Francis Galton 
387 
— ((?^ — 1), (n — 3), (n — 5)). Then I took from the Probability Integral Tables 
the corresponding values of hx. 
As an example of the complete process let n = 10, then the most probable 
values of @ (hx) for the ten competitors will, according to my assumption, be 
-0-9, -0-7, -0-5, -0-.3, -0-1, +01, +0-3, +0'5, +0-7, +0-9. 
They are separated by equal distances from one another and by the half of 
those distances from the septa, including the terminals, that enclose them. 
Confining ourselves to the first three terms on the positive side, that is to 
+ 0 9, + 0"7 and + 0"5, we find from the Probability Integral Tables that the 
corresponding values of hx are + 1-1631, + 07329, +0-4770. 
The percentage values of (a — c) and (b — c) (as described above in (2)), are 
quickly derived from these. We will call them X and F, and their sum S. 
/ia = 1-1631 h (a - c) = 0-6861 
hb = 0-7329 h (6 - c) = 0-2559 
he = 0-4770 hS =0-9420 
X -.100 :: h (a -c):h8; 7: 100 : : /i (6 - c) : hS. 
Whence Z = 72-8, F=27-2. 
Thus h disappears from the result while m, the Mean, does not come under 
consideration. If it had been taken into consideration by writing m + a for a, 
m + 6 for b, and m + c for c, it would have been eliminated by the subtractions, 
as h was by the divisions. 
Similarly if n be taken = 1000, the values of 0 (hx) for [A], [B], and [G] would 
be +0-9990, + 0 9970, and + 0-9950 which give ha= + 2 3268, hb ^ + 2-0985, and 
/ic = + 1-9849. 
Proceeding in this way for many widely different values of ?i, I found to my 
astonishment that the resultant X and F values for those of n=10 and above, 
came out curiously alike, as is shown in Table I. 
TABLE I. 
n 
X 
Y 
X+Y 
3 
66-7 
33-3 
100-0 
5 
71-0 
29-0 
10 
72-8 
27-2 
20 
73-8 
26-2 
50 
74-3 
25-7 
100 
74-5 
25-5 
1,000 
75-1 
24-9 
10,000 
75-3 
24-7 
100,000 
75-4 
24-6 
