Gumbel analysis from 100 and 50-year floods

a) Mean and standard deviation

$\bar y_n=0.5362$, $S_n=1.1124$, $y_T=-\ln[\ln(T/(T-1))]$, $K=(y_T-\bar y_n)/S_n$, $x_T=\bar x+K\sigma$.

Subtract the two equations:

$$\sigma=\frac{1200-1060}{3.653-3.026}=223.1 \text{ m}^3/\text{s}$$ $$\bar x=1200-3.653\times223.1=385.1 \text{ m}^3/\text{s}$$

Answer: mean = 385.1 m³/s, standard deviation = 223.1 m³/s.

b) 500-year flood

$$y_{500}=-\ln[\ln(500/499)]=6.214,\qquad K=\frac{6.214-0.5362}{1.1124}=5.104$$ $$x_{500}=385.1+5.104\times223.1=1524 \text{ m}^3/\text{s}$$

Answer: $x_{500}$ ≈ 1524 m³/s.

c) 95% confidence limits of the 500-year flood

$$b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(5.104)+1.1(5.104)^2}=6.024$$ $$S_e=\frac{b\sigma}{\sqrt N}=\frac{6.024\times223.1}{\sqrt{30}}=245.3 \text{ m}^3/\text{s}$$ $$x_T'=x_T\pm f(c)S_e=1524\pm1.96\times245.3=1524\pm481$$

Answer: 95% confidence limits are about 1043 m³/s and 2004 m³/s.

d) Probability of three floods of 500-year size (or larger) in the next 10 years

The annual probability is $p=1/500=0.002$. Floods in different years are independent, so use the binomial distribution with n = 10, r = 3:

$$P=\binom{10}{3}p^3(1-p)^7=120\times(0.002)^3\times(0.998)^7 \approx 9.47\times10^{-7}$$

Answer: probability ≈ $9.47\times10^{-7}$ (about 1 in a million), practically negligible.