Flood of return period T using log-normal distribution

Log-normal distribution. The logarithm of the annual flood is assumed to be normally distributed. The flood of a given return period is then found using the standard normal variate in place of the frequency factor.

Procedure (N years of annual maxima, T > N)

1. Arrange the annual flood series $x_1,\dots,x_N$ and take logarithms, $y_i=\log_{10}x_i$ (or ln).
2. Compute the mean and standard deviation of the log series:

$$\bar y=\frac1N\sum y_i,\qquad S_y=\sqrt{\frac{\sum(y_i-\bar y)^2}{N-1}}$$

3. Compute the exceedance probability $P=1/T$ and the non-exceedance probability $F=1-1/T$.
4. Find the standard normal variate $z_T$ corresponding to $F$ from the standard normal table (the frequency factor of the log-normal distribution, $K_T=z_T$). A good approximation is $z=w-\dfrac{2.515517+0.802853w+0.010328w^2}{1+1.432788w+0.189269w^2+0.001308w^3}$ with $w=\sqrt{\ln(1/P^2)}$ for $0<P\le0.5$. For example, $z=2.054$ for T = 50 years and $z=2.326$ for T = 100 years.
5. Compute the log of the flood of return period T:

$$y_T=\bar y+z_T\,S_y$$

6. Convert back to get the flood:

$$x_T=10^{\,y_T}\ \ (\text{or }e^{y_T}\text{ if natural logs are used})$$

7. As a check, plot the points with Weibull positions $T=(N+1)/m$ on log-normal probability paper; the data should be close to a straight line, which can be extended to T for a graphical result.

The result is an extrapolation beyond the record (T > N), so confidence limits should be considered; the log-normal distribution is a special case of log-Pearson III with zero skew.