MR. K. PEARSON ON THE MATHEMATICAL THEORY OF EVOLUTION. 371 
Then 
, , , a eS 
Ti = Mara yet Mal. ty Vee pr eal, £1 
and we have at once 

eo, onG— » _ (2-1) ad-—m) 
i, 2S, Nig se mg 
DAL = 0% Tat VA) 
Then a,/a, = (m’, — 1)/(m’, — 1), and a, + a@ = b give a, and a, Finally yp is 
given as before by 
a my Me" T'(m, + mg + 2) 
b (my + mgmt T(m, + 1) Mm, + 1) 

Yo = 
(16.) A perhaps still more interesting and usual case arises when one end of the 
range is given, 7.e., when p’,, but not b, is known. For example, a curve of distri- 
bution of disease with age, the liability to the disease starting with birth. Here we 
require to calculate from the observations a, py, @’, and p's The solution is as 
follows : 
Let 
H's/h a? = Xoy H's) (fe of) = Xs 3 
then 
2 Gear) (m’, + m’) eae 
Xe = omy (mi, +m’, i071) TF iate 
_— (m+ 2) m+ m,) _ 1 + 20 
a m’, (mM, + m',+ 2) ~ 1+ 2u 

2 

if vy = 1/m', and uv = 1/(m’, + m’,). 




Solving * 
1 + x3 — 2x2 2N3 — Xo — XoXs 
= ae v= : 
2 (X2 — Xs) 2 (%_ — Xs) 
Thus, : opie 
2 (yy — 2 (x2 — x3) (X%s — 1) 1 = x0 
m ae (X2 — Xs) nil. = (Xo — x3) (x3 )¢ No) 
2N3— X2— XN? (1 + x3 — 2X2) (23 — X2 — KoXs) 
b= pw) (m+ m,)/m’, = p) v/u 
= rie 2x3 Sam X OEE XCOXS ; 
; 2 (x2 — Xs) 

determines the range. 
Hence, since 
m,—1 
Ce ARG Os) ING) CxO, = — 
os WM by 
we have, with the aid of the previous expression for yg, the complete solution of the 
problem, 
a) 15) WY 
