T−1))]/T. T = 1: 1000/1 = $1000·00. T = 2: (1000 + 3 × 100)/2 = $650·00. T = 3: (1000 + 3 × 260)/3 = $593·33. T = 4: (1000 + 3 × 541)/4 = $655·75. T = 5: (1000 + 3 × 818·1)/5 = $690·86. Step 4: The cost is lowest at T = 3, and it is below the individual-replacement-only cost of 3 × 1000/3·45 = $869·57 per week. The optimum policy is to replace all 1000 bulbs together every 3 weeks and replace failed bulbs individually in between, at an average cost of about $593 per week.