23-Ind-B1 Reliability and Maintainability · December 2016

Question 1 of 11: Joint Probability Distribution — Marginals, Mean/SD, Covariance, Correlation, Independence

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2016 — 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: three sections — Section A: do 2 of 4 (30 marks); Section B: do 2 of 3 (30 marks); Section C: do 2 of 4 (40 marks) — a 6-question, 100-mark paper as printed. All eleven questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — joint distributions and covariance (ch. 5), point/interval estimation (ch. 8), hypothesis testing incl. two-sample and goodness-of-fit tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor and factorial experiments (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — two-way factorial ANOVA and $2^k$ designs (ch. 5, 6–7).

Question 1 (Section A.1): Joint Probability Distribution — Marginals, Mean/SD, Covariance, Correlation, Independence (15 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. 100 joint observations of $(X,Y)$, $X,Y\in\{1,2,3\}$:

$Y\backslash X$	1	2	3
1	15	10	10
2	10	15	20
3	5	10	5

Find. (a) $p_X(x)$; (b) $p_Y(y)$; (c) $\mu_X,\sigma_X$; (d) $\mathrm{Cov}(X,Y)$; (e) $\rho_{XY}$; (f) independence.

Approach. Divide every cell by $N=100$ to get the joint pmf, sum rows/columns for the marginals, then apply the definitional moment formulas.

1. (a)/(b) Marginal pmfs. Summing columns for $X$ and rows for $Y$ (each count $/100$): $$p_X(1)=0.30,\quad p_X(2)=0.35,\quad p_X(3)=0.35$$ $$p_Y(1)=0.35,\quad p_Y(2)=0.45,\quad p_Y(3)=0.20$$
2. (c) Mean and SD of $X$. $\mu_X=\sum x\,p_X(x)=1(0.30)+2(0.35)+3(0.35)=2.05$. $E[X^2]=1(0.30)+4(0.35)+9(0.35)=2.7975$, so $\sigma_X^2=2.7975-2.05^2=0.6475$ and $$\sigma_X=\sqrt{0.6475}=\boxed{0.8047}$$
3. (d) Covariance. $\mu_Y=1(0.35)+2(0.45)+3(0.20)=1.85$. $E[XY]=\sum_{x,y}xy\,p(x,y)=3.85$ (summing all nine cells). $$\mathrm{Cov}(X,Y)=E[XY]-\mu_X\mu_Y=3.85-(2.05)(1.85)=\boxed{0.0575}$$
4. (e) Correlation. $E[Y^2]=1(0.35)+4(0.45)+9(0.20)=4.15$, $\sigma_Y^2=4.15-1.85^2=0.5275$, $\sigma_Y=0.7263$. $$\rho_{XY}=\frac{\mathrm{Cov}(X,Y)}{\sigma_X\sigma_Y}=\frac{0.0575}{0.8047\times 0.7263}=\boxed{0.0984}$$
5. (f) Independence. Under independence every cell must equal $p_X(x)p_Y(y)$. Cell $(X{=}1,Y{=}1)$: observed $p(1,1)=15/100=0.15$, but $p_X(1)p_Y(1)=0.30\times 0.35=0.105\ne 0.15$. $$\boxed{\text{Not independent}}$$ (several other cells disagree the same way — e.g. $p(2,3)=0.20$ vs. $p_X(2)p_Y(3)=0.35\times0.20=0.07$).

Summary

Quantity	Value
$p_X(1,2,3)$	0.30, 0.35, 0.35
$p_Y(1,2,3)$	0.35, 0.45, 0.20
$\mu_X$, $\sigma_X$	2.05, 0.8047
$\mathrm{Cov}(X,Y)$	0.0575
$\rho_{XY}$	0.0984
Independent?	No

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