Gumbel analysis, 80-year and 1.4-year floods

Gumbel's equation is linear in the reduced variate: $x_T=\bar x+K\sigma=a+b\,y_T$, with $y_T=-\ln[\ln(T/(T-1))]$. The two given floods fix the line.

$$b=\frac{620000-215000}{4.3757-(-0.2254)}=88023 \qquad\Rightarrow\qquad x_T=215000+88023\,[y_T-(-0.2254)]$$

(i) Probability of a flood ≥ 440 000

$$y=-0.2254+\frac{440000-215000}{88023}=2.3308$$ $$P(X\ge440000)=1-e^{-e^{-y}}=1-e^{-e^{-2.3308}}=0.0926$$

Answer: probability ≈ 0.0926 (about 9.3%; return period ≈ 10.8 years).

(ii) Flood with a 40-year recurrence interval

$$y_{40}=-\ln\left[\ln\frac{40}{39}\right]=3.6762$$ $$x_{40}=215000+88023\,(3.6762-(-0.2254))=558429 \text{ units}$$

Answer: $x_{40}$ ≈ 558429 units.

(iii) Probability of a flood ≥ 575 000 in the next 25 years

$$y=-0.2254+\frac{575000-215000}{88023}=3.8645$$ $$p=1-e^{-e^{-3.8645}}=0.02075 \quad(T=48.2 \text{ years})$$

The probability of at least one such flood in n = 25 years is

$$1-(1-p)^{25}=1-(1-0.02075)^{25}=0.4080$$

Answer: probability ≈ 0.4080 (about 40.8%).