This is a problem in scientific estimateion. Since the problem isn't very specific we will have to use reasonable numbers.

First we need to estimate how much power this array might possibly generate. Since there is no size given for the panels we will estimate that each panel is 2 sq. meters in area. This gives 2x12500=25,000 sq.meters of collecting area. Panels are typically 10%-15% efficicent, meaning that only a fraction of the solar energy striking the panel is converted to electrical work. We will use the higher 15% estimate. Solar energy flow (power) is 1000 kW per sq. meter. The best we can get out of this array is:

25,000 x 1000 x .15=3.75x10^6 kW

We will use the usual electrical work unit of kW-hr. Now all we need to do is estimate how many hours per day the panels are in operation over the course of a year. 55.33 cm per year is not a high number of inchs of precipitation per year (roughly 20 inches/year) and 18.6 C average temperature would place our panel in the drier midwest climates. We will assume that on the 365-84=281 days of the year with no precipitation the panels are working all day. The angle the Sun strikes the panels is important as the power absorbed goes down with the sine of the strike angle. Also the latitude plays a part in that the Sun is not high overhead for much of the day or the year. We can estimate that all these factors together might reduce the power output by 50% without further information. This means that the total hours of operation are 281x12x.5=1686 hours per year. This gives a yearly estimate of electrical work as:

1686x3.75x10^6=6.3x10^9 kW-hr

The per capita power consumption in the US in 2007 was 13,652 kw-hr per year. This means that this array can supply yearly power for

6.3x10^9/13652=463,000 households for an entire year. That's seems pretty impressive.