129 



Remark. We hère found the value of z corresponding 

 to any value of x from the figure. So for 



X -- OH = + 2.00 we find 2 = OH = - 0.70. 

 There is no difficulty in thus finding the values of the z 

 corresponding to as many values of x as we please. But 

 the same thing may be done even more conveniently and 

 easily by a little table. Such a table will be found at the 

 end of this paper (table 11). 



With the argument z it gives the value of S (scheme). 

 For the présent purpose we hâve to enter this table with 

 the given values of S (scheme) and to take out the cor- 

 responding values of the z. In fact the values z of table 2 

 hâve been thus computed. 



14, Question b. It remains to find the way in which 

 the déviations — therefore the growth and the average 

 fluctuation — of the x dépend on the size of the x. 



The z being normally distributed, we know (see begin- 

 ning of art. 13) that the déviations are independent of the 

 size of the z. The growth of the bigger and of the smaller 

 individuals are the same and the individual divergences 

 from this mean growth are also the same. 



From this it is easy to dérive the growth of the x. 

 This is perhaps most easily seen from the example 

 summarised in table 2. 



if X grows from 0.0 to 0.5 z grows from — 1.500 to 

 — 1.485 that is 



if at the average value 0.25 of x, x grows 0.500 

 z grows 0.015; similarly 



if at the average value 0.75 of x, x grows 0.500 

 z grows 0.085; 



if at the average value 1.25 of x, x grows 0.500 

 z grows 0.238 etc. 



From this we find at once that 



0-500 a 

 ir, near x = 0.25, z grows a, x grows fT/^Tc"' 



