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23-Ind-B1 Reliability and Maintainability · December 2014

Question 2 of 9: Joint Distribution — Covariance and Correlation

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Notes on this paper

National Exams — December 2014 — 98-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: four sections — Section A: do 2 of 3 questions (20 marks); Section B: do 1 of 2 (25 marks); Section C: do 1 of 2 (25 marks); Section D: do 1 of 2 (30 marks) — a 5-question, 100-mark paper as printed. All nine questions across the four sections are solved below for completeness. Page 1's own NOTES list is mis-numbered (two items both labelled "4."), transcribed as printed.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — point/interval estimation (ch. 8–9), hypothesis testing (ch. 9–10), simple/multiple linear regression (ch. 11–12), single- and two-factor ANOVA (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).

Question 2 (Section A.1): Joint Distribution — Covariance and Correlation (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The joint pmf $f(x,y)=P(X=x,Y=y)$, $x,y\in\{1,2,3\}$:

$y \backslash x$123
10.050.050.10
20.050.100.35
30.000.200.10

Find. (a) $\mathrm{Cov}(X,Y)$; (b) $\rho_{X,Y}=\mathrm{Corr}(X,Y)$.

Approach. Sum the table to confirm it is a valid joint pmf, reduce to the marginals $P(X)$ and $P(Y)$, then build $\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]$ and $\rho_{X,Y}=\mathrm{Cov}(X,Y)/(\sigma_X\sigma_Y)$.

  1. Part (a) — marginals and means. The nine cells sum to $1.00$, confirming a valid joint pmf. Summing columns gives $P(X{=}1)=0.10$, $P(X{=}2)=0.35$, $P(X{=}3)=0.55$; summing rows gives $P(Y{=}1)=0.20$, $P(Y{=}2)=0.50$, $P(Y{=}3)=0.30$. Then $$E[X]=1(0.10)+2(0.35)+3(0.55)=2.45,\qquad E[Y]=1(0.20)+2(0.50)+3(0.30)=2.10.$$
  2. Part (a) — second moments and covariance. $E[X^2]=1^2(0.10)+2^2(0.35)+3^2(0.55)=6.45$ gives $\mathrm{Var}(X)=6.45-2.45^2=0.4475$; $E[Y^2]=1^2(0.20)+2^2(0.50)+3^2(0.30)=4.90$ gives $\mathrm{Var}(Y)=4.90-2.10^2=0.4900$. Summing $xy\cdot f(x,y)$ over all nine cells gives $E[XY]=5.15$, so $$\mathrm{Cov}(X,Y) = E[XY]-E[X]E[Y] = 5.15-(2.45)(2.10) = \boxed{0.0050}.$$
  3. Part (b) — correlation. $$\rho_{X,Y} = \frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)\,\mathrm{Var}(Y)}} = \frac{0.0050}{\sqrt{(0.4475)(0.4900)}} = \boxed{0.0107}.$$
QuantityResult
$E[X]$, $E[Y]$2.45, 2.10
$\mathrm{Var}(X)$, $\mathrm{Var}(Y)$0.4475, 0.4900
$\mathrm{Cov}(X,Y)$0.0050
$\rho_{X,Y}$0.0107
Check: $\rho_{X,Y}=0.0107$ is very close to zero, meaning $X$ and $Y$ are almost uncorrelated (no meaningful linear association) even though the joint table is clearly not the product of independent marginals (e.g. $f(1,3)=0.00 \ne P(X{=}1)P(Y{=}3)=0.10\times0.30=0.030$) — a near-zero correlation does not imply independence, only the absence of a linear relationship.