Question 3 of 6: Log-Normal Flood-Frequency Analysis

Question 3: Log-Normal Flood-Frequency Analysis (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) The 2-year and 10-year return-period discharges, $Q_2$ and $Q_{10}$, assuming a log-normal population. (b) A check on the log-normality assumption itself.

Approach. Transform each discharge to $y=\log_{10}Q$, compute the sample mean $\bar y$ and standard deviation $s_y$ of the transformed data, then back-transform using $Q_T=10^{\bar y + z_T s_y}$, where $z_T$ is the standard-normal variate for return period $T$ ($z_T=0$ at $T=2$ yr, since 50% non-exceedance corresponds to the median; $z_T=1.282$ at $T=10$ yr, the standard-normal deviate for 90% non-exceedance).

1. Log-transform and sample statistics. $$y=\log_{10}Q:\ \ 1.000,\ 1.342,\ 1.447,\ 1.176,\ 0.903$$ $$\bar y=\frac{\sum y}{n}=\frac{5.869}{5}=1.174\qquad s_y=\sqrt{\frac{\sum(y-\bar y)^2}{n-1}}=0.227$$
2. 2-year return period ($z_2=0$, the median). $$Q_2=10^{\bar y+z_2 s_y}=10^{1.174+0}=\boxed{14.9\ \text{m}^3/\text{s}}$$
3. 10-year return period ($z_{10}=1.282$, the standard-normal deviate at 90% non-exceedance). $$Q_{10}=10^{\bar y+z_{10}s_y}=10^{1.174+1.282(0.227)}=10^{1.465}=\boxed{29.2\ \text{m}^3/\text{s}}$$ Discussion: with only 5 years of record the fitted mean and standard deviation are themselves uncertain, so both estimates — especially $Q_{10}$, which extrapolates beyond the two largest observed flows — should be treated as indicative rather than precise; a longer record would tighten both figures.

(b) Checking the log-normality assumption. The standard graphical check is to rank the five discharges, assign each an empirical exceedance probability with a plotting-position formula (Weibull, $p=m/(n+1)$, $m=1$ for the largest), convert each probability to its standard-normal variate $z$, and plot $\log_{10}Q$ against $z$ on ordinary (or, equivalently, $Q$ directly on log-normal probability paper). If the population is genuinely log-normal, these points should scatter closely about a straight line, since a log-normal variable is by definition one whose logarithm plots as a straight line against a normal probability scale. A quantitative companion check is the sample skewness coefficient of the log-transformed data: a true normal (hence log-normal, in logs) population has zero skew, so a computed skewness for $y=\log_{10}Q$ close to zero supports the assumption, while a strongly non-zero value would argue against it.