24-hr maximum rainfall for 50-year return period

Method. Rank the annual maxima, find the return period of each by a plotting-position formula, plot rainfall against T on semi-log paper (T on the log axis), fit a straight line and extend it to T = 50 years.

Weibull formula (N = 16 years of data, m = rank in descending order):

$$T=\frac{N+1}{m}=\frac{17}{m}, \qquad P=\frac1T$$

Plotting. Plot the 16 points (T on the logarithmic horizontal axis, 24-h rainfall on the arithmetic vertical axis). Gumbel-distributed data fall close to a straight line on this paper, with equation:

$$P_{24}=a+b\log_{10}T$$

A line fitted through the points gives $b=7.49$ cm per log-cycle and $a=7.33$ cm.

 P24 (cm)
  20 |                        o  <- T=50 (extrapolated)
  16 |                  o
  12 |          o  o
   8 |   o o o
   4 +--+------+-------+--------- T (yr, log)
     1.1      2       10      50

Reading at T = 50 years:

$$P_{24,50}= 7.33 + 7.49\times\log_{10}(50)=20.05 \text{ cm}$$

Answer: 24-h maximum rainfall for the 50-year return period ≈ 20 cm. As a check, Gumbel's analytical method with $\bar y_n=0.5157$, $S_n=1.0316$ (N = 16), mean = 10.31 cm and $\sigma$ = 2.59 cm gives 18.8 cm, which is of the same order.