Paper II — Q1
Explain Single Sampling Plan with the help of an example. Also, write the importance of an Operating Characteristics Curve in a…
Explain Single Sampling Plan with the help of an example. Also, write the importance of an Operating Characteristics Curve in a sampling plan. 10 marks
Solve the above assignment problem.
Depot I II III IV V A 160 130 175 190 200 Town B 135 120 130 160 175 C 140 110 145 170 185 D 50 50 80 80 110 E 55 35 80 80 105 10 marks
Use algebraic method to solve the above game.
Player B B₁ B₂ B₃ B₄ A₁ 0·25 0·20 0·14 0·30 Player A A₂ 0·27 0·16 0·12 0·14 A₃ 0·35 0·08 0·15 0·19 A₄ −0·02 0·08 0·13 0·00 10 marks
Consider the Markov Chain with transition probability matrix:
0 1 2 0 (0 1 0) 1 (½ 0 ½) 2 (0 1 0)
Show that the states are periodic and persistent non-null. 10 marks
State the importance of the hazard function. If the hazard rate of a component is given by:
h(t) = { 0.015, t ≤ 200 { 0.025, t > 200
then find an expression for the reliability function of the component. 10 marks
हिंदी में प्रश्न पढ़ें
एक उदाहरण की सहायता से, एकल प्रतिचयन आयोजना की व्याख्या कीजिए। एक प्रतिचयन आयोजना में संकारक अभिलक्षण वक्र के महत्व को भी लिखिए। (10 अंक)
निम्नलिखित नियतन समस्या को हल कीजिए :
डिपो I II III IV V A 160 130 175 190 200 शहर B 135 120 130 160 175 C 140 110 145 170 185 D 50 50 80 80 110 E 55 35 80 80 105 (10 अंक)
बीजीय विधि का उपयोग करके निम्नलिखित खेल को हल कीजिए :
खिलाड़ी B B₁ B₂ B₃ B₄ A₁ 0·25 0·20 0·14 0·30 खिलाड़ी A A₂ 0·27 0·16 0·12 0·14 A₃ 0·35 0·08 0·15 0·19 A₄ −0·02 0·08 0·13 0·00 (10 अंक)
संक्रमण प्रायिकता आव्यूह
0 1 2 0 (0 1 0) 1 (½ 0 ½) 2 (0 1 0)
के साथ एक मार्कोव श्रृंखला पर विचार कीजिए। दर्शाइए कि अवस्थाएँ आवर्ती और सततावृत अनिराकरणीय हैं। (10 अंक)
संकटप्रस्तता फलन के महत्व को बताइए। यदि किसी घटक की संकटप्रस्तता दर इस प्रकार दी गई है :
h(t) = { 0.015, t ≤ 200 { 0.025, t > 200
तो घटक के विश्वसनीयता फलन के एक व्यंजक को प्राप्त कीजिए। (10 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(b) Depot: I, II, III, IV, V Town A: 160, 130, 175, 190, 200 Town B: 135, 120, 130, 160, 175 Town C: 140, 110, 145, 170, 185 Town D: 50, 50, 80, 80, 110 Town E: 55, 35, 80, 80, 105
(c) Player B: B1, B2, B3, B4 Player A A1: 0.25, 0.20, 0.14, 0.30 Player A A2: 0.27, 0.16, 0.12, 0.14 Player A A3: 0.35, 0.08, 0.15, 0.19 Player A A4: -0.02, 0.08, 0.13, 0.00
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) Single sampling plan and the OC curve
A single sampling plan (N, n, c) draws one random sample of n items from a lot, counts the defectives d, accepts the lot if d ≤ c and rejects it otherwise. Example: n = 50, c = 1 for a large lot. The probability of acceptance for lot fraction defective p is Pa(p) = (1 − p)⁵⁰ + 50p(1 − p)⁴⁹, so Pa(0.01) ≈ 0.91 and Pa(0.05) ≈ 0.28.
The Operating Characteristic (OC) curve plots Pa(p) against p. Its importance: it shows how well the plan discriminates good lots from bad; it gives the producer's risk α = 1 − Pa(AQL) and the consumer's risk β = Pa(LTPD); it allows comparison of plans and helps choose n and c to balance inspection cost against protection.
(b) Assignment problem (minimisation)
Row reduction (subtracting 130, 120, 110, 50, 35) and then column reduction (subtracting 0, 0, 10, 30, 55) give zeros in A: II; B: II, III, V; C: II; D: I, II, IV; E: II. Three lines (rows B and D, column II) cover all zeros, which is fewer than 5, so take the smallest uncovered element 15, subtract it from uncovered cells and add it at the intersections of lines. The zero positions then give the unique complete assignment: A → V, B → III, C → II, D → I, E → IV. Minimum cost = 200 + 130 + 110 + 50 + 80 = 570.
(c) Game by the algebraic method
The matrix is the payoff to A. There is no saddle point (maximin 0.14, minimax 0.15). Using dominance: A₃ dominates A₄; B₂ dominates B₁ and B₃ dominates B₄; then A₁ dominates A₂. The reduced game is A₁: (0.20, 0.14), A₃: (0.08, 0.15) against B₂, B₃.
For A playing A₁ with probability p, equalising the payoffs against B₂ and B₃: 0.20p + 0.08(1 − p) = 0.14p + 0.15(1 − p), so p = 7/13 and 1 − p = 6/13. For B playing B₂ with probability q: 0.20q + 0.14(1 − q) = 0.08q + 0.15(1 − q), so q = 1/13 and 1 − q = 12/13. Value v = 0.08 + 0.12 × 7/13 = 47/325 ≈ 0.1446. Optimal strategies: A (7/13, 0, 6/13, 0); B (0, 1/13, 12/13, 0).
(d) Markov chain: periodic and persistent non-null
From 0 and 2 the chain moves only to 1, and from 1 only to 0 or 2. Hence a return to any state occurs only in an even number of steps, and two-step returns have positive probability (0→1→0, 1→0→1, 2→1→2), so every state has period 2. All states communicate (0 ↔ 1 ↔ 2), so the chain is irreducible, and a finite irreducible chain is persistent. Solving πP = π with Σπ = 1 gives π = (1/4, 1/2, 1/4). A positive stationary distribution means the mean return times 1/πᵢ are finite, so all states are persistent non-null.
(e) Hazard function and reliability
Importance: the hazard h(t) = f(t)/R(t) is the instantaneous failure rate given survival to t. It shows whether failure risk is increasing, constant or decreasing (wear-out, random failure, early failure), and determines R(t) = exp(−∫₀ᵗ h(u)du); it is used in maintenance, warranty and survival analysis.
For t ≤ 200: R(t) = e^(−0.015t). For t > 200: ∫₀ᵗ h = 0.015(200) + 0.025(t − 200) = 0.025t − 2, so R(t) = e^(2 − 0.025t). R(t) = e^(−0.015t) for 0 ≤ t ≤ 200 and R(t) = e^(2 − 0.025t) for t > 200 (continuous at t = 200, where R = e⁻³).
How this answer will be evaluated
What earns full marks: Rigorous derivations, correct algebraic solutions, and clear interpretation of statistical concepts.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (1(a)) Explain a single sampling plan with an example and state the importance of the OC curve, 10 marks 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Definition of the plan with N, n, c and the accept/reject rule
- A worked example with Pa(p)
- Importance of the OC curve: discrimination, producer's and consumer's risks, comparing plans
Loses marks
- Confusing single and double plans
- Wrong acceptance rule
Earns more
- Correct Pa values such as 0.91 and 0.28
- Use of AQL and LTPD
Extra mark
- Distinction between binomial and hypergeometric calculation
- (1(b)) Solve the 5×5 assignment problem, 10 marks 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Row and column reduction
- Covering lines and adjustment by 15
- Assignment A–V, B–III, C–II, D–I, E–IV with cost 570
Loses marks
- Treating it as maximisation
- Arithmetic errors in the total
Earns more
- Statement that cost is to be minimised
- Row minima 130, 120, 110, 50, 35 and column minima 0, 0, 10, 30, 55
Extra mark
- Check that the assignment is unique
- (1(c)) Solve the 4×4 game by the algebraic method, 10 marks 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- No saddle point and reduction by dominance to a 2×2 game
- Equalising equations giving p = 7/13 and q = 1/13
- Value 47/325 and optimal strategies
Loses marks
- Wrong dominance direction for the minimising player
- Wrong value
Earns more
- Correct dominance order: A3 over A4, B2 over B1, B3 over B4, A1 over A2
- Verification of the value
Extra mark
- Cross-check by linear programming
- (1(d)) Show that the states of the Markov chain are periodic and persistent non-null, 10 marks 10 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Period 2 argument from bipartite structure
- Irreducibility and finite chain implying persistence
- Stationary distribution (1/4, 1/2, 1/4) implying non-null
Loses marks
- Claiming aperiodicity
- Calling the chain transient
Earns more
- Two-step return paths shown
- Mean return times 1/πᵢ
Extra mark
- Remark that no limiting distribution exists because of periodicity
- (1(e)) State the importance of the hazard function and find the reliability function for the given piecewise hazard, 10 marks 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Hazard h(t) = f(t)/R(t) as instantaneous failure rate and its role in describing ageing
- R(t) = exp(−∫h) with the piecewise integration
- R(t) = e^(−0.015t) for t ≤ 200 and e^(2 − 0.025t) for t > 200
Loses marks
- Using a single exponential for all t
- Not adding the first-interval contribution
Earns more
- Continuity at t = 200
- Applications in maintenance and survival analysis
Extra mark
- Classification as increasing, constant or decreasing failure rate
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
Practice this exact question
Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
Evaluate my answer →More from Statistics 2021 Paper II
- Q1 Explain Single Sampling Plan with the help of an example. Also, write the importance of a…
- Q2 A manufacturer finds that on the average, a television set is used 1.8 hours per day. A o…
- Q3 (a) Explain M|G|1 queuing system. Obtain Pollaczek-kinchine formula. (15 marks) (b) Use M…
- Q4 (a) Solve the following linear programming problem: Maximize z = 3x₁ + 5x₂ subject to the…