23-Ind-B1 Reliability and Maintainability · December 2018

Question 4 of 9: Continuous-Random-Variable PDF — Validity, Mean, Variance

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

Notes on this paper

National Exams — December 2018 — 17-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 (20 marks); Section B: do 2 of 3 (40 marks); Section C: do 1 of 2 (20 marks) — a 9-question, printed-total-80-mark paper as actually structured (the front page's own summary line states “Exam: 5 Questions. Total marks: 100”, but 20+40+20=80 by the section instructions and per-question mark values printed beside each question — the front page's 100 is internally inconsistent with its own section table; the 80-mark reading is used throughout, consistent with the section-by-section split printed on pages 2, 4 and 6). All nine questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — normal distribution and sums of normals (ch. 4–5), point/interval estimation (ch. 8), two-sample hypothesis testing (ch. 9–10), simple linear regression (ch. 11), single-factor ANOVA and multiple comparisons (ch. 13). 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) — Bartlett's test, $2^k$ factorial designs (ch. 3, 6).

Question 4 (Section A.4): Continuous-Random-Variable PDF — Validity, Mean, Variance (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. $f(x)=\dfrac{2x}{R^2}$ on $0<x<R$, and $0$ elsewhere, for a constant $R>0$.

Find. (a) Confirm $f$ is a valid pdf; (b) $E[X]$; (c) $\mathrm{Var}(X)$.

Approach. A valid pdf must be non-negative everywhere and integrate to 1 over its support; the mean and variance follow from $E[X]=\int x f(x)\,dx$ and $\mathrm{Var}(X)=E[X^2]-(E[X])^2$.

(a) Non-negativity and total probability. On $0<x<R$, $f(x)=2x/R^2\ge0$ since $x>0,R^2>0$; $f(x)=0$ elsewhere is trivially non-negative. Total probability: $$\int_0^R \frac{2x}{R^2}\,dx=\frac{2}{R^2}\left[\frac{x^2}{2}\right]_0^R=\frac{2}{R^2}\cdot\frac{R^2}{2}=\boxed{1}$$ Both conditions hold, so $f(x)$ is a valid probability density function.
(b) Mean. $$E[X]=\int_0^R x\cdot\frac{2x}{R^2}\,dx=\frac{2}{R^2}\int_0^R x^2\,dx=\frac{2}{R^2}\cdot\frac{R^3}{3}=\boxed{\dfrac{2R}{3}}$$
(c) Variance. First find $E[X^2]$: $$E[X^2]=\int_0^R x^2\cdot\frac{2x}{R^2}\,dx=\frac{2}{R^2}\int_0^R x^3\,dx=\frac{2}{R^2}\cdot\frac{R^4}{4}=\frac{R^2}{2}$$ Then $$\mathrm{Var}(X)=E[X^2]-(E[X])^2=\frac{R^2}{2}-\left(\frac{2R}{3}\right)^2=\frac{R^2}{2}-\frac{4R^2}{9}=\frac{9R^2-8R^2}{18}=\boxed{\dfrac{R^2}{18}}$$

A linearly-increasing triangular density like this one is a natural model whenever a bounded physical quantity is more likely to take larger values than smaller ones within its range — for instance, the depth of a random surface flaw on a component inspected up to a maximum detectable depth $R$, or a wear dimension that grows roughly proportionally to exposure time before the part is retired at $R$. The result $E[X]=2R/3$, sitting two-thirds of the way to the upper bound, quantifies exactly that upward skew, and the closed-form variance $R^2/18$ makes it easy to compare this shape's spread against a uniform distribution on the same interval ($\mathrm{Var}=R^2/12$ for $U(0,R)$) — the triangular shape is less spread out, since probability mass concentrates increasingly near $R$ rather than spreading evenly across the whole interval.

Final Results
Quantity	Value
$\int_0^R f(x)\,dx$	1 (valid pdf)
$E[X]$	$2R/3$
$\mathrm{Var}(X)$	$R^2/18$

Topic: Continuous probability density functions — validity and moments.

Key relations: $f(x)\ge0$ and $\int_{-\infty}^{\infty}f(x)\,dx=1$; $E[X]=\int xf(x)\,dx$; $\mathrm{Var}(X)=E[X^2]-(E[X])^2$.

Why this works: This triangular-shaped density (linearly increasing from 0 at $x=0$ to $2/R$ at $x=R$) weights larger values of $x$ more heavily than smaller ones, which is why its mean, $2R/3$, sits above the midpoint $R/2$ of the support.

Common pitfall: Computing $\mathrm{Var}(X)=E[(X-E[X])^2]$ by direct expansion and mis-tracking a sign, or forgetting to square $E[X]$ before subtracting — always compute $E[X^2]$ independently as a cross-check ($E[X^2]$ must exceed $(E[X])^2$ for the variance to be positive).

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th ed., §4.1–4.2 (continuous random variables, mean and variance).

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