PROF. K. PEARSON ON THE MATHEMATICAL THEORY OF EVOLUTION. 85 
Given that a frequency-curve is compounded of two normal curves, equations 
(29), (28), (27), (14), (15), (16), and (17) form the complete solution of the problem. 
We may throw the whole solution into the following form :— 
Stage I. —Find the centroid of the frequency-curve and calculate ^, 3 , /xg, [x^, g-, 
and X-. 
Stage II. —Solve (29) for and find the corresponding values of from (27). 
Stage III. —Find the positions of the axes of the component normal curves 
from (28). 
Stage IV. —The fractions and % that the areas of the normal curves are of 
the area of the frequency-curve are the roots of the quadratic ; 
lb 
PC - 4p, 
- = 0 
(30). 
Stage V. —Since a-Jli = and ajh = -v/co, the standard-deviations are given 
at once on substituting in (18) and (19). . 
(9.) The whole method may be illustrated by the following numerical example:— 
Breadth of ^‘‘Forehead” of Crabs. —Professor W. F. II. Weldon has very kindly 
given me the following statistics from among his measurements on crabs. They are 
for 1000 individuals from Naples. The abscissm of the curve are the ratio of “fore¬ 
head ” to body-length, and one unit of abscissa = '004 of body-length. No. 1 of the 
abscissae corresponds to ‘580 — ‘583 of body-length. The ordinates represent the 
number of individual crabs corresponding to each set of ratios of forehead to body- 
length. Thus there was one crab fell into the range ‘580 — ‘583, three fell into the 
range ’584 — ‘587, five into the range ‘588 — ‘591, and so on. The average length 
of animals measured 35 millims., and measurements were recorded to ‘1 millim. 
Abscissae. 
Ordinates. 
Abscissae. 
Ordinates. 
I 
1 
16 
74 
•A 
3 
17 
84 
3 
5 
18 
86 
4 
0 
19 
96 
5 
7 
20 
85 
6 
10 
21 
75 
7 
13 
22 
47 
8 
19 
23 
43 
9 
20 
24 
24 
10 
25 
25 
19 
II 
40 
26 
9 
12 
31 
27 
5 
13 
60 
28 
0 
u 
15 
62 
54 
29 
1 
j 
