GELE Adjustment Computations (Least Squares) — Theory of Errors, Weights and Most Probable ValueExam Answer Templates

Marks

1

Topic

Types of Errors

Difficulty

easy

Template Id

T1

Examiner Tip

The examiner expects at least two distinguishing traits (small magnitude AND normal distribution or sign-varying). One trait alone may not be sufficient for the full mark.

Model Answer

A random error is a small, unpredictable error that remains after blunders and systematic errors have been eliminated; it follows the normal (Gaussian) distribution, is equally likely to be positive or negative, and decreases in effect as the number of observations increases.

Question Type

very_short_answer

Answer Structure

• One sentence: correct definition with at least two distinguishing characteristics [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing random errors as 'mistakes' or confusing them with blunders — no mark awarded.
• Omitting that they follow the normal distribution — partial answer, examiner may still award the mark if another key characteristic is stated clearly.
• Vague answer such as 'errors due to human error' with no distinguishing detail.

Key Phrases To Include

• normal distribution
• equally likely positive or negative
• small and unpredictable
• remains after blunders and systematic errors are removed

Marks

1

Topic

Most Probable Value

Difficulty

easy

Template Id

T2

Examiner Tip

At 1-mark level, the single word 'arithmetic mean' paired with the formula is a complete, safe answer. Do not over-explain — time is limited.

Model Answer

The Most Probable Value (MPV) of a quantity from equally-reliable (equal-weight) direct observations is the arithmetic mean: x̄ = Σx / n, where n is the number of observations.

Question Type

very_short_answer

Answer Structure

• State that MPV equals the arithmetic mean [1 mark]
• Write the formula x̄ = Σx / n (optional but reinforces the mark)

Scoring Breakdown

Common Mark Deductions

• Stating 'median' or 'mode' instead of arithmetic mean.
• Omitting the qualifier 'equally-reliable' — acceptable at 1-mark level but becomes critical in longer answers.
• Writing the weighted mean formula when the question specifies equal reliability.

Key Phrases To Include

• arithmetic mean
• equally-reliable
• x̄ = Σx / n

Marks

1

Topic

Weights

Difficulty

easy

Template Id

T3

Examiner Tip

This is a classic one-mark trap. The board exam frequently tests whether you know it is σ² (variance) not σ. Write 'w = 1/σ²' explicitly to secure the mark.

Model Answer

The weight of an observation is inversely proportional to its variance (square of the standard deviation): w = 1/σ². A more precise observation (smaller σ) thus carries a larger weight.

Question Type

very_short_answer

Answer Structure

• State the inverse proportionality to variance (not σ alone) [1 mark]
• Write the formula w = 1/σ²

Scoring Breakdown

Common Mark Deductions

• Writing w ∝ 1/σ instead of 1/σ² — this is the most common single-mark error in this topic and earns zero marks.
• Stating 'directly proportional to standard deviation' — completely incorrect.

Key Phrases To Include

• inversely proportional to variance
• w = 1/σ²
• not 1/σ

Marks

2

Topic

Types of Errors

Difficulty

easy

Template Id

T4

Examiner Tip

Use parallel structure: one paragraph per error type with the same elements (definition → key characteristic → example). Examiners can then award marks item-by-item efficiently.

Model Answer

Blunder (Mistake): A gross error caused by carelessness, inexperience, or equipment malfunction (e.g., misreading a vernier, recording 50.10 m as 51.10 m). Blunders do not follow any statistical law and must be detected and eliminated before adjustment. Systematic Error: An error that follows a definite mathematical or physical law, has a consistent sign and magnitude under the same conditions (e.g., a tape that is 0.005 m too short always makes measured distances appear too long). Systematic errors can be modelled, calibrated, and corrected.

Question Type

short_answer

Answer Structure

• Define blunder with one characteristic and one example [1 mark]
• Define systematic error with one characteristic and one example [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing systematic errors with random errors (both have 'small' errors — wrong; systematic can be large).
• No examples given — examiner may still award the mark if the definition is precise, but examples make the answer safer.
• Stating systematic errors are 'unpredictable' — this is the definition of random errors.

Key Phrases To Include

• gross error
• carelessness
• must be eliminated
• definite mathematical or physical law
• consistent sign
• corrected by calibration

Marks

2

Topic

Most Probable Value and Standard Deviation

Difficulty

medium

Template Id

T5

Examiner Tip

Show residuals as a column table; it reduces arithmetic errors and makes the working easy for the examiner to check line-by-line.

Model Answer

Given: n = 5 observations x₁ = 125.32, x₂ = 125.28, x₃ = 125.35, x₄ = 125.30, x₅ = 125.25 m Step 1 — Most Probable Value (arithmetic mean): x̄ = Σx / n = (125.32 + 125.28 + 125.35 + 125.30 + 125.25) / 5 x̄ = 626.50 / 5 x̄ = 125.30 m Step 2 — Residuals v = x - x̄: v₁ = 125.32 − 125.30 = +0.02 v₂ = 125.28 − 125.30 = −0.02 v₃ = 125.35 − 125.30 = +0.05 v₄ = 125.30 − 125.30 = 0.00 v₅ = 125.25 − 125.30 = −0.05 Step 3 — Standard deviation of a single observation: Σv² = (0.02)² + (0.02)² + (0.05)² + (0.00)² + (0.05)² = 0.0004 + 0.0004 + 0.0025 + 0.0000 + 0.0025 = 0.0058 σ = √(Σv² / (n−1)) = √(0.0058 / 4) = √0.00145 = 0.0381 m Answer: MPV = 125.30 m; σ = 0.038 m (to nearest millimetre)

Question Type

numerical

Answer Structure

• State all given values [no separate mark, but establishes work]
• Compute x̄ = Σx/n correctly [1 mark]
• Compute residuals, Σv², and σ = √(Σv²/(n−1)) correctly [1 mark]

Scoring Breakdown

Common Mark Deductions

• Dividing Σv² by n instead of (n−1) — loses the second mark.
• Rounding x̄ prematurely, leading to incorrect residuals.
• Omitting the unit 'metres' from the final answers.
• Computing |v| instead of v² — leads to wrong σ.

Key Phrases To Include

• x̄ = Σx / n
• v = x − x̄
• σ = √(Σv² / (n−1))
• n−1 (not n)

Marks

2

Topic

Weights in Leveling

Difficulty

easy

Template Id

T6

Examiner Tip

Always state the governing rule before substituting numbers. In this case, 'w ∝ 1/K' is a scorable statement, not just a tool for calculation.

Model Answer

Rule: For leveling, weight is inversely proportional to route length K, i.e., w ∝ 1/K. Compute weights (using the shortest route as reference): w₁ ∝ 1/3, w₂ ∝ 1/6, w₃ ∝ 1/2 Express as simple ratio (multiply each by 6, the LCM): w₁ : w₂ : w₃ = 2 : 1 : 3 Answer: The relative weights are Route 1 (3 km) = 2, Route 2 (6 km) = 1, Route 3 (2 km) = 3.

Question Type

short_answer

Answer Structure

• State the rule w ∝ 1/K for leveling [1 mark]
• Correctly compute and express relative weights [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using w ∝ K (directly proportional) — conceptual error, loses both marks.
• Computing correct numerical values but failing to state the rule — loses the first mark.
• Leaving weights as 0.333 : 0.167 : 0.5 without simplifying or verifying consistency — acceptable if examiner confirms equivalence, but ratio form is cleaner.

Key Phrases To Include

• w ∝ 1/K
• inversely proportional to distance
• shorter route has higher weight

Marks

2

Topic

Standard Deviation

Difficulty

medium

Template Id

T7

Examiner Tip

This is a conceptual question often ignored during review. Examiners reward the phrase 'degrees of freedom' — it demonstrates statistical maturity.

Model Answer

When only a sample of n observations is available, the true population mean μ is unknown and is estimated by the sample mean x̄. Because x̄ is computed from the same data, the residuals vᵢ = xᵢ − x̄ are not fully independent — they must satisfy the constraint Σv = 0, which effectively uses up one degree of freedom. Dividing by (n−1) corrects for this and produces an unbiased estimator of the population variance σ². This denominator (n−1) is called Bessel's correction.

Question Type

short_answer

Answer Structure

• Identify that using x̄ instead of the true mean μ reduces the degrees of freedom by one [1 mark]
• State that dividing by (n−1) gives an unbiased estimate of population variance (Bessel's correction) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Simply stating 'because it is more accurate' without explaining degrees of freedom — too vague for marks.
• Confusing sample standard deviation with population standard deviation.
• Omitting the phrase 'unbiased' or its equivalent — the examiner looks for this specific concept.

Key Phrases To Include

• degrees of freedom
• Bessel's correction
• unbiased estimator
• constraint Σv = 0
• sample mean estimated from the data

Marks

3

Topic

Weighted Mean

Difficulty

medium

Template Id

T8

Examiner Tip

Notice that Obs 2 and 3 have the same σ (same weight = 2500), each four times the weight of Obs 1 (weight = 625). The answer should therefore lie closer to the average of 250.18 and 250.10 than to 250.12 — use this as a sanity check.

Model Answer

Given: x₁ = 250.12 m, σ₁ = 0.04 m x₂ = 250.18 m, σ₂ = 0.02 m x₃ = 250.10 m, σ₃ = 0.02 m Step 1 — Compute weights (w = 1/σ²): w₁ = 1/(0.04)² = 1/0.0016 = 625 w₂ = 1/(0.02)² = 1/0.0004 = 2500 w₃ = 1/(0.02)² = 1/0.0004 = 2500 Step 2 — Compute weighted products (wx): w₁x₁ = 625 × 250.12 = 156,325.00 w₂x₂ = 2500 × 250.18 = 625,450.00 w₃x₃ = 2500 × 250.10 = 625,250.00 Step 3 — Weighted mean: Σw = 625 + 2500 + 2500 = 5625 Σwx = 156,325.00 + 625,450.00 + 625,250.00 = 1,407,025.00 x̄_w = Σwx / Σw = 1,407,025.00 / 5625 = 250.138 m Answer: The weighted mean distance is 250.138 m ≈ 250.14 m.

Question Type

numerical

Answer Structure

• List all given values [setup, no mark]
• Correctly compute each weight w = 1/σ² [1 mark]
• Correctly compute each weighted product wx [1 mark]
• Correctly compute x̄_w = Σwx / Σw with correct answer and unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing w = 1/σ instead of 1/σ² — loses the first mark and cascades errors to subsequent steps.
• Arithmetic errors in wx products — loses the second mark.
• Rounding Σwx prematurely — leads to incorrect final answer.
• Omitting the unit from the final answer.

Key Phrases To Include

• w = 1/σ²
• x̄_w = Σwx / Σw
• weighted mean
• metres

Marks

2

Topic

Probable Error

Difficulty

medium

Template Id

T9

Examiner Tip

Memorise the coefficient 0.6745 exactly — it appears directly in board exam MCQ options and any deviation will select the wrong answer.

Model Answer

The probable error of a single observation (PE) is the value of error such that exactly 50% of all errors in a normally distributed set are numerically less than PE in absolute value. It defines the error band within which half the observations are expected to fall. Relationship to standard deviation: PE = 0.6745σ For the mean: PE_mean = 0.6745 × (σ/√n)

Question Type

short_answer

Answer Structure

• Define probable error as the 50th-percentile error band [1 mark]
• State PE = 0.6745σ and optionally the mean form [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating PE = 0.6745/σ (inverted) — wrong formula.
• Describing PE as 'the average error' — incorrect; it is a percentile concept.
• Using a coefficient other than 0.6745 (some students write 0.6745σ² — wrong).

Key Phrases To Include

• 50% of all errors
• PE = 0.6745σ
• normal distribution
• probable error of the mean = 0.6745σ/√n

Marks

3

Topic

Standard Deviation of the Mean

Difficulty

medium

Template Id

T10

Examiner Tip

Part (b) is a classic board exam question on the limitation of averaging: to halve σ_x̄, you need four times the observations, not twice. Always square the ratio.

Model Answer

Given: n = 4, σ = 0.030 m Part (a) — Standard deviation of the mean: σ_x̄ = σ / √n = 0.030 / √4 = 0.030 / 2 = 0.015 m Part (b) — Required number of observations for σ_x̄ = 0.010 m: σ_x̄ = σ / √n → √n = σ / σ_x̄ = 0.030 / 0.010 = 3 n = 3² = 9 observations Answer: (a) σ_x̄ = 0.015 m (b) n = 9 total measurements are required.

Question Type

numerical

Answer Structure

• Correctly apply σ_x̄ = σ/√n for Part (a) [1 mark]
• Obtain σ_x̄ = 0.015 m with unit [1 mark]
• Rearrange formula and solve for n = 9 in Part (b) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Dividing σ by n (not √n) — formula error, loses Part (a) mark.
• In Part (b), solving for √n = 3 then writing n = 3 instead of n = 9 — algebraic error loses the third mark.
• Forgetting that precision improves as √n: some students write n = 3 (tripling the measurements) instead of n = 9.

Key Phrases To Include

• σ_x̄ = σ/√n
• standard deviation of the mean
• n = 9
• improves as √n not n

Marks

3

Topic

Weighted Mean in Leveling

Difficulty

medium

Template Id

T11

Examiner Tip

Notice that Route 2 (shortest, K = 2 km) has the highest weight (0.50) and pulls the weighted mean (15.223) closest to its value (15.218) compared to the simple mean (15.225). Use this check to validate your answer direction.

Model Answer

Given: ΔH₁ = 15.232 m, K₁ = 4 km → w₁ = 1/K₁ = 1/4 = 0.25 ΔH₂ = 15.218 m, K₂ = 2 km → w₂ = 1/K₂ = 1/2 = 0.50 ΔH₃ = 15.225 m, K₃ = 5 km → w₃ = 1/K₃ = 1/5 = 0.20 Rule applied: For differential leveling, w ∝ 1/K Weighted products: w₁ΔH₁ = 0.25 × 15.232 = 3.8080 w₂ΔH₂ = 0.50 × 15.218 = 7.6090 w₃ΔH₃ = 0.20 × 15.225 = 3.0450 Sums: Σw = 0.25 + 0.50 + 0.20 = 0.95 ΣwΔH = 3.8080 + 7.6090 + 3.0450 = 14.4620 Weighted mean: ΔH_w = ΣwΔH / Σw = 14.4620 / 0.95 = 15.223 m Answer: The most probable elevation difference is ΔH = 15.223 m.

Question Type

numerical

Answer Structure

• State the rule w ∝ 1/K and compute each weight [1 mark]
• Compute weighted products wΔH and their sums [1 mark]
• Apply weighted mean formula and state answer with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using w ∝ K (direct proportion) — all subsequent values are wrong, loses all marks.
• Arithmetic error in the division 14.462/0.95 — carry more decimal places to avoid.
• Treating all three routes as equally reliable (simple arithmetic mean) — conceptual error.

Key Phrases To Include

• w ∝ 1/K for leveling
• weighted mean
• ΔH_w = ΣwΔH / Σw
• shorter route has higher weight

Marks

5

Topic

Theory of Errors and Least Squares

Difficulty

hard

Template Id

T12

Examiner Tip

A 5-mark long-answer should fill at least one full exam page with four clear, numbered or headed sections. Use section headings (1. Types, 2. MPV, 3. Least Squares, 4. Conclusion) so the examiner can award marks section-by-section without hunting through a wall of text.

Model Answer

THEORY OF ERRORS IN GEODETIC SURVEYING 1. Classification of Errors Surveying observations are never perfectly accurate. Errors are classified into three types: (a) Blunders (Mistakes) — Gross errors caused by carelessness, incompetence, or equipment failure. Examples: misreading a rod, booking 52.10 m instead of 25.10 m. Blunders have no statistical law and must be detected (by independent checks or gross-error tests) and eliminated before any adjustment is performed. (b) Systematic Errors — Errors that follow a definite mathematical or physical law and have a consistent sign and magnitude. Examples: a steel tape that is 5 mm too short, atmospheric refraction in trigonometric leveling. Systematic errors are corrected by calibration, applying correction formulas, or improving field procedures. (c) Random Errors — Small, residual errors that remain after blunders and systematic errors are removed. They are equally likely to be positive or negative, are normally (Gaussian) distributed, and their magnitude decreases statistically as the number of observations increases. Random errors are the subject of adjustment computations. 2. Most Probable Value (MPV) The MPV is the value of an unknown that is most likely to be the true value, given the available observations. For n equally-reliable direct observations, the MPV is the arithmetic mean: x̄ = Σxᵢ / n For observations of unequal reliability (different weights), the MPV is the weighted mean: x̄_w = Σwᵢxᵢ / Σwᵢ where wᵢ = 1/σᵢ² Residuals are defined as vᵢ = xᵢ − x̄. Note that Σvᵢ = 0 for equally-weighted observations, which serves as an arithmetic check. 3. Principle of Least Squares The method of least squares (introduced by Gauss and Legendre) states that the Most Probable Value of an observed quantity is that value for which the sum of the squares of the residuals is a minimum: Σvᵢ² = Σ(xᵢ − x̄)² = minimum For weighted observations: Σwᵢvᵢ² = minimum. This principle is the theoretical basis for all rigorous geodetic adjustment, including the processing of GNSS baselines on the PRS92 reference frame and the adjustment of PPCS/UTM survey networks. 4. Conclusion Adjustment computations are applicable only after blunders are eliminated and systematic errors are corrected. The remaining random errors are then distributed optimally among the redundant observations by the least squares principle, yielding the MPV of each unknown with a quantifiable precision measure (standard deviation).

Question Type

long_answer

Answer Structure

• Classify and define all three error types with examples [2 marks — 1 for blunders/systematic, 1 for random]
• Define MPV with the arithmetic mean formula for equal weights and weighted mean for unequal weights [1 mark]
• State the principle of least squares: Σv² = minimum (and weighted form) [1 mark]
• Correct use of professional terminology and logical organisation throughout [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the random error discussion or treating it identically to systematic error — loses 1 mark.
• Writing the wrong MPV formula (e.g., median) — loses the MPV mark.
• Defining least squares as 'Σv = minimum' instead of 'Σv² = minimum' — loses the principle mark.
• No examples for any error type — examiner may deduct a mark for insufficient illustration.
• Writing in bullet points without any explanatory sentences — for a 5-mark question, full-sentence prose is expected.

Key Phrases To Include

• blunders must be eliminated
• systematic errors corrected by calibration
• random errors — normally distributed
• MPV — arithmetic mean for equal weights
• MPV — weighted mean for unequal weights
• w = 1/σ²
• Σv² = minimum (principle of least squares)
• Σwv² = minimum

Marks

5

Topic

Standard Deviation, Standard Error of the Mean, and Probable Error

Difficulty

hard

Template Id

T13

Examiner Tip

Always work in the most convenient unit (seconds here, not the full DMS angle). Convert to the full angle only at the final answer step. This reduces arithmetic errors and saves time.

Model Answer

Given: n = 5 angle measurements (convert seconds for computation) For convenience, let x represent the seconds part; the degrees-minutes portion is constant at 45°12'. x₁ = 20", x₂ = 24", x₃ = 18", x₄ = 22", x₅ = 16" Step 1 — MPV (arithmetic mean): x̄ = Σx / n = (20 + 24 + 18 + 22 + 16) / 5 = 100 / 5 = 20" MPV = 45°12'20" Step 2 — Residuals v = x − x̄: v₁ = 20 − 20 = 0" v₂ = 24 − 20 = +4" v₃ = 18 − 20 = −2" v₄ = 22 − 20 = +2" v₅ = 16 − 20 = −4" Check: Σv = 0 + 4 − 2 + 2 − 4 = 0 ✓ Step 3 — Σv²: Σv² = 0² + 4² + (−2)² + 2² + (−4)² = 0 + 16 + 4 + 4 + 16 = 40 Step 4 — Standard deviation of a single observation: σ = √(Σv² / (n−1)) = √(40 / 4) = √10 = 3.162" Step 5 — Standard deviation of the mean: σ_x̄ = σ / √n = 3.162 / √5 = 3.162 / 2.236 = 1.414" Step 6 — Probable error of the mean: PE_mean = 0.6745 × σ_x̄ = 0.6745 × 1.414 = 0.954" Final Answers: (a) MPV = 45°12'20" (b) σ = 3.16" (standard deviation, single observation) (c) σ_x̄ = 1.41" (standard deviation of the mean) (d) PE_mean = 0.95" (probable error of the mean)

Question Type

numerical

Answer Structure

• Compute MPV = 45°12'20" correctly [1 mark]
• Correct residuals with Σv = 0 check and correct Σv² = 40 [1 mark]
• Correct σ = √(Σv²/(n−1)) = 3.16" [1 mark]
• Correct σ_x̄ = σ/√n = 1.41" [1 mark]
• Correct PE_mean = 0.6745 × σ_x̄ = 0.95" [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not showing the Σv = 0 check — minor deduction in some marking schemes.
• Using n = 5 instead of (n−1) = 4 in the σ formula — loses the σ mark.
• Computing PE_mean = 0.6745 × σ (not σ_x̄) — loses the PE mark.
• Rounding σ too early and propagating a rounding error to σ_x̄ and PE_mean.
• Forgetting to attach arcsecond (") units to all final answers.

Key Phrases To Include

• MPV = arithmetic mean
• v = x − x̄
• Σv = 0 (check)
• σ = √(Σv²/(n−1))
• σ_x̄ = σ/√n
• PE_mean = 0.6745 × σ_x̄

Marks

2

Topic

Standard Deviation and Standard Error

Difficulty

easy

Template Id

T14

Examiner Tip

Illustrate the answer with a brief numerical example if space allows (e.g., 'If σ = 0.04 m and n = 4, then σ_x̄ = 0.02 m'). This demonstrates understanding beyond formula recall.

Model Answer

σ (standard deviation of a single observation) measures the spread or precision of individual measurements around the mean. It quantifies how much any single observation may deviate from the true value. σ_x̄ (standard deviation of the mean) measures the precision of the mean value itself. It is given by σ_x̄ = σ/√n. σ_x̄ is always smaller than σ because averaging n observations reduces the random scatter: extreme values in either direction tend to cancel each other out, making the mean a more stable and precise estimator than any single observation. The improvement is proportional to √n.

Question Type

short_answer

Answer Structure

• Define σ (single observation spread) and σ_x̄ (precision of the mean) [1 mark]
• Explain why σ_x̄ < σ using the formula σ_x̄ = σ/√n and the statistical reasoning [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating σ_x̄ = σ/n — wrong formula, loses the first mark.
• Saying only 'because we take more measurements' without explaining the √n relationship.
• Confusing variance with standard deviation.

Key Phrases To Include

• precision of individual observations
• precision of the mean
• σ_x̄ = σ/√n
• random errors tend to cancel
• improves as √n

Marks

3

Topic

Weights Proportional to Number of Observations

Difficulty

medium

Template Id

T15

Examiner Tip

This question rewards examinees who understand where the rules come from. Write the two-step derivation (σ_x̄ = σ/√n, then w = 1/σ_x̄² = n/σ²) explicitly — this is what separates a 3-mark answer from a 1-mark answer.

Model Answer

Principle: When the same instrument and field conditions are used, the standard deviation of the mean of n observations is σ_x̄ = σ/√n. The weight of the mean is inversely proportional to its variance: w ∝ 1/σ_x̄² = n/σ² For equal field precision (same σ per observation), weight is directly proportional to the number of observations: w ∝ n Practical Example: A closed traverse has Side AB measured 4 times (σ_per obs = 5 mm) and Side BC measured only once (σ_per obs = 5 mm). Side AB: σ_x̄ = 5/√4 = 2.5 mm → w_AB = 1/(2.5)² = 1/6.25 = 0.16 (or relative weight = 4) Side BC: σ_x̄ = 5/√1 = 5.0 mm → w_BC = 1/(5.0)² = 1/25 = 0.04 (or relative weight = 1) Relative weights: w_AB : w_BC = 4 : 1 Conclusion: The mean of 4 repeated measurements of AB carries four times the weight of the single measurement of BC, because its mean is twice as precise (σ_x̄ reduced by √4 = 2), and weight scales as 1/σ², hence 2² = 4.

Question Type

short_answer

Answer Structure

• State the principle: w ∝ n when σ per observation is constant [1 mark]
• Derive or state the relationship through σ_x̄ = σ/√n and w = 1/σ_x̄² [1 mark]
• Apply to a concrete numerical example showing correct weight ratio [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating w ∝ √n instead of w ∝ n — incorrect, loses the principle mark.
• Not showing the derivation through σ_x̄ — the rule w ∝ n seems arbitrary without it.
• Using a vague example without actual numbers — loses the third mark.

Key Phrases To Include

• w ∝ n for equal per-observation precision
• σ_x̄ = σ/√n
• w = 1/σ_x̄²
• weight ratio 4:1