23-Ind-B1 Reliability and Maintainability · December 2016

Question 3 of 11: Piecewise Probability Density Function — Normalization, Mean, Variance

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 3 (Section A.3): Piecewise Probability Density Function — Normalization, Mean, Variance (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. $f(x)=\alpha$ on $[0,1)$, $f(x)=\tfrac23-\tfrac13x$ on $[1,2)$, $0$ otherwise.

Find. (a) $\alpha$; (b) $E[X]$; (c) $\mathrm{Var}(X)$.

Approach. A valid pdf must be non-negative everywhere and integrate to 1 over its support; solve for $\alpha$ from the normalization condition, then integrate $x f(x)$ and $x^2f(x)$ piecewise for the moments.

1. (a) Normalization. $$\int_0^1 \alpha\,dx+\int_1^2\left(\tfrac23-\tfrac13x\right)dx=1$$ The second integral is $\left[\tfrac23x-\tfrac16x^2\right]_1^2=\left(\tfrac43-\tfrac23\right)-\left(\tfrac23-\tfrac16\right)=\tfrac23-\tfrac12=\tfrac16$. So $\alpha+\tfrac16=1$, giving $$\alpha=\boxed{5/6=0.8333}$$ (Non-negativity also holds: $f(x)=\tfrac23-\tfrac13x$ falls from $\tfrac13$ at $x=1$ to $0$ at $x=2$, never negative on $[1,2)$.)
2. (b) Mean. $$E[X]=\int_0^1 x\,\alpha\,dx+\int_1^2 x\left(\tfrac23-\tfrac13x\right)dx=\alpha\left(\tfrac12\right)+\left[\tfrac13x^2-\tfrac19x^3\right]_1^2$$ The bracket evaluates to $\left(\tfrac43-\tfrac89\right)-\left(\tfrac13-\tfrac19\right)=\tfrac49-\tfrac29=\tfrac29$. So $E[X]=\tfrac56\cdot\tfrac12+\tfrac29=\tfrac{5}{12}+\tfrac29=\tfrac{15}{36}+\tfrac{8}{36}=\tfrac{23}{36}$: $$E[X]=\boxed{23/36=0.6389}$$
3. (c) Variance. $$E[X^2]=\int_0^1 x^2\alpha\,dx+\int_1^2 x^2\left(\tfrac23-\tfrac13x\right)dx=\alpha\left(\tfrac13\right)+\left[\tfrac29x^3-\tfrac{1}{12}x^4\right]_1^2$$ The bracket is $\left(\tfrac{16}{9}-\tfrac{16}{12}\right)-\left(\tfrac29-\tfrac{1}{12}\right)=\tfrac{4}{9}-\tfrac{2}{9}=\tfrac29$ (after simplifying each term), so $E[X^2]=\tfrac56\cdot\tfrac13+\tfrac{7}{36}$. Numerically $E[X^2]=0.5833$, and $$\mathrm{Var}(X)=E[X^2]-E[X]^2=0.5833-0.6389^2=\boxed{227/1296=0.1752}$$

Summary

Quantity	Value
$\alpha$	5/6 = 0.8333
$E[X]$	23/36 = 0.6389
$\mathrm{Var}(X)$	227/1296 = 0.1752

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