492 Prof. Karl Pearson. 



VH Vi, ^3 v* their coefficients of variation, i.e., t^/m^ <r 2 /m 2 , 

 <r 4 /W4 respectively ; n 2 , r 23 , r 34 , r 41 , r 24 , r 13 , the six coefficients of corre- 

 lation ; EX, e 2 , 6 3 , e 4 the deviations of the four organs from, their means, 

 i.e, Xi = !! + !, ic 2 = ^2 + e 2) # 3 = w 3 +c 3 , 0*4 = m4-f e 4 ; ?' 13 the mean 

 value of the index a?i/aj s , and ^ 24 the mean value of # 2 /#4 ; 2i, 2 2 the 

 standard deviations of the indices x : /x s and # 2 /#4 respectively ; and n 

 the total number of groups of organs. 



We shall suppose the ratios of the deviations to the mean absolute 

 values of the organs are so small that their cubes may be neglected. 



Then 



w 3 



= S( Cl ) 8(63) S(6 lC3 ) 



if we neglect quantities of the third order in e/wi. But S(e x ) = 

 S(e 3 ) = 0, S(e 1 e 3 ) = nr w triff^ and S(e 3 2 ) = na^. 



Hence: = ^(l + ^-r,^^) (i). 



Similarly ^ = 2* (i+^-ww.) . . () 



Thus we see that the mean of an index is not the ratio of the means 

 of the corresponding absolute measurements, but differs by a quan- 

 tity depending on the correlation and variation coefficients of the 

 absolute measurements. 



(3) Proposition II. To find the standard deviation of an index in 

 terms of the coefficients of variation, and coefficient of correlation of 

 the two absolute meastirements. 



= ^ { S /^-^+ square terms) 2 V 

 ma 2 I \m! m 3 / J 



if we neglect cubic terms. 



.-. 2 18 = i l ^(v l *+vj-2r l3 v l v 3 ). .......... (iii). 



