Question 4 of 4: Flood-Risk Probability, Log-Pearson III Return Period, and a Triangular Rainfall PDF

Question 4: Flood-Risk Probability, Log-Pearson III Return Period, and a Triangular Rainfall PDF (33.3 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. River carrying capacity, its return period, and a 20-year planning horizon (Part 4.1); log-Pearson Type III flood statistics and one observed peak flow (Part 4.2); a symmetric triangular rainfall PDF (Part 4.3):

Triangular PDF, base 0–60 cm, $f(x)=x/900$ for $0\le x\le30$ cm (mirror-symmetric about $x=30$).

Find. (4.1) $P(\text{park floods}\ge3\text{ times in 20 yr})$. (4.2) The return period of the 875 m³/s peak flow. (4.3a) The complete PDF and a proof it integrates to 1. (4.3b) $P(X\le20\ \text{cm})$. (4.3c) The mean summer rainfall.

Approach. Part 4.1 treats yearly flooding as an independent Bernoulli trial with exceedance probability $p=1/T$ and applies the binomial distribution. Part 4.2 converts the observed flow to a standardized log-Pearson Type III frequency factor and inverts the Wilson–Hilferty skew-adjustment to recover an exceedance probability. Part 4.3 completes the triangle by symmetry and integrates piecewise.

1. Part 4.1 — Binomial flood-risk probability. The annual exceedance probability is $p=1/T=1/10=0.10$; over $n=20$ independent years, the number of floods $X\sim\text{Binomial}(n=20,\,p=0.10)$. Rather than sum the (long) tail directly, use the complement: $$P(X\ge3)=1-P(X=0)-P(X=1)-P(X=2)$$ $$\begin{aligned}P(X=0)&=\binom{20}{0}(0.1)^0(0.9)^{20}=0.1216\\P(X=1)&=\binom{20}{1}(0.1)^1(0.9)^{19}=0.2702\\P(X=2)&=\binom{20}{2}(0.1)^2(0.9)^{18}=0.2852\end{aligned}$$ $$P(X\ge3)=1-(0.1216+0.2702+0.2852)=1-0.6769=\boxed{0.323\ (32.3\%)}$$
2. Part 4.2 — Log-Pearson Type III return period. Transform the observed flow and standardize it as an LP3 frequency factor: $$y_{2008}=\log_{10}(875)=\boxed{2.9420}$$ $$K=\frac{y_{2008}-\bar y}{S_y}=\frac{2.9420-2.09}{0.4439}=\boxed{1.919}$$ The Wilson–Hilferty approximation relates a standard-normal variate $z$ to the Pearson frequency factor $K$ through the skew $C_s$ (with $k=C_s/6$): $$K(z)=z+(z^2-1)k+\tfrac13(z^3-6z)k^2-(z^2-1)k^3+zk^4+\tfrac13k^5$$ Solving $K(z)=1.919$ numerically for $k=-0.244/6=-0.0407$ gives $z=\boxed{2.052}$. The non-exceedance probability is $\Phi(z)$, so the annual exceedance probability and return period are $$p=1-\Phi(2.052)=\boxed{0.0201\ (2.01\%)}$$ $$T=\frac{1}{p}=\boxed{49.8\ \text{years}\approx50\text{-year event}}$$
3. Part 4.3(a) — Complete the PDF and prove unit area. The PDF is a symmetric isosceles triangle on $[0,60]$, so by mirror symmetry about the peak at $x=30$: $$f(x)=\begin{cases}x/900, & 0\le x\le30\\ (60-x)/900, & 30\le x\le60\end{cases}$$ Its peak value is $f(30)=30/900=1/30\approx0.0333\ \text{cm}^{-1}$ — consistent with the general triangular-PDF identity $f_{peak}=2/\text{base}=2/60=1/30$, itself required for unit area (area of a triangle $=\tfrac12\times\text{base}\times\text{height}$). Integrating directly confirms it: $$\begin{aligned}\int_0^{60}f(x)\,dx&=\int_0^{30}\frac{x}{900}\,dx+\int_{30}^{60}\frac{60-x}{900}\,dx\\&=\left[\frac{x^2}{1800}\right]_0^{30}+\left[\frac{60x-x^2/2}{900}\right]_{30}^{60}\\&=0.5+0.5=\boxed{1.0}\ \checkmark\end{aligned}$$
4. Part 4.3(b) — Probability rainfall does not exceed 20 cm. Since $20<30$, only the rising branch is needed: $$\begin{aligned}P(X\le20)&=\int_0^{20}\frac{x}{900}\,dx=\left[\frac{x^2}{1800}\right]_0^{20}\\&=\frac{400}{1800}=\boxed{0.222\ (22.2\%)}\end{aligned}$$
5. Part 4.3(c) — Mean summer rainfall. By the PDF's mirror symmetry about $x=30$, the mean must equal the axis of symmetry, $E[X]=\boxed{30\ \text{cm}}$; direct integration confirms it: $$\begin{aligned}E[X]&=\int_0^{30}x\cdot\frac{x}{900}\,dx+\int_{30}^{60}x\cdot\frac{60-x}{900}\,dx\\&=10+20=\boxed{30\ \text{cm}}\end{aligned}$$