XXVI 



Tables for Statisticians and Biometricians [IX — X 



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



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

 = 398,0682. 

 Hence I X I N = ^tt 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 



sV = - 3-9214, ^' = 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 2 the standard deviation of the tail about its 



mean : 



d = 95 - 3-9214 = 5-5786 sees., 



2* = „ 2 ' _ „/* = 17-168,288, 



and accordingly 2 2 /e£ l = "5517. 



If this value be compared with those for fa in Table XT, 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. 



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

 Vol. n. p. 3. 



