The rainfall year begins on July 1 and ends on June 30 since the rainy season is primarily in the winter and surrounding few months. The plot above shows the cumulative rainfall in inches as recorded by my rain gauge sitting in an open field at my house in the Winterwarm Area of Fallbrook, CA. The current year's rainfall is shown in bold.
The rainfall for the previous years alone is plotted with labels identifying the years and also unlabeled.
Here are the statistics for each of the winter rainy months from 1994-5 to 2005-6:
| Month | Min | Max | Mean | Median |
|---|---|---|---|---|
| Nov | 0.00 | 4.29 | 1.53 | 1.16 |
| Dec | 0.00 | 4.41 | 1.67 | 1.18 |
| Jan | 0.04 | 12.48 | 3.94 | 2.28 |
| Feb | 0.12 | 12.48 | 4.65 | 4.49 |
| Mar | 0.00 | 10.08 | 2.40 | 1.65 |
| Apr | 0.31 | 2.60 | 1.39 | 1.42 |
| Nov-Apr | 4.84 | 29.13 | 15.59 | 12.80 |
| Jul-Jun | 4.84 | 37.05 | 17.22 | 14.00 |
The mean rainfall is the average of all values. The median rainfall is the "one in the middle"; half of all years are wetter and half of all years are drier.
Given enough measurements, the mean rainfall is always larger since rainfall is distributed such that it is equally probable to get, for example, twice as much rainfall as half as much rainfall as the median. The average of those two extremes is not the median rainfall; it is (2 + 0.5) / 2 = 1.25, 25% more than the median. All years different from the median rainfall by a constant multiplicative amount contribute toward making the average greater than the median, with the factor continuously being larger with larger deviations from the median. (For example, the years with "1.1 times more" and "1.1 times less" rainfall average to (1.100 + 1/1.100) / 2 = 1.005, 0.5% more than the average.)
In mathematical terms, the rainfall distribution is not distributed according to the normal Gaussian bell curve; instead, the logarithm of the rainfall is distributed in that way. The technical term for the rainfall distribution is that it is log-normally distributed. Again, this is just a mathematically-precise way of saying that one has equal chances of getting x times the median rainfall as getting 1/x times the median rainfall. In a normal distribution, one has equal chances of getting +x inches and -x inches from the median rainfall.
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Copyright © 1997-2007 by Tom Chester.
Permission is freely granted to reproduce any or all of this page as long as credit is given to me at this source:
http://sd.znet.com/~schester/fallbrook/weather/rainfall.html
Comments and feedback: Tom Chester
Last update: 23 February 2007.
The plot may have been updated later - see the date at the top of the plot.
