GELE Adjustment Computations (Least Squares) — Error Propagation, Variance-Covariance and Error EllipsesExam Answer Templates

Marks

1

Topic

Law of Propagation of Variances

Difficulty

easy

Template Id

T1

Examiner Tip

The examiner checks for the squared partial derivative (∂f/∂xᵢ)² — this is the most commonly dropped element. Write the formula exactly as given and include the summation sign.

Model Answer

For a function y = f(x₁, x₂, …, xₙ) of statistically independent measured quantities x₁, x₂, …, xₙ, the variance of y is: σ_y² = (∂f/∂x₁)²σ₁² + (∂f/∂x₂)²σ₂² + … + (∂f/∂xₙ)²σₙ² where ∂f/∂xᵢ are the partial derivatives of f with respect to each measured quantity.

Question Type

very_short_answer

Answer Structure

• State the formula σ_y² = Σ(∂f/∂xᵢ)²σᵢ² with correct notation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing σ_y = Σ(∂f/∂xᵢ)σᵢ instead of squaring the partial derivatives and the standard deviations
• Omitting the independence condition, which justifies the sum form
• Confusing variance (σ²) with standard deviation (σ) in the formula

Key Phrases To Include

• Law of Propagation of Variances
• partial derivative
• σ_y² = Σ(∂f/∂xᵢ)²σᵢ²
• independent measured quantities

Marks

1

Topic

Error Propagation — Sum of Quantities

Difficulty

easy

Template Id

T2

Examiner Tip

For a 1-mark numerical, write the formula and the final answer — no elaborate working needed. The mark is for both the correct method (quadrature) and correct number.

Model Answer

Given: σ₁ = σ₂ = 5" For the sum y = x₁ + x₂ (independent quantities): σ_y = √(σ₁² + σ₂²) = √(5² + 5²) = √(25 + 25) = √50 σ_y = 7.07"

Question Type

numerical

Answer Structure

• State the formula for the sum [0.5 mark]
• Substitute values and compute correct numerical answer with unit ["] [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Adding standard deviations linearly: 5 + 5 = 10" (wrong — this assumes perfect correlation)
• Omitting the arc-second (") unit on the final answer
• Not writing the formula before substituting

Key Phrases To Include

• add in quadrature
• √(σ₁² + σ₂²)
• 7.07"
• independent

Marks

2

Topic

Error Propagation — Sum vs. Mean

Difficulty

medium

Template Id

T3

Examiner Tip

The examiner specifically looks for both formulas AND the context where each applies — a formula alone without context earns only half the mark at this level.

Model Answer

Variance of the sum (y = x₁ + x₂ + … + xₙ): σ_sum = σ√n This applies when the total value is the accumulation of n independent measurements, e.g., the total length of a traverse leg taped in n equal bays. Variance of the mean (ȳ = (x₁ + x₂ + … + xₙ)/n): σ_mean = σ/√n This applies when a single quantity is measured n times and averaged to improve precision, e.g., repeated distance measurements with an EDM.

Question Type

short_answer

Answer Structure

• State the formula σ_sum = σ√n with a correct example [1 mark]
• State the formula σ_mean = σ/√n with a correct example, clearly distinguishing from the sum [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing the two formulas — this is the most critical conceptual error in this chapter
• Writing one formula and claiming it applies to both situations
• No engineering example to contextualise each formula

Key Phrases To Include

• σ√n (sum of n equal observations)
• σ/√n (mean of n observations)
• accumulation
• repeated measurement
• independent

Marks

2

Topic

Error Propagation — Product of Quantities

Difficulty

medium

Template Id

T4

Examiner Tip

Always compute and label ∂A/∂L and ∂A/∂W explicitly before substituting — examiners award a mark for the correct partial derivatives alone, independent of the arithmetic.

Model Answer

Given: L = 80.00 m, σ_L = 0.03 m; W = 40.00 m, σ_W = 0.02 m Function: A = L × W Partial derivatives: ∂A/∂L = W = 40.00 m ∂A/∂W = L = 80.00 m Law of Propagation of Variances: σ_A² = (∂A/∂L)²σ_L² + (∂A/∂W)²σ_W² σ_A² = (40.00)²(0.03)² + (80.00)²(0.02)² σ_A² = 1600(0.0009) + 6400(0.0004) σ_A² = 1.44 + 2.56 = 4.00 m⁴ σ_A = √4.00 = 2.00 m²

Question Type

numerical

Answer Structure

• Write the function A = L × W and compute both partial derivatives [0.5 mark]
• Write the propagation formula with both terms [0.5 mark]
• Substitute numerical values correctly [0.5 mark]
• State correct final answer 2.00 m² with proper unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using σ_L and σ_W as the partial derivatives instead of W and L
• Writing the unit as m instead of m² for area standard deviation
• Computing σ_A² = (∂A/∂L)σ_L + (∂A/∂W)σ_W (forgetting to square)

Key Phrases To Include

• ∂A/∂L = W
• ∂A/∂W = L
• Law of Propagation of Variances
• σ_A² = (W)²σ_L² + (L)²σ_W²
• 2.00 m²

Marks

2

Topic

Variance-Covariance Matrix

Difficulty

medium

Template Id

T5

Examiner Tip

Examiners specifically check for σ₀² as the multiplier — its absence loses a half-mark. Also identify each matrix symbol to show you understand the least-squares framework.

Model Answer

The variance-covariance matrix of adjusted parameters, denoted Σ_x̂, is a square symmetric matrix that fully describes the precision of and correlation among the adjusted unknown parameters from a least-squares adjustment. Formula: Σ_x̂ = σ₀² (AᵀPA)⁻¹ where: • σ₀² = a posteriori reference variance (estimated unit-weight variance) • A = design matrix (coefficient matrix of observation equations) • P = weight matrix of the observations • (AᵀPA)⁻¹ = inverse of the normal equation matrix Diagonal elements = variances of individual adjusted parameters (σ_x², σ_y², …) Off-diagonal elements = covariances between pairs of adjusted parameters (σ_xy, …)

Question Type

short_answer

Answer Structure

• Define Σ_x̂ in words as a symmetric matrix describing precision and correlation [0.5 mark]
• Write the formula Σ_x̂ = σ₀²(AᵀPA)⁻¹ [0.5 mark]
• Identify each symbol in the formula [0.5 mark]
• State what diagonal and off-diagonal elements represent [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing Σ_x̂ = (AᵀPA)⁻¹ without the σ₀² scalar multiplier
• Not identifying what the diagonal and off-diagonal elements represent
• Confusing A (design matrix) with the normal equation matrix AᵀPA

Key Phrases To Include

• variance-covariance matrix
• Σ_x̂ = σ₀²(AᵀPA)⁻¹
• a posteriori reference variance
• design matrix
• weight matrix
• diagonal = variances
• off-diagonal = covariances

Marks

2

Topic

Error Propagation — Sum of n Equal Observations

Difficulty

easy

Template Id

T6

Examiner Tip

State explicitly that you are computing the error of a SUM (not a mean) before writing the formula — this explanation is what separates a 2/2 from a 1/2 answer.

Model Answer

Given: n = 6 segments, σ per segment = 0.010 m The total distance D = d₁ + d₂ + … + d₆ is a sum of n independent equal observations. Applying the propagation formula for a sum of n equal independent observations: σ_D = σ√n = 0.010 × √6 = 0.010 × 2.4495 σ_D = 0.0245 m ≈ 24.5 mm

Question Type

numerical

Answer Structure

• Identify that total distance is a SUM of n independent segments (not a mean) [0.5 mark]
• Write the formula σ_D = σ√n [0.5 mark]
• Substitute: σ√n = 0.010√6 [0.5 mark]
• State correct answer 0.0245 m with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using σ/√n = 0.010/√6 = 0.00408 m — this is the error of the mean, not the total
• Not explaining why σ√n applies (it is a sum, not an average)
• Incorrect arithmetic for √6

Key Phrases To Include

• sum of n independent equal observations
• σ_D = σ√n
• 0.0245 m
• error accumulates

Marks

3

Topic

Error Ellipse

Difficulty

medium

Template Id

T7

Examiner Tip

For a 3-mark question divided into three parts, allocate one clear paragraph per part — examiners mark by part, so a well-structured answer with three distinct blocks maximises marks even if one part is weak.

Model Answer

(a) Geometric meaning: The standard error ellipse is an ellipse centred on a computed 2-D position (e.g., a UTM coordinate pair in PRS92/PPCS) that represents the locus of equal uncertainty of that position. Its semi-major axis a and semi-minor axis b are derived from the eigenvalues of the 2×2 position sub-matrix of the variance-covariance matrix Σ_x̂. The orientation angle θ of the major axis (measured from the North or Easting axis) is determined by the eigenvectors of that sub-matrix. (b) Probability level: The standard error ellipse corresponds to approximately 39% probability — meaning there is a 39% chance that the true position lies within the ellipse. For practical engineering and cadastral survey reporting (e.g., under PD 1529 or RA 8560 standards), the ellipse is scaled by a factor k = 2.45 to achieve the 95% confidence ellipse. (c) Elongated ellipse: An elongated error ellipse (a >> b) indicates weak survey geometry in the direction of the major axis. This means the network configuration — the arrangement of control points and observation directions — provides poor constraint in that direction. In a PPCS-based horizontal control network, this typically arises from unfavorable azimuth distribution of baselines or a lack of redundant observations in a particular direction.

Question Type

short_answer

Answer Structure

• (a) Define the standard error ellipse: centre, semi-axes from eigenvalues of covariance sub-matrix, orientation from eigenvectors [1 mark]
• (b) State 39% probability for standard ellipse; mention 95% confidence ellipse with k = 2.45 scale factor [1 mark]
• (c) Interpret elongated ellipse as weak geometry/poor constraint in the direction of the major axis [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the standard ellipse represents 68% or 95% probability — both are incorrect; it is ~39%
• Confusing error ellipse with error circle (circle only applies when σ_x = σ_y and σ_xy = 0)
• No reference to eigenvalues/eigenvectors — describing the ellipse without its mathematical origin loses the first mark
• Not linking an elongated ellipse to network geometry in an engineering context

Key Phrases To Include

• eigenvalues of covariance sub-matrix
• semi-major axis a, semi-minor axis b
• orientation angle θ
• 39% probability
• 95% confidence ellipse
• scale factor k = 2.45
• weak geometry
• poor constraint

Marks

3

Topic

Matrix Form of Error Propagation

Difficulty

medium

Template Id

T8

Examiner Tip

The most common error is reversing J and Jᵀ. Remember: J is on the left, Jᵀ is on the right — the formula is analogous to the change-of-basis rule in linear algebra.

Model Answer

Matrix Form of Error Propagation: Formula: Σ_y = J Σ_x Jᵀ where: • Σ_y = variance-covariance matrix of the output quantities y (m×m matrix for m outputs) • J = Jacobian matrix of partial derivatives, with Jᵢⱼ = ∂yᵢ/∂xⱼ (m×n matrix for m outputs and n inputs) • Σ_x = variance-covariance matrix of the input quantities x (n×n matrix) • Jᵀ = transpose of the Jacobian Derivation logic: If y = Jx (linearised relation), then by covariance propagation rules, Σ_y = J Σ_x Jᵀ. Advantage over scalar form: 1. Handles correlated inputs automatically — the scalar form assumes independent inputs (σ_xy = 0), but the matrix form propagates off-diagonal covariance terms through the product J Σ_x Jᵀ. 2. Simultaneously propagates multiple output quantities, yielding a complete Σ_y matrix that includes both the output variances (diagonal) and output covariances (off-diagonal). 3. More compact and computationally efficient for large systems, as used in least-squares network adjustments.

Question Type

short_answer

Answer Structure

• State the formula Σ_y = J Σ_x Jᵀ with correct dimensions/notation [1 mark]
• Define all terms: J (Jacobian with ∂yᵢ/∂xⱼ elements), Σ_x, Σ_y, Jᵀ [1 mark]
• State at least two clear advantages over the scalar form, particularly the handling of correlations [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing Σ_y = Jᵀ Σ_x J (wrong order of transpose — must be J on left, Jᵀ on right)
• Not defining J as the Jacobian of partial derivatives
• Stating only one advantage without elaborating on the correlation-handling capability

Key Phrases To Include

• Σ_y = J Σ_x Jᵀ
• Jacobian matrix
• Jᵢⱼ = ∂yᵢ/∂xⱼ
• correlated inputs
• off-diagonal covariance
• multiple output quantities

Marks

3

Topic

Matrix Form of Error Propagation with Correlation

Difficulty

hard

Template Id

T9

Examiner Tip

The presence of a non-zero covariance σ_LW in the problem statement is a signal to use the full matrix form — examiners include it specifically to test whether you can handle correlated inputs.

Model Answer

Given: L = 60 m, W = 30 m; σ_L = 0.04 m, σ_W = 0.03 m, σ_LW = +0.001 m² Function: A = f(L, W) = L × W Step 1 — Jacobian (row vector): J = [∂A/∂L ∂A/∂W] = [W L] = [30 60] Step 2 — Input covariance matrix: Σ_x = | σ_L² σ_LW | = | 0.0016 0.001 | | σ_LW σ_W² | | 0.001 0.0009| Step 3 — Apply Σ_A = J Σ_x Jᵀ: First compute J Σ_x: [30 60] × |0.0016 0.001 | = [30(0.0016) + 60(0.001) , 30(0.001) + 60(0.0009)] |0.001 0.0009| [0.048 + 0.060 , 0.030 + 0.054 ] = [0.108 , 0.084] Then Σ_A = (J Σ_x) × Jᵀ = [0.108 0.084] × |30| = 0.108(30) + 0.084(60) |60| = 3.240 + 5.040 = 8.280 m⁴ Step 4 — Standard deviation: σ_A = √8.280 = 2.877 m²

Question Type

numerical

Answer Structure

• Compute Jacobian J = [W L] = [30 60] [0.5 mark]
• Write the 2×2 input covariance matrix Σ_x with σ_LW in off-diagonal [0.5 mark]
• Apply Σ_A = J Σ_x Jᵀ with correct matrix multiplication showing intermediate steps [1 mark]
• State σ_A = √8.280 = 2.877 m² with correct unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Ignoring σ_LW = 0.001 and computing as if inputs are independent (this changes the answer to √(σ_L²W² + σ_W²L²))
• Placing σ_LW on the diagonal instead of the off-diagonal positions
• Using J as a column vector instead of a row vector, leading to wrong matrix dimensions

Key Phrases To Include

• J = [∂A/∂L ∂A/∂W] = [W L]
• Σ_A = J Σ_x Jᵀ
• off-diagonal covariance σ_LW
• 2.877 m²

Marks

5

Topic

Error Ellipse — Eigenvalue Method

Difficulty

hard

Template Id

T10

Examiner Tip

Show the quadrant determination of θ explicitly — write the signs of the numerator and denominator of tan(2θ) and state which quadrant 2θ falls in. Examiners deduct for a correct magnitude but wrong quadrant.

Model Answer

Given: Σ = |σ_x² σ_xy| = |0.0025 0.0010| |σ_xy σ_y²| |0.0010 0.0049| (a) Semi-axes of the standard error ellipse: The semi-axes are derived from the eigenvalues λ of the covariance matrix: Characteristic equation: λ² - (σ_x² + σ_y²)λ + (σ_x²σ_y² - σ_xy²) = 0 Trace: σ_x² + σ_y² = 0.0025 + 0.0049 = 0.0074 Determinant: σ_x²σ_y² - σ_xy² = (0.0025)(0.0049) - (0.0010)² = 0.00001225 - 0.0000100 = 0.00000225 λ² - 0.0074λ + 0.00000225 = 0 Using the quadratic formula: λ = [0.0074 ± √(0.0074² - 4 × 0.00000225)] / 2 λ = [0.0074 ± √(0.00005476 - 0.00000900)] / 2 λ = [0.0074 ± √0.00004576] / 2 λ = [0.0074 ± 0.006765] / 2 λ₁ = (0.0074 + 0.006765) / 2 = 0.014165 / 2 = 0.007082 m² λ₂ = (0.0074 - 0.006765) / 2 = 0.000635 / 2 = 0.000318 m² Semi-major axis: a = √λ₁ = √0.007082 = 0.0841 m ≈ 84.1 mm Semi-minor axis: b = √λ₂ = √0.000318 = 0.0178 m ≈ 17.8 mm (b) Orientation angle of the major axis: Using the formula: tan(2θ) = 2σ_xy / (σ_x² - σ_y²) = 2(0.0010) / (0.0025 - 0.0049) = 0.0020 / (-0.0024) = -0.8333 2θ = arctan(-0.8333) = -39.81° → adjusted to the correct quadrant Since σ_xy > 0 and (σ_x² - σ_y²) < 0, the major axis lies in the second/fourth quadrant. 2θ = 180° - 39.81° = 140.19° θ = 70.1° (orientation of major axis from the x-axis / Easting direction) Conclusion: Standard error ellipse: a = 84.1 mm, b = 17.8 mm, θ = 70.1° from Easting

Question Type

numerical

Answer Structure

• Identify σ_x², σ_y², σ_xy from the given matrix [0.5 mark]
• Write the characteristic equation for eigenvalues [0.5 mark]
• Compute trace and determinant correctly [0.5 mark]
• Solve quadratic and obtain λ₁ = 0.007082 m² and λ₂ = 0.000318 m² [1 mark]
• Compute semi-axes: a = √λ₁ ≈ 84.1 mm, b = √λ₂ ≈ 17.8 mm [1 mark]
• Apply tan(2θ) formula and compute orientation angle θ ≈ 70.1° with quadrant check [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using a = λ₁ and b = λ₂ (eigenvalues directly as axes) instead of taking the square roots
• Omitting the quadrant check for θ — arctan gives values in (-90°, 90°) and may need adjustment
• Swapping σ_x² - σ_y² and σ_y² - σ_x² in the tan(2θ) formula
• Not stating the units (metres or millimetres) of the semi-axes in the final answer
• Computing the 95% confidence ellipse axes without being asked (standard ellipse uses eigenvalues directly)

Key Phrases To Include

• eigenvalues of the covariance matrix
• characteristic equation λ² - (trace)λ + (determinant) = 0
• semi-major axis a = √λ₁
• semi-minor axis b = √λ₂
• tan(2θ) = 2σ_xy/(σ_x² - σ_y²)
• orientation angle θ
• quadrant check

Marks

1

Topic

Error Ellipse — Probability and Confidence Level

Difficulty

easy

Template Id

T11

Examiner Tip

This is a factual 1-mark question — memorise both values (39% and k = 2.45) as a pair. The examiner expects both in a single sentence.

Model Answer

The standard error ellipse covers approximately 39% probability. To obtain the 95% confidence ellipse, the semi-axes a and b are multiplied by the scale factor k = 2.45, giving: a₉₅ = 2.45a and b₉₅ = 2.45b.

Question Type

very_short_answer

Answer Structure

• State 39% for the standard ellipse AND state k = 2.45 for the 95% ellipse [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 68% (the 1-sigma probability for a 1-D normal distribution, not the 2-D ellipse)
• Stating 95% as the probability of the standard error ellipse without the scaling step
• Providing only the probability without the scale factor, or only the scale factor without the probability

Key Phrases To Include

• 39% probability
• standard error ellipse
• k = 2.45
• 95% confidence ellipse

Marks

1

Topic

Error Ellipse — Geometric Interpretation

Difficulty

easy

Template Id

T12

Examiner Tip

Always frame your answer in terms of geometry and observation distribution — these are the engineering-context terms that earn the mark.

Model Answer

A near-circular error ellipse (a ≈ b) indicates well-conditioned geometry — the network provides approximately equal precision in all directions. An elongated ellipse (a >> b) indicates weak geometry and poor constraint in the direction of the major axis, meaning observations are insufficient or poorly distributed in that direction.

Question Type

very_short_answer

Answer Structure

• Contrast near-circular (well-conditioned, equal precision all directions) vs. elongated (weak geometry, poor constraint in major-axis direction) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating a circular ellipse is 'perfect' without linking to geometry or observation distribution
• Only describing one case (either circular or elongated) without comparing both

Key Phrases To Include

• well-conditioned geometry
• equal precision in all directions
• weak geometry
• poor constraint
• major axis direction

Marks

2

Topic

Error Propagation — Sum of Unequal Observations

Difficulty

medium

Template Id

T13

Examiner Tip

When σᵢ values are unequal, always use the full √(Σσᵢ²) formula — the σ√n shortcut only applies when all partial standard deviations are identical.

Model Answer

(a) Total distance: D = D₁ + D₂ + D₃ = 200 + 150 + 250 = 600.00 m (b) Standard deviation of D: Since D₁, D₂, D₃ are independent: σ_D = √(σ₁² + σ₂² + σ₃²) σ_D = √(0.02² + 0.03² + 0.04²) σ_D = √(0.0004 + 0.0009 + 0.0016) σ_D = √0.0029 σ_D = 0.0539 m ≈ 53.9 mm

Question Type

numerical

Answer Structure

• (a) Compute D = 200 + 150 + 250 = 600.00 m [0.5 mark]
• (b) Write formula σ_D = √(σ₁² + σ₂² + σ₃²) [0.5 mark]
• Substitute and compute correctly: σ_D = 0.0539 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding standard deviations linearly: 0.02 + 0.03 + 0.04 = 0.09 m
• Using σ√n formula (valid only when all σᵢ are equal, which they are not here)
• Forgetting to take the square root of the sum of squares

Key Phrases To Include

• independent segments
• √(σ₁² + σ₂² + σ₃²)
• add in quadrature
• 0.0539 m

Marks

3

Topic

Error Propagation — Scaled Product

Difficulty

medium

Template Id

T14

Examiner Tip

Always differentiate the full function, including any scalar multipliers like the 1/2 in the triangle area formula — a common mistake is to drop this coefficient when computing partial derivatives.

Model Answer

Given: A = (1/2)bh; b = 40.00 m (σ_b = 0.05 m), h = 25.00 m (σ_h = 0.04 m) Step 1 — Partial derivatives: ∂A/∂b = (1/2)h = (1/2)(25.00) = 12.50 m ∂A/∂h = (1/2)b = (1/2)(40.00) = 20.00 m Step 2 — Law of Propagation of Variances (independent quantities): σ_A² = (∂A/∂b)²σ_b² + (∂A/∂h)²σ_h² σ_A² = (12.50)²(0.05)² + (20.00)²(0.04)² σ_A² = 156.25 × 0.0025 + 400.00 × 0.0016 σ_A² = 0.390625 + 0.640000 σ_A² = 1.030625 m⁴ Step 3 — Standard deviation: σ_A = √1.030625 = 1.015 m² Check: Area = (1/2)(40)(25) = 500.00 m²; σ_A/A = 1.015/500 = 0.203% (relative precision)

Question Type

numerical

Answer Structure

• Compute both partial derivatives: ∂A/∂b = h/2 = 12.50 m and ∂A/∂h = b/2 = 20.00 m [1 mark]
• Write and apply propagation formula: σ_A² = (12.50)²(0.05)² + (20.00)²(0.04)² [1 mark]
• Compute σ_A = 1.015 m² with correct unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing ∂A/∂b = h instead of (1/2)h (forgetting the 1/2 coefficient in the function)
• Reporting the unit as m instead of m²
• Squaring the standard deviations but forgetting to square the partial derivatives

Key Phrases To Include

• ∂A/∂b = (1/2)h
• ∂A/∂h = (1/2)b
• σ_A² = (∂A/∂b)²σ_b² + (∂A/∂h)²σ_h²
• 1.015 m²

Marks

5

Topic

Variance-Covariance Matrix and Correlation

Difficulty

hard

Template Id

T15

Examiner Tip

PRC board exam questions on this topic frequently require computation of ρ_xy — memorise ρ_xy = σ_xy/(σ_x σ_y) and note that |ρ| ≤ 1. A ρ_xy near ±1 means the coordinate errors are highly linearly dependent; near 0 means nearly independent.

Model Answer

Given: σ₀² = 4 mm² = 4 × 10⁻⁶ m²; Q = |0.50 0.15| |0.15 0.80| (a) Variance-covariance matrix: Σ = σ₀² × Q = (4 × 10⁻⁶) × |0.50 0.15| |0.15 0.80| Σ = |(4×10⁻⁶)(0.50) (4×10⁻⁶)(0.15)| = |2.00×10⁻⁶ 0.60×10⁻⁶| |(4×10⁻⁶)(0.15) (4×10⁻⁶)(0.80)| |0.60×10⁻⁶ 3.20×10⁻⁶| In mm²: Σ = |2.00 0.60| mm² |0.60 3.20| (b) Standard deviations: σ_x = √(Σ₁₁) = √(2.00 mm²) = 1.414 mm ≈ 1.4 mm σ_y = √(Σ₂₂) = √(3.20 mm²) = 1.789 mm ≈ 1.8 mm (c) Correlation coefficient: σ_xy = Σ₁₂ = 0.60 mm² (positive off-diagonal) Since σ_xy = +0.60 mm² > 0, x and y are positively correlated. ρ_xy = σ_xy / (σ_x × σ_y) = 0.60 / (1.414 × 1.789) = 0.60 / 2.529 = 0.237 Conclusion: Station P₁ has σ_x = 1.4 mm, σ_y = 1.8 mm, and the coordinates are weakly positively correlated (ρ_xy = 0.237).

Question Type

numerical

Answer Structure

• (a) Multiply σ₀² × Q element-by-element to get Σ with correct values and units [1.5 marks]
• (b) Compute σ_x = √(2.00 mm²) = 1.414 mm and σ_y = √(3.20 mm²) = 1.789 mm [1 mark]
• (c) State positive correlation from sign of σ_xy, and compute ρ_xy = 0.237 using the formula [2.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not multiplying by σ₀² — treating Q directly as the covariance matrix
• Taking σ_x = Σ₁₁ (the variance) instead of σ_x = √Σ₁₁ (the standard deviation)
• Confusing the sign of σ_xy for correlation direction — off-diagonal is always the covariance, not the correlation coefficient
• Not stating the engineering interpretation (weak/strong, positive/negative) alongside the numerical ρ value

Key Phrases To Include

• Σ = σ₀² × Q
• diagonal elements = variances
• off-diagonal = covariance σ_xy
• σ_x = √(Σ₁₁)
• ρ_xy = σ_xy/(σ_x × σ_y)
• positively correlated
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