Question 1 of 10: Continuous Distribution — Valid pdf, Mean and Variance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2019 — 17-Ind-A6, Systems Simulation, 3 hours duration. Section A: do 3 of 5 (30 marks); Section B: do 1 of 2 per the front-page summary table — the Part B section instructions on the page itself read “do 2 of 3” instead, a genuine inconsistency in the source paper's own front matter (flagged below); Section C: do 1 of 2 (20 marks). This study resource answers all ten questions across the three parts.
Check — front-page/section instructions disagree. The cover page's summary table states “Section B: Do 1 of 2 Questions, Total marks: 20,” but Part B's own section banner (page 4) reads “Complete two of the following three sets of questions,” and Part B in fact contains three questions worth 15 marks each. The cover page's own overall total (70 marks across “5 Questions”) is likewise only consistent with the “2 of 3” reading (30+30+... does not match 70 either way exactly, since 30(A)+2×15(B)+20(C)=80, one section choice short of 70); this is an internal inconsistency in the exam's own printed materials. All three Part B questions are answered in full below regardless.
Reference texts. Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation (5th ed.) — primary text for random-variate generation, input/output analysis, variance reduction, verification & validation, and design of simulation experiments.
Question 1 (Part A.1): Continuous Distribution — Valid pdf, Mean and Variance (10 marks)
Given. $f(x)=x^2+\tfrac23x+\tfrac13$ on $0\le x\le c$, with $c$ to be determined.
Find. (a) the value of $c$ that makes $f$ a valid pdf; (b) $E(X)$ and $\operatorname{Var}(X)$.
Approach. Impose $\int_0^c f(x)\,dx=1$ and solve for $c$; with $c=1$ fixed, compute $E(X)=\int xf(x)\,dx$ and $E(X^2)=\int x^2f(x)\,dx$, then $\operatorname{Var}(X)=E(X^2)-E(X)^2$.
Solve for c. Since $f(x)\ge0$ on $[0,c]$ for any $c>0$, the only condition to check is total probability:
$$\int_0^c\left(x^2+\frac23x+\frac13\right)dx=\left[\frac{x^3}{3}+\frac{x^2}{3}+\frac{x}{3}\right]_0^c=\frac{c^3+c^2+c}{3}=1\ \Rightarrow\ c^3+c^2+c-3=0.$$
Factoring, $c^3+c^2+c-3=(c-1)(c^2+2c+3)$; the quadratic factor has discriminant $2^2-4(3)=-8<0$, so it has no real root. The only real solution is
$$\boxed{c=1}.$$
Compute E(X). With $c=1$, $f(x)=x^2+\tfrac23x+\tfrac13$ on $[0,1]$:
$$E(X)=\int_0^1 x\,f(x)\,dx=\int_0^1\left(x^3+\frac23x^2+\frac13x\right)dx=\left[\frac{x^4}{4}+\frac{2x^3}{9}+\frac{x^2}{6}\right]_0^1=\frac14+\frac29+\frac16=\frac{23}{36}.$$
$$\boxed{E(X)=\frac{23}{36}\approx0.6389}$$
Compute Var(X). First $E(X^2)$:
$$E(X^2)=\int_0^1 x^2f(x)\,dx=\int_0^1\left(x^4+\frac23x^3+\frac13x^2\right)dx=\left[\frac{x^5}{5}+\frac{x^4}{6}+\frac{x^3}{9}\right]_0^1=\frac15+\frac16+\frac19=\frac{43}{90}\approx0.4778.$$
Then
$$\operatorname{Var}(X)=E(X^2)-E(X)^2=\frac{43}{90}-\left(\frac{23}{36}\right)^2\approx0.4778-0.4082=\boxed{0.0696}.$$
The corresponding standard deviation is $\sigma_X=\sqrt{0.0696}\approx0.2638$, so the distribution concentrates roughly two-thirds of its mass within about $\pm0.26$ of the mean $0.6389$ — a spread that fits comfortably inside the support $[0,1]$, which is a quick sanity check that neither moment has been mis-integrated. It is also worth noting why the mean sits above the midpoint of the support: the density $f(x)=x^2+\tfrac23x+\tfrac13$ is increasing on $[0,1]$, rising from $f(0)=\tfrac13$ to $f(1)=2$, so probability mass is weighted toward the upper end of the interval and $E(X)=0.6389$ correctly exceeds $0.5$.
