Statistics 2023 Paper I 50 marks Compulsory Solve

Q1

(a) Out of 1000 persons born, only 900 reach the age of 15 years, and out of every 1000 who reach the age of 15 years, 950 reach the age of 50 years. Out of every 1000 who reach the age of 50 years, 40 die in one year. Accordingly, what is the probability that a person would attain the age of 51 years ? (10 marks) (b) Let X be a continuous random variable with probability density function : f(x) = $$ \begin{cases} \frac{x}{2}, & 0 \leq x < 1 \\\ \frac{1}{2}, & 1 \leq x < 2 \\\ \frac{3-x}{2}, & 2 \leq x < 3 \\\ 0, & \text{elsewhere} \end{cases} $$ Obtain the cumulative distribution function of X and hence find the value of $P\left(X > \frac{3}{2}\right)$. (10 marks) (c) Let {Xₙ, n ≥ 1} be a sequence of mutually independent random variables such that P(Xₙ = nᵅ) = P(Xₙ = – nᵅ) = 0·5, for any α > 0. Derive the condition on α under which the sequence {Xₙ, n ≥ 1} obeys WLLNs. (10 marks) (d) Apply Run Test to test the randomness of the following sequence of H and T at 5% level of significance : HHHHHHTHHHHHTHTHHHH TTHHHHTHHHTTHHHHHH THHTTHHTHHH Given : Z₍₀·₀₂₅₎ = 1·96 Z₍₀·₀₅₎ = 1·645 (10 marks) (e) Differentiate between prior and posterior distributions. In case of squared error loss function, find out the Bayes estimator for unknown parameter. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) 1000 जन्म लेने वाले व्यक्तियों में से, केवल 900, 15 वर्ष तक की आयु तक पहुँच पाते हैं, तथा प्रति 1000 व्यक्ति जो 15 वर्ष की आयु तक पहुँचते हैं, उनमें से 950 व्यक्ति 50 वर्ष की आयु तक पहुँचते हैं। प्रति 1000 व्यक्तियों में जो 50 वर्ष की आयु तक पहुँचते हैं, उनमें से 40 व्यक्तियों की एक वर्ष में मृत्यु हो जाती है। तदनुसार एक व्यक्ति के 51 वर्ष की आयु तक पहुँचने की प्रायिकता क्या है ? (10 अंक) (b) माना X एक सतत यादृच्छिक चर है जिसका प्रायिकता घनत्व फलन है : f(x) = $$ \begin{cases} \frac{x}{2}, & 0 \leq x < 1 \\\ \frac{1}{2}, & 1 \leq x < 2 \\\ \frac{3-x}{2}, & 2 \leq x < 3 \\\ 0, & \text{अन्यथा} \end{cases} $$ X का संचयी वितरण फलन निकालिए तथा इससे $P\left(X > \frac{3}{2}\right)$ का मान ज्ञात कीजिए। (10 अंक) (c) माना {Xₙ, n ≥ 1} परस्पर स्वतंत्र यादृच्छिक चरों की श्रृंखला इस प्रकार है कि P(Xₙ = nᵅ) = P(Xₙ = – nᵅ) = 0·5, किसी भी α > 0 के लिए । α पर उस प्रतिबंध को निकालिए जिसके अंतर्गत, श्रृंखला {Xₙ, n ≥ 1} निबल बृहत् संख्याओं के नियम (WLLNs) का पालन करती है । (10 अंक) (d) H एवं T के निम्नलिखित अनुक्रम की यादृच्छिकता जाँचने के लिए परम्परा (रन) परीक्षण, 5% सार्थकता स्तर, पर प्रयुक्त कीजिए : HHHHHHTHHHHHTHTHHHH TTHHHHTHHHTTHHHHHH THHTTHHTHHH दिया गया है : Z₍₀·₀₂₅₎ = 1·96 Z₍₀·₀₅₎ = 1·645 (10 अंक) (e) पूर्व एवं पश्च बंटनों में विभेद कीजिए । वर्ग-त्रुटि हानि फलन की स्थिति में अज्ञात प्राचल का बेज़ आकलक ज्ञात कीजिए । (10 अंक)

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Directive word: Solve

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How this answer will be evaluated

Approach

Solve each sub-part systematically with clear mathematical working. For (a), apply conditional probability using survival data; (b) integrate piecewise to find CDF and evaluate tail probability; (c) apply Khinchin's WLLN condition checking variance behavior; (d) count runs and apply normal approximation for hypothesis testing; (e) state Bayes theorem and minimize posterior expected loss. Allocate approximately 2 minutes per mark, presenting each solution with clear labeling and logical flow from given information to final answer.

Key points expected

  • (a) Correct application of chain rule for conditional probability: P(age 51) = P(survive to 15) × P(survive to 50 | 15) × P(survive 50-51 | 50) = 0.9 × 0.95 × 0.96
  • (b) Proper piecewise integration of f(x) to obtain F(x) with continuity checks at x=1 and x=2, then P(X > 3/2) = 1 - F(3/2) = 5/8
  • (c) Derivation that E(Xₙ) = 0, Var(Xₙ) = n^(2α), and application of Khinchin's theorem requiring (1/n²)ΣVar(Xᵢ) → 0, yielding condition α < 1/2
  • (d) Correct counting of runs (r=12), expected runs μᵣ = 2n₁n₂/(n₁+n₂) + 1, variance σᵣ², and Z-test showing |Z| < 1.96 so randomness not rejected
  • (e) Clear distinction: prior π(θ) represents pre-sample belief, posterior π(θ|x) ∝ L(x|θ)π(θ) updates belief; Bayes estimator under squared error loss is posterior mean E[θ|x]

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10Correctly identifies all given parameters for each sub-part: survival probabilities in (a), piecewise domain boundaries in (b), i.i.d. structure and moment conditions in (c), run counting rules and hypotheses in (d), prior-posterior relationship in (e)Identifies most parameters correctly but misses one boundary condition or misstates a probability; some confusion between prior and likelihood in (e)Major errors in identifying given information, wrong probability statements, incorrect domain specification for piecewise function, or fundamental misunderstanding of Bayesian setup
Method choice20%10Selects optimal methods: chain rule of conditional probability (a), piecewise integration with proper limits (b), Khinchin's WLLN with variance analysis (c), Wald-Wolfowitz run test with normal approximation (d), Bayes theorem with squared error loss minimization (e)Uses correct general approach but suboptimal method (e.g., Chebyshev instead of Khinchin in c), or correct method with unnecessary stepsWrong methodological approach entirely, such as using simple probability instead of conditional in (a), or ignoring piecewise nature in (b), or wrong test selection in (d)
Computation accuracy20%10All calculations precise: (a) 0.8208 or 821/1000; (b) F(x) correctly integrated with values 0, x²/4, (2x-1)/4, (-x²+6x-5)/4, 1 in respective intervals; (c) algebraic manipulation leading to α < 1/2; (d) n₁=35, n₂=15, μᵣ=22, σᵣ²≈6.86, Z≈-3.82; (e) clean derivation of posterior mean formulaMinor arithmetic slips in one sub-part, or correct final answers with some intermediate steps omitted or unclearMultiple calculation errors, wrong integration limits, incorrect run counting, or algebraic mistakes leading to wrong conditions
Interpretation20%10Clear interpretation: survival probability context in (a), verification of CDF properties (non-decreasing, right-continuous, limits 0 and 1) in (b), explanation of why α < 1/2 ensures sample mean consistency in (c), explicit hypothesis testing conclusion in (d), decision-theoretic justification of posterior mean in (e)Some interpretation present but missing key insights like CDF verification or why the specific α condition emergesNo interpretation provided, or meaningless statements; fails to connect mathematical results to statistical meaning
Final answer & units20%10All five final answers clearly stated with appropriate precision: probability values (a, b), condition on α (c), test conclusion with significance level (d), Bayes estimator formula (e); proper mathematical notation throughoutFinal answers present but with inconsistent precision, missing units where relevant, or one sub-part answer omittedMissing or incorrect final answers, answers without supporting working, or failure to state conclusions for hypothesis test in (d)

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