23-Ind-B1 Reliability and Maintainability · May 2015
Question 1 of 10: Joint Probability Function — Correlation Coefficient and Independence
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 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 questions (10 marks); Section B: do 2 of 3 (20 marks); Section C: do 1 of 3 (20 marks) — a 5-question, 50-mark paper as printed. All ten 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 (ch. 5), binomial/geometric probability (ch. 3), normal distribution and sampling distributions (ch. 4–7), point/interval estimation (ch. 8–9), hypothesis testing incl. sample-size design (ch. 9–10), simple linear regression and ANOVA (ch. 11–13), Bartlett's and Tukey's tests, single-degree-of-freedom contrasts and 23 factorial designs (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).
Question 1 (Section A.1): Joint Probability Function — Correlation Coefficient and Independence (5 marks)
Find. The correlation coefficient $\rho_{xy}$, and what it says about independence.
Approach. Compute $E[X]$, $E[Y]$, $\mathrm{Var}(X)$, $\mathrm{Var}(Y)$ and $E[XY]$ from the table, then $\rho_{xy}=\mathrm{Cov}(X,Y)/\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}$.
Marginal means and variances. $E[X]=\sum x\,p_X(x)=0(0.35)+1(0.25)+2(0.40)=1.05$, $E[X^2]=0+0.25+1.60=1.85$, so $\mathrm{Var}(X)=1.85-1.05^2=0.7475$. Likewise $E[Y]=0(0.35)+1(0.35)+2(0.30)=0.95$, $E[Y^2]=0+0.35+1.20=1.55$, so $\mathrm{Var}(Y)=1.55-0.95^2=0.6475$.
Joint moment. $E[XY]=\sum_{x,y}xy\,p(x,y)$ — only the $(1,1),(1,2),(2,1),(2,2)$ cells contribute: $1(1)(0.15)+1(2)(0.05)+2(1)(0.10)+2(2)(0.05)=0.15+0.10+0.20+0.20=0.65$.
Covariance and correlation. $\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]=0.65-(1.05)(0.95)=-0.3475$. $$\rho_{xy}=\frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)\,\mathrm{Var}(Y)}}=\frac{-0.3475}{\sqrt{0.7475\times 0.6475}}=\boxed{-0.499}$$
Independence check. A quick direct check confirms it: if $X,Y$ were independent, every cell would satisfy $p(x,y)=p_X(x)p_Y(y)$, e.g. cell $(0,0)$ would need $0.35\times 0.35=0.1225$, but the table gives $0.05$ — already a contradiction. The moderate negative correlation is also visible qualitatively in the table itself: probability mass is concentrated where $X$ is large and $Y$ is small (cell $(2,0)=0.25$) and where $X$ is small and $Y$ is large (cell $(0,2)=0.20$), which is exactly the pattern a negative $\rho_{xy}$ describes.
