(a) Distinguish between process control and product control. Explain the various sources of variation encountered in a process…

(a) Process control and product control; sources of variation

Process control checks quality while production is going on, mainly with Shewhart control charts (x̄, R, p, c); it is preventive and keeps the process in statistical control. Product control inspects finished or intermediate lots by acceptance sampling (Dodge–Romig plans) and accepts or rejects them; it only sorts the product and does not improve the process.

Sources of variation:

• Chance (common) causes: many small, random, inherent causes such as vibration or voltage and material fluctuations. They give the natural spread, cannot be economically removed, and are reduced only by redesigning the process.
• Assignable (special) causes: identifiable causes in men (untrained operator), machines (worn tool), materials (bad batch), methods, measurement (uncalibrated gauge) and environment.

Elimination: a point outside the 3σ limits, a run or a trend signals an assignable cause. Trace it with a cause-and-effect diagram, correct it at source (retrain, repair, change supplier, recalibrate) and recompute the limits.

(b) Simulation of the project profit

Cumulative probabilities give two-digit intervals. Price: 00–19 → ₹3, 20–69 → ₹4, 70–99 → ₹5. Variable cost: 00–29 → ₹1, 30–89 → ₹2, 90–99 → ₹3. Volume: 00–29 → 2000, 30–59 → 3000, 60–99 → 5000. Each trial uses three consecutive numbers (price, cost, volume); profit = (price − cost) × volume − 4000.

1. 81, 32, 60: (5−2)×5000−4000 = 11000
2. 04, 46, 31: (3−2)×3000−4000 = −1000
3. 67, 25, 24: (4−1)×2000−4000 = 2000
4. 10, 40, 02: (3−2)×2000−4000 = −2000
5. 39, 68, 08: (4−2)×2000−4000 = 0
6. 59, 66, 90: (4−2)×5000−4000 = 6000
7. 12, 64, 79: (3−2)×5000−4000 = 1000
8. 31, 86, 68: (4−2)×5000−4000 = 6000
9. 82, 89, 25: (5−2)×2000−4000 = 2000
10. 11, 98, 16: (3−3)×2000−4000 = −4000

Total = 21000, so the simulated average profit = 21000/10 = ₹2100 (the exact expected profit is (4.1 − 1.8) × 3500 − 4000 = ₹4050).

(c) Conditional distribution of N(s) given N(t) = n

Let N(t) be a Poisson process with rate λ. Its increments N(s) and N(t) − N(s) are independent, with Poisson(λs) and Poisson(λ(t − s)) laws. For k = 0, 1, …, n, P(N(s) = k | N(t) = n) = P(N(s) = k) P(N(t) − N(s) = n − k) / P(N(t) = n) = [e^(−λs)(λs)^k/k!] [e^(−λ(t−s))(λ(t−s))^(n−k)/(n−k)!] / [e^(−λt)(λt)ⁿ/n!] = C(n, k) (s/t)^k (1 − s/t)^(n−k), and the probability is 0 for k > n.

Comment: the conditional law is Binomial(n, s/t), free of λ, so N(t) carries all the information about λ. Given n events in [0, t], each event falls in [0, s] independently with probability s/t; the event times behave like n independent uniform points on [0, t]. Hence E[N(s) | N(t) = n] = ns/t and Var = n(s/t)(1 − s/t).

(d) Theory of games

Assumptions: two rational players with conflicting interests; each has a finite number of strategies; the payoffs are known to both; choices are simultaneous, without collusion; in a zero-sum game one player's gain is the other's loss.

Maximin (maximising player A): take each row minimum and choose the largest, maximin = maxᵢ minⱼ aᵢⱼ, the payoff A can guarantee. Minimax (minimising player B): take each column maximum and choose the smallest, minimax = minⱼ maxᵢ aᵢⱼ, the loss B can limit. Always maximin ≤ minimax; equality means a saddle point and pure strategies; otherwise mixed strategies.

Algebraic method (2×2, no saddle point, entries a₁₁, a₁₂, a₂₁, a₂₂): A plays row 1 with probability p so that the expected payoff is equal against both columns, p·a₁₁+(1−p)·a₂₁ = p·a₁₂+(1−p)·a₂₂. With D = a₁₁+a₂₂−a₁₂−a₂₁: p = (a₂₂−a₂₁)/D, B plays column 1 with q = (a₂₂−a₁₂)/D, and V = (a₁₁a₂₂−a₁₂a₂₁)/D. For larger games, reduce by dominance, then solve Σᵢ aᵢⱼpᵢ = V, Σpᵢ = 1.

(e) Censoring and type-II estimation for the exponential law

Importance: waiting for every item to fail is slow and costly, especially for highly reliable items; censoring stops the test early, saves time and cost, copes with items lost or withdrawn, and still uses the survival time of unfailed items. Type-I censoring stops at a fixed time (number of failures random); type-II stops at the r-th failure (time random, r fixed).

Estimation: n items are tested with f(x) = θ⁻¹e^(−x/θ), stopped at the r-th failure x(1) ≤ … ≤ x(r). The likelihood is L = [n!/(n − r)!] θ^(−r) exp(−T/θ), T = Σᵢ₌₁ʳ x(i) + (n − r)x(r) (total time on test). d ln L/dθ = −r/θ + T/θ² = 0 gives θ̂ = T/r. 2T/θ ~ χ²(2r), so E(θ̂) = θ and Var(θ̂) = θ²/r, which equals the Cramér–Rao bound (information r/θ²); θ̂ is the minimum-variance unbiased estimator. A 100(1 − α)% confidence interval is 2T/χ²_α/2(2r) < θ < 2T/χ²₁−α/2(2r), using upper-tail points.