Paper II — Q7
7.(a) Derive, by starting from a suitable functional form for lₓ, the formula (i) Lₓ = (lₓ + lₓ₊₁)/2 and (ii) Lₓ = (lₓ …
7.(a) Derive, by starting from a suitable functional form for lₓ, the formula Lₓ = (lₓ + lₓ₊₁)/2 and (ii) Lₓ = (lₓ - lₓ₊₁)/((log lₓ - log lₓ₊₁)) = -(dₓ)/(log pₓ)
eₓ⁰ = 1/2 + Σlimitsᵢ₌₁^∞ (i dₓ₊ᵢ)/(lₓ)
where
lₓ = members of the cohort alive at age x
Lₓ = number of years lived, in the aggregate, by the cohort of l₀ persons between age x and (x+1)
dₓ = number of persons dying between age x and (x+1) = lₓ - lₓ₊₁
pₓ = probability that a person of age x will survive till age (x+1)
eₓ⁰ = 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 |
|---|
Convert the ten scores 1, 2, ..., 10 into standard scores with mean 50 and standard deviation 10.
7.(c) On the life table with lₓ = (100-x)/190, 5 ≤ x ≤ 100,
Find
the chance that a child who has reached age 5 will live to age 60.
the chance that a man of age 30 will live until age 80.
the probability of dying within 5 years for a man aged 40.
the expectation of life at age 40.
the chance that of the three men aged 30 at least one survives till age 80.
हिंदी में प्रश्न पढ़ें
7.(a) lₓ के लिए एक उपयुक्त फलनिक रूप से शुरू करके निम्नलिखित सूत्र को व्युत्पन्न कीजिए
Lₓ = (lₓ + lₓ₊₁)/2 और (ii) Lₓ = (lₓ – lₓ₊₁)/(log lₓ – log lₓ₊₁) = – dx/log pₓ
e°ₓ = 1/2 + Σᵢ₌₁^∞ (i dₓ₊ᵢ)/lₓ
जहाँ
lₓ = जत्था (कोहोर्ट) के सदस्य जो आयु x तक जीवित हैं
Lₓ = जितने वर्ष जीवित रहे, सकल में l₀ व्यक्तियों के जत्थों द्वारा आयु x और आयु (x+1) के बीच
dₓ = व्यक्तियों की संख्या जिनकी मृत्यु आयु x और (x+1) के बीच में होती है = lₓ - lₓ₊₁
pₓ = आयु x के एक व्यक्ति के आयु (x+1) तक जीवित रहने की प्रायिकता है
eₓ⁰ = 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 |
|---|
दस समंकों 1, 2, ..., 10 को मानक समंकों में बदलो जिनका माध्य 50 और मानक विचलन 10 है।
7.(c) वय-सारणी में lₓ = (100-x)/190 के साथ, 5 ≤ x ≤ 100,
ज्ञात कीजिए
प्रायिकता कि एक बच्चा जो आयु 5 पर पहुँच गया है, वह आयु 60 तक जीवित रहेगा।
प्रायिकता कि एक व्यक्ति जिसकी आयु 30 वर्ष है वह आयु 80 तक जीवित रहेगा।
प्रायिकता कि एक व्यक्ति जिसकी आयु 40 वर्ष है, वह 5 वर्ष के अन्दर मर जायेगा।
आयु 40 पर जीवन की प्रत्याशा।
प्रायिकता कि 30 वर्ष की आयु वाले तीन व्यक्तियों में से कमसे कम एक आयु 80 तक जीवित रहे।
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) Derivation of the three life-table formulae
Notation: lₓ is the number alive at age x, Lₓ the years lived by the cohort between x and x+1, dₓ = lₓ − lₓ₊₁, pₓ = lₓ₊₁/lₓ, and eₓ⁰ the complete expectation of life. (i) Linear form: assume lₓ₊ₜ = lₓ + (lₓ₊₁ − lₓ)t for 0 ≤ t ≤ 1. Step 1: Lₓ = ∫₀¹ lₓ₊ₜ dt = lₓ + (lₓ₊₁ − lₓ)/2. Step 2: Hence Lₓ = (lₓ + lₓ₊₁)/2. (ii) Exponential form: assume constant force of mortality, so lₓ₊ₜ = lₓ pₓᵗ. Step 1: Lₓ = ∫₀¹ lₓ pₓᵗ dt = lₓ(pₓ − 1)/log pₓ = (lₓ₊₁ − lₓ)/log pₓ = −dₓ/log pₓ. Step 2: Since log lₓ − log lₓ₊₁ = −log pₓ, this equals (lₓ − lₓ₊₁)/(log lₓ − log lₓ₊₁). It holds for pₓ ≠ 1. (iii) Complete expectation: eₓ⁰ = (1/lₓ)∫₀^∞ lₓ₊ₜ dt = (1/lₓ)(Lₓ + Lₓ₊₁ + Lₓ₊₂ + ⋯). Step 1: Using (i), the integral is lₓ/2 + lₓ₊₁ + lₓ₊₂ + ⋯. Step 2: Since lₓ₊ᵢ = dₓ₊ᵢ + dₓ₊ᵢ₊₁ + ⋯, collecting terms gives lₓ₊₁ + lₓ₊₂ + ⋯ = dₓ₊₁ + 2dₓ₊₂ + 3dₓ₊₃ + ⋯ = Σ i dₓ₊ᵢ. Step 3: Therefore eₓ⁰ = 1/2 + (1/lₓ)Σ i dₓ₊ᵢ, the sum running over i = 1, 2, ….
(b) Z-scores and standard scores
(i) Z = (X − 60)/12 and the standard score is T = 50 + 10Z. Step 1: For raw scores 84, 78, 72, 66, 60, 54, 48, 42, 36, Z = 2, 1·5, 1, 0·5, 0, −0·5, −1, −1·5, −2. Step 2: The standard scores are 70, 65, 60, 55, 50, 45, 40, 35, 30. (ii) Scores 1 to 10, treated as a population. Step 1: Mean = 5·5 and Σ(X − 5·5)² = 82·5, so σ = √(82·5/10) = √33/2 = 2·872. Step 2: Standard score = 50 + 10(X − 5·5)/σ = 50 + 20(X − 5·5)/√33. Step 3: For X = 1 to 10 the standard scores are 34·33, 37·81, 41·30, 44·78, 48·26, 51·74, 55·22, 58·70, 62·19, 65·67. Their mean is 50 and standard deviation 10.
(c) Probabilities and expectation from lₓ = (100 − x)/190, 5 ≤ x ≤ 100
(i) P(age 5 to 60) = l₆₀/l₅ = 40/95 = 8/19 = 0·4211. (ii) P(age 30 to 80) = l₈₀/l₃₀ = 20/70 = 2/7 = 0·2857. (iii) P(a man aged 40 dies within 5 years) = 1 − l₄₅/l₄₀ = 1 − 55/60 = 1/12 = 0·0833. (iv) Expectation of life at 40: e₄₀⁰ = (1/l₄₀)∫₀⁶⁰ l₄₀₊ₜ dt = (1/60)∫₀⁶⁰ (60 − t) dt = (1/60)(3600 − 1800) = 30 years. (v) Each man aged 30 survives to 80 with p = 2/7, independently. Step 1: P(none survives) = (5/7)³ = 125/343. Step 2: P(at least one survives) = 1 − 125/343 = 218/343 = 0·6356.
How this answer will be evaluated
What earns full marks: Rigorous derivation in (a); error-free computation in (b) and (c) with clear interpretation.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Derive Lx=(lx+lx+1)/2, Lx=-dx/log px and ex0=1/2+(1/lx)·sum i dx+i from suitable forms of lx. 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Linear lx assumption and integration for Lx
- Exponential lx=lx px^t giving Lx=-dx/log px
- ex0 as sum of Lx using the linear result
- Rearrangement of sum of lx+i into sum i dx+i
Loses marks
- Quoting formulas without derivation; using the same assumption for (i) and (ii).
Earns more
- Correct assumption stated and each integral/step derived.
Extra mark
- Note that (ii) needs px not equal to 1.
- (b) Find Z-scores and standard scores for the raw scores and convert scores 1 to 10 to mean 50, SD 10. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Z=(X-60)/12 and T=50+10Z
- Standard scores 70 to 30
- Mean 5.5 and SD sqrt(33)/2 for scores 1 to 10
- Standard scores 34.33 to 65.67
Loses marks
- Using sample SD without stating it; arithmetic slips in values.
Earns more
- Correct formula, all nine and ten values computed.
Extra mark
- Check that the converted scores have mean 50 and SD 10.
- (c) Compute the survival probabilities and the expectation of life from lx=(100-x)/190. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- 8/19 and 2/7
- 1/12 for dying within 5 years
- e40=30 years by integration
- 218/343 for at least one of three surviving
Loses marks
- Using l40/l45 inverted; treating the three men as dependent.
Earns more
- All five parts correct with working.
Extra mark
- Mention that the model is uniform death between ages 5 and 100.
What "Derive" is asking you to do
Reach the stated expression from a starting relation, justifying every step. The destination is printed in the question, so only the route earns marks, and the assumptions you work under are part of that route.
Structure that answers it
Assumptions and notation defined → starting relation or governing equation → each step with its justification → the required expression → limiting case or boundary check
Where marks are lost
Writing the standard result first and fitting three lines to it, which an examiner reads at a glance. Marks also go on assumptions left unstated — lossless medium, small amplitude, errors independent with zero mean — and on symbols used before they are defined, even when the question says usual notations.
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.
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