23-Ind-B1 Reliability and Maintainability · December 2017

Question 4 of 10: Piecewise-Linear Density — Normalization, Mean, Variance

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

Notes on this paper

National Exams — December 2017 — 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 3 (40 marks) — a 6-question, 100-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) — discrete/continuous distributions (ch. 3–4), joint distributions (ch. 5), point/interval estimation and sample size (ch. 8), hypothesis testing incl. two-sample tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor experiments (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — multi-factor and $2^k$ factorial designs (ch. 5–6).

Question 4 (Section A.4): Piecewise-Linear Density — 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. $f(x)=2(x+2)/c$ on $1\lt x\lt2$, and $0$ elsewhere.

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

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

1. (a) Normalization. On $(1,2)$, $2(x+2)>0$ for all $x$, so non-negativity requires only $c>0$. Setting the integral to 1: $$\int_1^2\frac{2(x+2)}{c}\,dx=1\ \Longrightarrow\ \int_1^2 2(x+2)\,dx=c$$ $$\int_1^2 2(x+2)\,dx=\Big[x^2+4x\Big]_1^2=(4+8)-(1+4)=12-5=7\ \Longrightarrow\ \boxed{c=7}$$
2. (b) Mean. With $c=7$, $f(x)=\tfrac{2}{7}(x+2)$: $$E[X]=\int_1^2 x\cdot\frac{2(x+2)}{7}\,dx=\frac{2}{7}\int_1^2(x^2+2x)\,dx=\frac{2}{7}\left[\frac{x^3}{3}+x^2\right]_1^2$$ The bracket is $\left(\tfrac83+4\right)-\left(\tfrac13+1\right)=\tfrac73+3=\tfrac{16}{3}$, so $$E[X]=\frac27\cdot\frac{16}{3}=\boxed{\frac{32}{21}=1.5238}$$
3. (c) Variance. $$E[X^2]=\int_1^2 x^2\cdot\frac{2(x+2)}{7}\,dx=\frac27\int_1^2(x^3+2x^2)\,dx=\frac27\left[\frac{x^4}{4}+\frac{2x^3}{3}\right]_1^2$$ The bracket is $\left(4+\tfrac{16}{3}\right)-\left(\tfrac14+\tfrac23\right)=\tfrac{28}{3}-\tfrac{11}{12}=\tfrac{101}{12}$, so $E[X^2]=\tfrac27\cdot\tfrac{101}{12}=\tfrac{101}{42}=2.4048$. $$\mathrm{Var}(X)=E[X^2]-\big(E[X]\big)^2=\frac{101}{42}-\left(\frac{32}{21}\right)^2=\boxed{\frac{73}{882}=0.0828}$$

Summary

Quantity	Value
$c$	7
$E[X]$	32/21 = 1.5238
$\mathrm{Var}(X)$	73/882 = 0.0828

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