Statistics 2023 Paper II 50 marks Derive

Q7

7.(a) Derive, by starting from a suitable functional form for $l_x$, the formula (i) $L_x = \dfrac{l_x + l_{x+1}}{2}$ and (ii) $L_x = \dfrac{l_x - l_{x+1}}{(\log l_x - \log l_{x+1})} = -\dfrac{d_x}{\log p_x}$ (iii) $e_x^0 = \dfrac{1}{2} + \sum\limits_{i=1}^{\infty} \dfrac{i d_{x+i}}{l_x}$ where $l_x$ = members of the cohort alive at age $x$ $L_x$ = number of years lived, in the aggregate, by the cohort of $l_0$ persons between age $x$ and $(x+1)$ $d_x$ = number of persons dying between age $x$ and $(x+1)$ $= l_x - l_{x+1}$ $p_x$ = probability that a person of age $x$ will survive till age $(x+1)$ $e_x^0$ = expectation of life at age $x$ 7.(b) (i) 400 students are given a test. The average is 60 and the standard deviation is 12. Obtain the Z-score and the standard scores equivalent to raw scores. The raw scores are given by | Raw scores | 84 | 78 | 72 | 66 | 60 | 54 | 48 | 42 | 36 | (ii) Convert the ten scores 1, 2, ..., 10 into standard scores with mean 50 and standard deviation 10. 7.(c) On the life table with $l_x = \dfrac{100-x}{190}$, $5 \leq x \leq 100$, Find (i) the chance that a child who has reached age 5 will live to age 60. (ii) the chance that a man of age 30 will live until age 80. (iii) the probability of dying within 5 years for a man aged 40. (iv) the expectation of life at age 40. (v) the chance that of the three men aged 30 at least one survives till age 80.

हिंदी में प्रश्न पढ़ें

7.(a) $l_x$ के लिए एक उपयुक्त फलनिक रूप से शुरू करके निम्नलिखित सूत्र को व्युत्पन्न कीजिए (i) $L_x = (l_x + l_{x+1})/2$ और (ii) $L_x = (l_x – l_{x+1})/(\log l_x – \log l_{x+1}) = – dx/\log p_x$ (iii) $e°_x = 1/2 + \sum_{i=1}^\infty (i d_{x+i})/l_x$ जहाँ $l_x$ = जत्था (कोहोर्ट) के सदस्य जो आयु $x$ तक जीवित हैं $L_x$ = जितने वर्ष जीवित रहे, सकल में $l_0$ व्यक्तियों के जत्थों द्वारा आयु $x$ और आयु $(x+1)$ के बीच $d_x$ = व्यक्तियों की संख्या जिनकी मृत्यु आयु $x$ और $(x+1)$ के बीच में होती है $= l_x - l_{x+1}$ $p_x$ = आयु $x$ के एक व्यक्ति के आयु $(x+1)$ तक जीवित रहने की प्रायिकता है $e_x^0$ = आयु $x$ पर जीवन की प्रत्याशा 7.(b) (i) 400 विद्यार्थियों ने एक परीक्षा दी है। औसत 60 है और मानक विचलन 12 है। Z-समंक और मानक समंकों को प्राप्त कीजिए जो कि यथाप्रास समंकों के तुल्य हैं। यथाप्रास समंक नीचे दिये गये हैं। | यथाप्रास समंक | 84 | 78 | 72 | 66 | 60 | 54 | 48 | 42 | 36 | (ii) दस समंकों 1, 2, ..., 10 को मानक समंकों में बदलो जिनका माध्य 50 और मानक विचलन 10 है। 7.(c) वय-सारणी में $l_x = \dfrac{100-x}{190}$ के साथ, $5 \leq x \leq 100$, ज्ञात कीजिए (i) प्रायिकता कि एक बच्चा जो आयु 5 पर पहुँच गया है, वह आयु 60 तक जीवित रहेगा। (ii) प्रायिकता कि एक व्यक्ति जिसकी आयु 30 वर्ष है वह आयु 80 तक जीवित रहेगा। (iii) प्रायिकता कि एक व्यक्ति जिसकी आयु 40 वर्ष है, वह 5 वर्ष के अन्दर मर जायेगा। (iv) आयु 40 पर जीवन की प्रत्याशा। (v) प्रायिकता कि 30 वर्ष की आयु वाले तीन व्यक्तियों में से कमसे कम एक आयु 80 तक जीवित रहे।

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Directive word: Derive

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How this answer will be evaluated

Approach

Begin with clear statement of assumptions for each derivation in 7(a), showing step-by-step integration for L_x formulas and summation manipulation for e_x^0. For 7(b), apply Z-score formula z = (X-μ)/σ systematically, then demonstrate linear transformation for standard scores with mean 50, SD 10. For 7(c), substitute given l_x = (100-x)/190 into survival probabilities, death probabilities, and life expectancy formulas, computing each numerical value with proper fraction handling. Allocate approximately 35% time to derivations in (a), 25% to standard score calculations in (b), and 40% to life table computations in (c) given its five sub-parts.

Key points expected

  • 7(a)(i): Assume l_x linear in [x, x+1], integrate l(t)dt from x to x+1 to obtain (l_x + l_{x+1})/2
  • 7(a)(ii): Assume l_x exponential (constant force of mortality), use l_t = l_x·e^{-μt} and integrate to derive harmonic mean form -d_x/log(p_x)
  • 7(a)(iii): Express T_x = Σ L_{x+i} using linear assumption, substitute L_x = (l_x + l_{x+1})/2, rearrange to obtain e_x^0 = 1/2 + Σ i·d_{x+i}/l_x
  • 7(b): Calculate Z-scores as (X-60)/12 for each raw score, then apply linear transformation 50 + 10·z for standard scores; for 1-10 scores, first find mean=5.5, SD=√8.25, then transform
  • 7(c): Compute survival probabilities as l_60/l_5, l_80/l_30; death probability as 1-l_45/l_40; e_40^0 = T_40/l_40 with T_40 = Σ L_{40+i}; binomial probability for at least one survivor among three men

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%9Explicitly states appropriate functional assumptions for each derivation (linearity for 7a-i, exponential/constant force of mortality for 7a-ii); correctly defines all life table symbols; proper setup of Z-score and linear transformation formulas; valid probability framework for 7c-vStates some assumptions but misses key ones (e.g., assumes linearity without stating it); minor symbol confusion; basic Z-score formula correct but transformation setup unclearMissing or incorrect assumptions; confuses l_x with L_x or d_x; wrong formula setup for standard scores; incorrect probability model for multiple lives
Method choice20%9Selects integration for L_x derivations with correct limits; uses summation by parts or careful algebraic manipulation for e_x^0; applies exact fractional arithmetic for 7(c) avoiding premature rounding; uses complementary probability for 'at least one' in 7c-vCorrect general approach but some inefficient steps; uses decimal approximations too early; correct method for standard scores but arithmetic errors in executionWrong method entirely (e.g., differentiation instead of integration); attempts to memorize without deriving; incorrect probability rule for multiple lives
Computation accuracy20%9Flawless integration yielding exact forms; correct algebraic simplification to target formulas; accurate Z-scores (z=2,1.5,1,0.5,0,-0.5,-1,-1.5,-2) and standard scores; precise values for 7(c): 35/95, 20/70, 5/95/(100-40)/190, e_40^0=30, 1-(50/70)^3Minor arithmetic slips in coefficients; one or two Z-score errors; correct survival ratios but arithmetic errors in final decimals; correct approach to e_x^0 but summation errorMajor computational errors in integration; incorrect Z-score calculations; wrong survival probabilities (reversing ratio); impossible values (>1 or <0)
Interpretation20%9Explains why linear assumption gives arithmetic mean and exponential gives harmonic mean; interprets e_x^0 formula showing 1/2 as average future lifetime of those dying in first year; relates standard scores to normative comparison; interprets 7c-v result in insurance/actuarial contextBrief interpretation of final answers; mentions what L_x represents without deeper insight; states standard scores allow comparison without explaining linear transformationNo interpretation provided; misinterprets life expectancy as median; confuses probability with odds; no contextual meaning for standard scores
Final answer & units20%10All 10+ required answers clearly boxed/labeled: three derived formulas, nine Z/standard scores, five 7(c) numerical answers with proper units (years for life expectancy, probability for chances); exact fractions preferred where possible; consistent decimal precisionMost answers present but some mislabeled or missing; mixed units (percentages vs proportions); inconsistent roundingMissing multiple final answers; no units; answers scattered without labels; order-of-magnitude errors

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