\.\\l 



for Xfn/isfiriiiiix it in/ 



[IX X 



and from the Table, p. 22, we have by formula (i), p. xiii : 



m,(4-0916) = -397,7378 + -916[3650] - $ (-91(i) (-084) [1043] 



= -398,0682. 

 Hence IJN = \/27r x "398,0682 = -9978. 



Thus the odds are 9978 to 22, say 454 to 1 against a deviation-complex as 

 great as or greater than this occurring in a French male skeleton, i.e. the bones very 

 improbably were those of a Frenchman. Actually they were those of a male of 

 the Aino race. 



Illustration (ii). The following are the ordinates of a frequency distribution 

 for the speed of American trotting horses*. It is assumed that they form a 

 truncated normal curve, and we require to determine (i) the mean of the whole 

 population, (ii) its standard deviation, and (iii) what fraction the ' tail ' is of the 

 whole population. 



The values of frequency in an arbitrary scale are : 



Taking the working origin at 20 19 seconds, we find 

 i/,' = -3-9214, v,' = 32-545,666 



for raw moment coefficients. Hence, if d be the distance from 29 seconds, i.e. the 

 stump of the tail from the mean, and S the standard deviation of the tail about its 

 mean: 



d = 9-5 - 3-9214 = 5-5786 sees., 



2 = Vf > _ Vi ' = 17-168,288, 

 and accordingly 2*/d J = '55l7. 



If this value be compared with those for i/r, in Table XI, p. 25, it will be seen 

 that we have got slightly more than the half of a normal curve, i.e. not a true 

 tail. We cannot therefore use Table XI, but must fall back on Table IX. 



* Gallon, R. S. Proc. Vol. 62, p. 310. See for another method of fitting, Pearson, Jiiomelrika, 

 Vol. n. p. 3. 





