Poisson Distribution
Answer 20

Absolutely positively Random; no doubt about it.

Explanation

We proceeded in this way. Counting "5 or more" as simply "5," we had a total of 535 hits over the 576 squares, for a rate r of 0.9288 hits per average square. Let's call it 0.9 so we can use the Table. We then copy the Table probabilities for r = 0.9 into our first column (combining the figures for p(5) and p(6) for greater accuracy), multiply by 576 to get the expected number of squares with that many hits and enter that result in the second or E column. Finally, we copy the data for actual hits into the A or Actual column:

	r = 0.9	E	A
p(0)	0.4066	234.20 ~ 234	229
p(1)	0.3659	210.76 ~ 211	211
p(2)	0.1647	94.87 ~ 95	93
p(3)	0.0494	28.45 ~ 28	35
p(4)	0.0111	6.39 ~ 6	7
p(5-6)	0.0023	1.32 ~ 1	1
TOT		575	576

There is a slight rounding error, yielding a total that is 1 too small; not important. We then eyeball the result. We find that the figures in the two columns are startlingly similar; every bit as close (to the subjective eye) as in the classic von Bortkiewicz comparison. There is thus no basis for suspecting anything other than a random generation process, hence the above answer.

Comment

One can readily see why statistics textbook writers in search of parallels for the classic, and thus too often quoted, von Bortkiewicz horse-kick data set have greeted these V-Bomb figures with squeals of glee. See however the next problem.

24 Aug 2007 / Contact The Project / Exit to Resources Page