Flood statistics and exceedance probability, 2000-2009

(i) Statistics (n = 10)

$$\bar x=\frac{\sum x}{n}=\frac{401.7}{10}=40.17 \text{ m}^3/\text{s}$$ $$s=\sqrt{\frac{\sum (x-\bar x)^2}{n-1}}=\sqrt{\frac{1016.92}{9}}=10.63 \text{ m}^3/\text{s}$$ $$C_s=\frac{n\sum (x-\bar x)^3}{(n-1)(n-2)\,s^3}=\frac{10\times6480.3}{9\times 8\times1201.1}=0.75$$

Answer: mean = 40.17 m³/s, standard deviation = 10.63 m³/s, skewness coefficient = 0.75 (positively skewed).

(ii) Probability of a 250 m³/s flood next year (Gumbel)

$$K=\frac{250-40.17}{10.63}=19.74$$ $$y_T=\bar y_n+K\sigma_n = 0.4952 + 19.74\times0.9497=19.24$$ $$P(X\ge 250)=1-e^{-e^{-y_T}}=1-e^{-e^{-19.24}} \approx 4.4\times10^{-9}$$

The return period is $T=1/P \approx 2.3\times10^{8}$ years.

Answer: probability ≈ $4.4\times10^{-9}$ (about $4.4\times10^{-7}$ %), i.e. practically zero. The value is far outside the range of the 10-year record (maximum 59.6 m³/s), so it is a very large extrapolation.