What is P(X >= mean)?

1 - exp(-exp(-0.5772)) = 0.4296

T = 1/P = 1/0.4296 = 2.33 years

Prove mean annual flood has 2.33-year return period

Short answer

For large N, reduced mean yn=0.5772 (Euler's constant). Exceedance probability P=1-e^{-e^{-0.5772}}=0.4296, so return period T=1/P≈2.33 years.

Prove that for a large sample as per Gumbel's distribution, the mean annual flood will have a return period of 2.33 years.

Proof. In Gumbel's method $x_T=\bar x+K\sigma$ , so the mean flood $x=\bar x$ corresponds to $K=0$ :

K=\frac{y_T-\bar y_n}{S_n}=0\ \Rightarrow\ y_T=\bar y_n

For a large sample ( $N\to\infty$ ), the reduced mean is $\bar y_n=0.5772$ (Euler's constant). The exceedance probability of this value is

P(X\ge\bar x)=1-e^{-e^{-y_T}}=1-e^{-e^{-0.5772}}

e^{-0.5772}=0.5615,\qquad e^{-0.5615}=0.5704,\qquad P=1-0.5704=0.4296

The return period is

T=\frac1P=\frac{1}{0.4296}\approx2.33 \text{ years}

Hence the mean annual flood has a return period of about 2.33 years for Gumbel's distribution (large sample). ∎

✓ Answer checked by Claude Sonnet 5.5