Question 4 of 4: Flood-Frequency Analysis — Lognormal, Log-Pearson III and Gamma Distributions

Question 4: Flood-Frequency Analysis — Lognormal, Log-Pearson III and Gamma Distributions (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. 4.1 – 12 years of maximum monthly precipitation (mm): 68.6, 54.1, 65.9, 64.9, 77.5, 77.8, 76.6, 90.5, 101.3, 92.1, 76.5, 67.1, plus the standard normal table below. 4.2 – log-space (base 10) statistics $\bar y=1.875$, $S_y=0.1$, $C_s=-0.25$. 4.3 – real-space statistics $\bar x=76.1$ mm, $S_x=14.8$ mm, fitted to a 2-parameter gamma distribution.

Find. 4.1(a) whether a 2-parameter lognormal is an appropriate fit; 4.1(b) $P(\text{2008 max monthly precip}\le 66\ \text{mm})$. 4.2 the 50-yr return-period monthly precipitation from the log-Pearson-III-type ($y=\log_{10}x$) statistics. 4.3 the 100-yr return-period monthly precipitation from the 2-parameter gamma fit.

Approach. 4.1 log-transforms the data and checks the skew of $y=\ln x$ for near-zero value (a 2-parameter lognormal is exactly a distribution whose log is NORMAL, i.e. $C_s(y)=0$), then uses the fitted normal to read a probability off the given $z$-table. 4.2 applies the Wilson–Hilferty approximation to invert the log-Pearson-III frequency factor $K_T$ from the standard normal variate and the skew. 4.3 exploits the fact that a 2-parameter gamma distribution's skew is fixed by its coefficient of variation ($C_s=2\,CV$), so the same Wilson–Hilferty $K_T$ machinery applies directly in real (untransformed) space.

1. Part 4.1(a) — Check the 2-parameter lognormal fit. Transforming to $y=\ln x$ and computing the sample moments (Bessel-corrected): $$\bar y = 4.318,\qquad S_y=0.1747,\qquad C_{s,y}=\boxed{0.003}$$ The raw data's own skew is $C_{s,x}=0.41$ (visibly right-skewed, as monthly rainfall maxima typically are), but once log-transformed the skew collapses to essentially zero. Since a 2-parameter lognormal distribution is, by definition, one whose logarithm is exactly NORMAL (symmetric, $C_s=0$), a log-skew this close to zero confirms the 2-parameter lognormal is an appropriate model for this station — no third (lower-bound shift) parameter is needed.
2. Part 4.1(b) — Probability 2008 maximum monthly precipitation ≤ 66 mm. Standardizing $\ln(66)=4.190$ against the fitted log-space distribution: $$z=\frac{\ln(66)-\bar y}{S_y}=\frac{4.190-4.318}{0.1747}=\boxed{-0.733}$$ Interpolating the given normal table between $z=-1.000\ (F=0.1587)$ and $z=-0.500\ (F=0.3085)$: $$F(-0.733)\approx0.1587+\frac{-0.733-(-1.000)}{-0.500-(-1.000)}\times(0.3085-0.1587)=\boxed{0.239\ (23.9\%)}$$
3. Part 4.2 — 50-yr return-period precipitation, log-Pearson-III statistics. For $T=50$ yr the non-exceedance probability is $1-1/50=0.98$; interpolating the given table between $z=2.000\ (F=0.9772)$ and $z=2.326\ (F=0.9900)$: $$z_{50}=2.000+\frac{0.98-0.9772}{0.9900-0.9772}\times(2.326-2.000)=\boxed{2.071}$$ The Wilson–Hilferty approximation converts this standard-normal variate to a skewed frequency factor $K_T$ for skew $C_s=-0.25$: $$K_T=\frac{2}{C_s}\left[\left(1+\frac{C_s z}{6}-\frac{C_s^2}{36}\right)^{3}-1\right]=\frac{2}{-0.25}\left[(0.9120)^3-1\right]=\boxed{1.932}$$ so $$y_{50}=\bar y+K_T S_y=1.875+1.932\times0.1=\boxed{2.068}$$ $$x_{50}=10^{y_{50}}=10^{2.068}=\boxed{117\ \text{mm}}$$
4. Part 4.3 — 100-yr return-period precipitation, 2-parameter gamma fit. Matching moments, the gamma shape and scale parameters are $$\alpha=\left(\frac{\bar x}{S_x}\right)^2=\left(\frac{76.1}{14.8}\right)^2=\boxed{26.4},\qquad \beta=\frac{S_x^2}{\bar x}=\frac{14.8^2}{76.1}=\boxed{2.88\ \text{mm}}$$ A 2-parameter gamma's own skew is fixed by its coefficient of variation, $C_s=2\,CV=2\,(S_x/\bar x)=\boxed{0.389}$. For $T=100$ yr, $1-1/100=0.99$ is an EXACT row of the given table, $z_{100}=2.326$. Applying the same Wilson–Hilferty formula with $C_s=0.389$: $$K_T=\frac{2}{0.389}\left[\left(1+\frac{0.389\times2.326}{6}-\frac{0.389^2}{36}\right)^3-1\right]=\boxed{2.61}$$ $$x_{100}=\bar x+K_T S_x=76.1+2.61\times14.8=\boxed{114.7\ \text{mm}}$$ (a direct numerical solution of the gamma distribution's own inverse-CDF at $\alpha=26.4$, $\beta=2.88$ mm gives $x_{100}=114.7$ mm as well, to three figures — confirming the Wilson–Hilferty approximation is essentially exact at this skew.)