Paper II — Q1
(a) State the significance of operating characteristic (OC) curves in control chart analysis. Obtain the general expression for…
State the significance of operating characteristic (OC) curves in control chart analysis. Obtain the general expression for the OC function corresponding to the mean (X̄) chart under the assumption of normal distribution for a quality characteristic. Using the expression, find the probability that a shift will be detected from μ₀ to μ₁ = μ₀ + 2σ, when an X̄ chart is used with 3σ limits, where the subgroup size is n = 6. (Standard normal table is provided.) 10 marks
What is meant by rectifying inspection? Explain the measures associated with rectifying inspection and derive the expressions of such measures in the case of a single sampling plan by attributes. 10 marks
The lifetime of a semiconductor laser has a log-normal distribution with parameters μ = 10 hours and σ = 1·5 hours. Find the probability that the lifetime exceeds 10000 hours.
What lifetime is exceeded by 99% of lasers? (Standard normal table is provided.) 5+5=10 marks
A stockist has to supply 400 units of a product every Monday to his customers. He gets the product at ₹ 50 per unit from the manufacturer. The cost of ordering and transportation from the manufacturer is ₹ 75 per order. The cost of carrying inventory is 7·5% per year of the cost of the product. Find (i) the economic lot size, (ii) the total optimal cost (including the capital cost) and (iii) the total weekly profit, if the item is sold for ₹ 55 per unit. 10 marks
On the average, 96 patients per 24-hour day require the service of an emergency clinic. Also, on the average, a patient requires 10 minutes of active attention. Assume that the facility can handle only one emergency at a time. Suppose that it costs the clinic ₹ 1,000 per patient treated to obtain an average serving time of 10 minutes, and that each minute of decrease in this average time would cost the clinic ₹ 100 per patient treated. How much would have to be budgeted by the clinic to decrease the average size of the queue from 1 1/3 patients to 1/2 patient? 10 marks
हिंदी में प्रश्न पढ़ें
नियंत्रण सांचित्र (चार्ट) विश्लेषण में संकारक अभिलक्षण (ओ. सी.) वक्रों के महत्व को बताइए। एक गुणवत्ता विशेषता के लिए प्रसामान्य बंटन की मान्यता के अंतर्गत, माध्य (X̄) चार्ट के तहत, ओ. सी. फलन का सामान्य व्यंजक प्राप्त कीजिए। व्यंजक का उपयोग करके एक शिफ्ट μ₀ से μ₁ = μ₀ + 2σ में खोजे जाने की प्रायिकता निकालिए, जबकि एक X̄ चार्ट का उपयोग 3σ सीमाओं के साथ किया जाता है, जहाँ उपसमूह का आमाप n = 6 है। (मानक प्रसामान्य तालिका प्रदान की गई है।) 10 अंक
सुधारात्मक निरीक्षण का क्या मतलब है? सुधारात्मक निरीक्षण से संबंधित मापों की व्याख्या कीजिए तथा गुणों के लिए एकल प्रतिदर्शन आयोजना के तहत ऐसे मापों के व्यंजक व्युत्पन्न कीजिए। 10 अंक
एक अर्धचालक लेजर के जीवन-काल का बंटन लघुगणकीय प्रसामान्य है, जिसके प्राचल μ = 10 घंटे तथा σ = 1·5 घंटे हैं। जीवन-काल 10000 घंटे से अधिक होने की प्रायिकता ज्ञात कीजिए।
99% लेजरों का जीवन-काल किस जीवन-काल से अधिक है? (मानक प्रसामान्य तालिका प्रदान की गई है।) 5+5=10 अंक
एक शेष व्यापारी को एक उत्पाद की 400 इकाइयाँ प्रत्येक सोमवार को अपने ग्राहकों को भेजनी होती हैं। वह उत्पादक से उत्पाद ₹ 50 प्रति इकाई के हिसाब से प्राप्त करता है। उत्पादक से आदेश तथा परिवहन की कीमत ₹ 75 प्रति ऑर्डर है। मालसूची (इन्वेंट्री) ले जाने की कीमत, उत्पाद की कीमत का 7·5% प्रति वर्ष है। ज्ञात कीजिए (i) मितव्ययी प्रचय परिमाण, (ii) कुल इष्टतम लागत (पूँजीगत लागत सम्मिलित) और (iii) कुल साप्ताहिक लाभ, यदि मद ₹ 55 प्रति इकाई के हिसाब से बेची जाती है। 10 अंक
औसतन 96 मरीजों को 24 घंटे प्रतिदिन आपातकालीन चिकित्सालय की सेवा की आवश्यकता है। औसतन एक मरीज को 10 मिनट के सक्रिय ध्यान की भी आवश्यकता है। मान लीजिए कि इस तरह की सुविधा एक समय में केवल एक आपातकालीन स्थिति को संभाल सकती है। मान लीजिए कि 10 मिनट का औसत सेवा समय प्राप्त करने के लिए इलाज किए गए प्रति रोगी पर चिकित्सालय ₹ 1,000 खर्च करता है, और इस औसत समय में कमी के प्रत्येक मिनट के लिए चिकित्सालय में इलाज किए गए प्रति रोगी पर ₹ 100 खर्च आता है। पंक्ति के औसत आमाप को 1 1/3 रोगियों से 1/2 रोगी तक कम करने के लिए चिकित्सालय द्वारा कितना बजट किया जाना चाहिए? 10 अंक
Model answer
Written 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.
(a) The OC curve of a control chart gives the probability that the chart does not signal a shift, i.e. the probability that a sample statistic falls inside the control limits when the process mean has actually shifted. Thus it measures the sensitivity and discriminating power of the chart: steeper OC curves mean quicker detection of shifts, while flatter curves mean poor detection. It helps choose subgroup size n and control-limit width, and is linked with Type II error β and average run length ARL = 1/(1−β).
Let the quality characteristic be X ~ N(μ, σ²), with independent observations and known σ. For an X̄ chart with centre μ₀ and 3-sigma limits, UCL = μ₀ + 3σ/√n, LCL = μ₀ − 3σ/√n. For a process mean μ, the OC function is L(μ) = P(LCL ≤ X̄ ≤ UCL | μ) = Φ((UCL − μ)/(σ/√n)) − Φ((LCL − μ)/(σ/√n)).
For a shift μ₁ = μ₀ + δσ, L(μ₁) = Φ(k − δ√n) − Φ(−k − δ√n), where k is the control-limit multiplier.
Here k = 3, δ = 2, n = 6. Hence L(μ₁) = Φ(3 − 2√6) − Φ(−3 − 2√6) = Φ(−1.89898) − Φ(−7.89898) ≈ 0.0288 − 0 ≈ 0.0288.
Therefore the probability of detecting the shift is 1 − L(μ₁) = 1 − 0.0288 = 0.9712.
(b) Rectifying inspection means an acceptance-sampling procedure in which rejected lots are completely screened, and all defective units found are removed or replaced by good units. Accepted lots are not fully inspected; only the sample is inspected and defectives in it are replaced. Its purpose is to improve outgoing quality while controlling inspection cost.
For a single sampling plan by attributes with lot size N, sample size n, acceptance number c, and incoming fraction defective p, let P_a be the probability of acceptance. Under the binomial model, P_a(p) = Σ_d=0^c C(n,d) p^d (1−p)^(n−d). If the finite-lot hypergeometric model is required, replace this by P_a(p) = Σ_d=0^c [C(Np,d) C(N−Np,n−d)]/C(N,n).
The main measures are:
- Average outgoing quality (AOQ). If a lot is accepted, the n sampled units are rectified, but the remaining N−n units are not inspected, so expected defectives leaving are p(N−n). If the lot is rejected, all defectives are removed, so outgoing defectives are zero. Hence AOQ(p) = [p P_a(p)(N−n)]/N. For large N, AOQ(p) ≈ p P_a(p).
- Average outgoing quality limit (AOQL). This is the worst outgoing quality after rectifying inspection: AOQL = max_0≤p≤1 AOQ(p).
- Average total inspection (ATI). Accepted lots require n inspections; rejected lots require N inspections. Thus ATI = n P_a + N(1−P_a) = n + (N−n)(1−P_a).
- The average fraction inspected is ATI/N.
Thus rectifying inspection improves outgoing quality at the cost of extra inspection, especially when P_a is small.
(c) Let X be the lifetime. Since X is log-normal, Y = ln X ~ N(μ, σ²) with μ = 10, σ = 1.5.
(i) We need P(X > 10000). Now P(X > 10000) = P(ln X > ln 10000) = P(Z > (ln 10000 − 10)/1.5). Using ln 10000 = 9.21034, z = (9.21034 − 10)/1.5 = −0.52644. Therefore P(X > 10000) = 1 − Φ(−0.52644) = Φ(0.52644) ≈ 0.7007. So the probability is about 0.7007.
(ii) Let x be the lifetime exceeded by 99% of lasers. Then P(X > x) = 0.99, so P(X ≤ x) = 0.01. Thus ln x = μ + σ z₀.₀₁, where z₀.₀₁ = −2.32635. Hence ln x = 10 + 1.5(−2.32635) = 6.51048. So x = e^6.51048 ≈ 672.15 hours. Thus 99% of lasers exceed about 672.15 hours.
(d) Assuming 52 weeks per year, annual demand D = 400 × 52 = 20800 units/year. Unit cost C = ₹50, ordering cost C₀ = ₹75/order, and carrying cost C_h = 7.5% of ₹50 = ₹3.75 per unit per year.
(i) Economic lot size: Q* = √(2 D C₀ / C_h) = √(2 × 20800 × 75 / 3.75) = √832000 = 80√130 ≈ 912.14 units. So the economic lot size is about 912 units.
(ii) Total optimal cost including capital cost: Annual purchase cost = D × C = 20800 × 50 = ₹1040000. At the EOQ, ordering cost + carrying cost is √(2 D C₀ C_h) = √(2 × 20800 × 75 × 3.75) = ₹300√130 ≈ ₹3420.53. Therefore total annual cost including capital cost is ₹1040000 + ₹3420.53 = ₹1043420.53 per year.
(iii) Total weekly profit: Annual revenue = D × selling price = 20800 × 55 = ₹1144000. Annual profit = ₹1144000 − ₹1043420.53 = ₹100579.47. Weekly profit = ₹100579.47 / 52 = ₹1934.22. Equivalently, weekly gross margin = 400(55 − 50) = ₹2000, and weekly inventory cost = ₹3420.53/52 ≈ ₹65.78, giving profit ≈ ₹1934.22. So the total weekly profit is about ₹1934.22.
(e) This is an M/M/1 queue, assuming Poisson arrivals, exponential service, and one server. Arrival rate λ = 96/24 = 4 patients per hour. Initial service time = 10 minutes = 1/6 hour, so initial service rate μ₀ = 6 patients per hour. Initial traffic intensity ρ₀ = λ/μ₀ = 4/6 = 2/3. Average queue length is L_q = ρ²/(1−ρ) = (4/9)/(1/3) = 4/3 patients, which matches the given value.
Let the required traffic intensity be ρ. To make L_q = 1/2, ρ²/(1−ρ) = 1/2. So 2ρ² + ρ − 1 = 0, giving ρ = 1/2 (positive root). Thus required service rate μ = λ/ρ = 4/(1/2) = 8 patients per hour. Required mean service time = 1/8 hour = 7.5 minutes. Decrease in service time = 10 − 7.5 = 2.5 minutes.
Cost per patient at the new service time: ₹1000 + 2.5 × ₹100 = ₹1250 per patient. With 96 patients per day, total required daily budget = 96 × ₹1250 = ₹120000. Additional budget over the original ₹1000 per patient arrangement = 96 × ₹250 = ₹24000 per day.
Therefore, to reduce the average queue from 4/3 patients to 1/2 patient, the clinic must budget an additional ₹24000 per day, i.e. a total of ₹120000 per day.
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.
How this answer will be evaluated
Approach
Framework: UPSC Statistics Paper 2. (a) calculate: given > formula > substitution > result with units > interpretation | (b) derive: given > assumptions > stepwise derivation > result > check | (c(i)) calculate: given > formula > substitution > result with units > interpretation | (c(ii)) calculate: given > formula > substitution > result with units > interpretation | (d) calculate: given > formula > substitution > result with units > interpretation | (e) calculate: given > formula > substitution > result with units > interpretation Full marks: All parts with complete derivations, correct calculations, and clear interpretations
Key points expected
- State significance of OC curves in control chart analysis
- Derive general OC function for X-bar chart (normal assumption)
- Substitute n=6, 3σ limits, and shift μ1 = μ0 + 2σ
- Calculate final probability using standard normal table
- Define rectifying inspection clearly
- Explain measures associated with rectifying inspection
- Derive expressions for single sampling plan by attributes
- Show stepwise derivation of the measures
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Significance of OC curves, general expression for X-bar chart, and detection probability for a 2σ shift. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- State significance of OC curves in control chart analysis
- Derive general OC function for X-bar chart (normal assumption)
- Substitute n=6, 3σ limits, and shift μ1 = μ0 + 2σ
- Calculate final probability using standard normal table
Loses marks
- Computation without stating normal distribution assumption
- Incorrect substitution of subgroup size n
Earns more
- Correct notation for estimator vs parameter
- Clean presentation of the derivation steps
Extra mark
- Sketch of the OC curve shape
- (b) Definition of rectifying inspection and derivation of associated measures for single sampling plan. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Define rectifying inspection clearly
- Explain measures associated with rectifying inspection
- Derive expressions for single sampling plan by attributes
- Show stepwise derivation of the measures
Loses marks
- Derivation without stating assumptions
- Confusing rectifying with non-rectifying inspection
Earns more
- Correct notation for sampling plan parameters
- Clear distinction between producer and consumer risk
Extra mark
- Example of a specific sampling plan
- (c(i)) Probability that semiconductor laser lifetime exceeds 10000 hours. 5 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- State log-normal distribution parameters μ=10, σ=1.5
- Convert to standard normal variable Z
- Use standard normal table for probability
- Interpret result in context of lifetime
Loses marks
- Using normal distribution directly instead of log-normal
- Incorrect standardization of the variable
Earns more
- Correct transformation formula for log-normal
- Clean table lookup with interpolation if needed
Extra mark
- Graphical representation of the tail probability
- (c(ii)) Lifetime exceeded by 99% of lasers. 5 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Identify 99th percentile from standard normal table
- Convert back to log-normal scale
- Calculate specific lifetime value
- State units (hours) in final answer
Loses marks
- Confusing 99% with 1% tail probability
- Forgetting to exponentiate when converting back
Earns more
- Correct inverse transformation from Z to X
- Clear statement of the percentile used
Extra mark
- Verification by plugging back into CDF
- (d) Economic lot size, total optimal cost, and total weekly profit for stockist. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate economic lot size (EOQ) using given parameters
- Compute total optimal cost including capital cost
- Determine total weekly profit at ₹55 selling price
- Show all cost components clearly
Loses marks
- Omitting capital cost from total optimal cost
- Incorrect annualization of weekly demand
Earns more
- Correct EOQ formula application
- Clear breakdown of ordering, carrying, and purchase costs
Extra mark
- Sensitivity analysis on lot size
- (e) Budget required to decrease average queue size from 1 1/3 to 1/2 patient. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Identify current and target queue sizes
- Calculate required decrease in average service time
- Determine cost per patient for the decrease
- Compute total budget for 96 patients per day
Loses marks
- Ignoring the 96 patients per day factor
- Incorrect calculation of time decrease needed
Earns more
- Clear relationship between service time and queue size
- Correct calculation of cost per minute decrease
Extra mark
- Discussion of practical constraints on service time
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