Statistics 2022 Paper II 50 marks Solve

Q2

(a) Samples of size n = 5 units are taken from a process every hour. The x̄ and R̄ values for a particular quality characteristic are determined. After 25 samples have been collected, we obtain x̄̄ = 20 and R̄ = 4·56. (i) What are the three-sigma control limits for x̄ and R? (ii) Estimate the process standard deviation if both the charts exhibit control. (iii) Assume that the process output is normally distributed. If the specifications are 19 ± 5, what are your conclusions regarding the process capability? (iv) If the process mean shifts to 24, what is the probability of not detecting this shift on the first subsequent sample? (d₂ = 2·326, D₁ = 0, D₂ = 4·918, D₃ = 0, D₄ = 2·114, A = 1·342, A₂ = 0·577, A₃ = 1·427, C₄ = 0·940, B₃ = 0, B₄ = 2·089) 15 marks (b) Define a Weibull distribution with scale parameter α and shape parameter β. Obtain the hazard function and reliability function of the model. Show also that the distribution satisfies increasing, constant and decreasing failure rate based on suitable choice of the shape parameter. 15 marks (c) A company uses the following acceptance-sampling procedure—A sample equal to 10% of the lot is taken. If 2% or less of the items in the sample are defective, the lot is accepted, otherwise it is rejected. If the submitted lot varies in size from 5000 units to 10000 units, what can you say about the protection by this plan? If 0·05 is the LTPD, does this scheme offer reasonable protection to the consumer? 20 marks

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(a) एक प्रक्रम से प्रत्येक घंटे आमाप n = 5 यूनिटों के प्रतिदर्श लिये जाते हैं। किसी विशेष गुणता-अभिलक्षण के लिए x̄ और R̄ के मानों को निकाला जाता है। 25 प्रतिदर्शों को एकत्रित करने के बाद, हम प्राप्त करते हैं x̄̄ = 20 और R̄ = 4·56। (i) x̄ और R के लिए तीन-सिग्मा नियंत्रण सीमाएँ क्या हैं? (ii) यदि दोनों संचित्र (चार्ट) नियंत्रण प्रदर्शित करते हैं, तो प्रक्रम मानक विचलन का आकलन कीजिए। (iii) मान लीजिए कि प्रक्रम उत्पादन प्रसामान्यतः बंटित है। यदि विनिर्देश 19 ± 5 हैं, तो प्रक्रम सामर्थ्य के बारे में आपके निष्कर्ष क्या हैं? (iv) यदि प्रक्रम माध्य 24 पर स्थानांतरित हो जाता है, तो प्रथम परवर्ती प्रतिदर्श पर इस स्थानांतरण को न पहचान पाने की प्रायिकता क्या है? (d₂ = 2·326, D₁ = 0, D₂ = 4·918, D₃ = 0, D₄ = 2·114, A = 1·342, A₂ = 0·577, A₃ = 1·427, C₄ = 0·940, B₃ = 0, B₄ = 2·089) 15 अंक (b) एक वेबुल बंटन को परिभाषित कीजिए जिसका मापक्रम प्राचल α और आकृति प्राचल β है। मॉडल का संकटप्रसूता फलन और विश्वसनीयता फलन प्राप्त कीजिए। यह भी दर्शाइए कि आकृति प्राचल के उपयुक्त विकल्प के आधार पर, बंटन वर्धमान, स्थिर और ह्रासमान विफलता दर को संतुष्ट करता है। 15 अंक (c) एक कंपनी निम्नलिखित स्वीकरण-प्रतिचयन कार्यविधि का प्रयोग करती है—एक प्रतिदर्श लिया गया है प्रचय के 10% के बराबर। यदि प्रतिदर्श में 2% या उससे कम मद दोषपूर्ण है, तो प्रचय स्वीकार किया जाता है, अन्यथा यह अस्वीकार कर दिया जाता है। यदि जमा किया गया प्रचय 5000 इकाइयों से 10000 इकाइयों के आमाप में बदलता है, तो आप इस आयोजना द्वारा सुरक्षा के बारे में क्या कह सकते हैं? यदि एल० टी० पी० डी० 0·05 है, तो क्या यह योजना उपभोक्ता को उचित सुरक्षा प्रदान करती है? 20 अंक

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Approach

This is a numerical problem requiring systematic calculation across three parts. Allocate approximately 35% time to part (a) with its four sub-parts on control charts (15 marks), 30% to part (b) on Weibull distribution derivations (15 marks), and 35% to part (c) on acceptance sampling analysis (20 marks). Begin each part with clear identification of given parameters, show all formulas before substitution, and conclude with explicit interpretation of results.

Key points expected

  • Part (a)(i): Correct application of x̄ chart limits using A₂R̄ and R chart limits using D₃R̄, D₄R̄ with n=5
  • Part (a)(ii): Estimation of process standard deviation using σ̂ = R̄/d₂ = 4.56/2.326
  • Part (a)(iii): Calculation of Cp and Cpk indices comparing process capability with specification limits 19±5
  • Part (a)(iv): Calculation of β-risk (Type II error) using normal distribution for shifted mean μ=24
  • Part (b): Derivation of Weibull hazard function h(t) = (β/α)(t/α)^(β-1) and reliability R(t) = exp[-(t/α)^β], with IFR/CFR/DFR classification based on β
  • Part (c): Analysis of variable sample size plan, calculation of acceptance probability using binomial/Poisson approximation, and evaluation against LTPD=0.05 for consumer protection

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness15%7.5Correctly identifies all given parameters (n=5, x̄̄=20, R̄=4.56, d₂=2.326, D₃=0, D₄=2.114, A₂=0.577) and selects appropriate constants from the table for each sub-part; for (c) correctly models the acceptance sampling plan with variable lot sizesIdentifies most parameters correctly but may confuse constants (e.g., uses A instead of A₂) or misidentify sample size relationship in part (c)Uses wrong parameters, confuses x̄̄ with x̄, or fails to recognize that sample size varies with lot size in part (c)
Method choice20%10Selects correct formulas: x̄ chart (x̄̄ ± A₂R̄), R chart (D₃R̄, D₄R̄), σ̂ = R̄/d₂, Cp/Cpk indices, β-risk calculation using standard normal; Weibull derivations from first principles; appropriate approximation for acceptance samplingUses correct general approach but may use wrong constant (e.g., uses d₂ for σ estimation in control limits) or applies normal approximation where binomial is neededUses completely wrong methods (e.g., uses s-chart formulas for R-chart, confuses producer/consumer risk, or fails to derive Weibull functions from PDF)
Computation accuracy25%12.5Accurate arithmetic: x̄ limits = 20 ± 2.63 → (17.37, 22.63), R limits = (0, 9.64), σ̂ = 1.96, Cp = 0.85, Cpk = 0.66; β = P(17.37 < x̄ < 22.63 | μ=24) = Φ(-1.29) - Φ(-4.71) ≈ 0.0985; Weibull derivations mathematically correct; acceptance probabilities calculated correctly for n=500 and n=1000Minor computational errors (e.g., arithmetic mistakes in limits, wrong z-values for β calculation, or algebraic slips in Weibull derivation) but method is soundMajor computational errors, wrong final values, or completely missing calculations for any sub-part
Interpretation25%12.5Clear interpretation: process is not capable (Cp<1, Cpk<1); β-risk interpretation; Weibull IFR when β>1, CFR when β=1, DFR when β<1; part (c) recognizes plan has poor discrimination and offers inadequate consumer protection at LTPD=0.05 due to variable sample size and loose criterionProvides some interpretation but may be vague or incomplete (e.g., states Cp value without concluding process capability, or describes Weibull shapes without linking to failure rates)No interpretation provided, or completely wrong conclusions (e.g., claims process is capable when Cp<1, or says plan offers good protection when it clearly doesn't)
Final answer & units15%7.5All final answers clearly boxed/labeled with appropriate units: control limits with same units as measurement, σ̂ in original units, Cp/Cpk dimensionless, β as probability, hazard and reliability functions properly expressed; part (c) explicitly states conclusion on consumer protectionMost answers present but may lack units or clear labeling; some final answers buried in workingMissing final answers, no units, or answers presented without context making them unidentifiable

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