K. Pearson and A. W. Young 
87 
(2) We are now in a position to set down the algebraical values of the product- 
moment coefficients. 
(a) s or i = 0. p^^ ,/ ?2/,o = 1-3.5 (2? - 1), 
Vo.o = 1> P2,o = 1> 2^4,0 = 3, ^6,0 = 15, 2^8,0 = 105> Pw,o = 945. 
These are of course the simple moment coefficients of the normal curve when 
the unit of abscissal length is the standard deviation. 
(b) SOX 1=1. 'p,i+i, 1 = 2,+i =1.3.5 (2? + 1) r, 
Pn = Ih. 1 = 3r, 2h. i = 15»'' T^?, i = 1 O^r, 2^9, j = 945r . 
Generally J>'it-i,i!lhuo = ^i^d provides a means of finding r and testing how 
far the correlation of two variates is normal. 
(c) s or / = 2. 
2^2,2* = 2^2.,2= ] -3.5 (2? - 1) {1 + 2/r2}. 
p,,,= l + 2r^ 2.,,4 = 3(l + 4f2), p^^_i5(i + 6,2)_ 
= 105 (1 + 8/2), 2h.,^ = 945 (1 + 10y2). 
{d) s OT t = 3. 
2)3, 2m = 2W3 = 1-3.5 (2M- 1) r{3 + 2/r2}. 
3 = 3r (3 + 2r2), ;/53,5 = 15>- (3 + 4/2), , = 105r (3 + 6/-2), p, ^ = 945r (3 f 8r). 
(e) s or / = 4. 
2'4,2. = 2^2^4 = 1-3.5 (2/ - 1) {3 + 12/^^ + _ ,4|^ 
7)4, 4 = 3 (3 + 24r2 + Sr") , 7)4, 0 = 15 (3 + .SGr^ + 24r*) , 
?J4, 8 = 105 (3 + 48r2 + 48}-"), 2J4, j„ = 945 (3 + GOr- + 80/-*). 
(/) s or / = 5. 
2^5, 2.+1 = 2^2.+!, 5=1.3.5 (2/ + 1 ) r {15 + 20^^^ + 4/ (/ - 1) r^}, 
7)5, 5 = 1 5f (15 + 40/2 j_ g,.4) ^ ^ _ 1 (25 + QOr- + 24r*) , 
2)5^ 9 = 94.5r (15 + 80^2 + 48r*). 
{g) s or t = 6. 
Pe,2f = V2f,G = 1.3.5 (2/- 1){15 + 90fr2+ G0/(/ - I)r* 8^ (f ~ ]) (^- 2)r«} 
Pe, 6 = 15 (15 + 270/2 + 360/4 _^ 43^0)^ ^ _ 105 (^5 ^ 300^.2 .|_ 7204 i92rG), 
y,,, j„ = 945 (15 + 450J-2 + 1200/" + 480/6). 
(h) s or t = 7. 
?'7,2m = 2'2/+i,7=1.3.5 (2/+ 1)/ 
{105 + 210//2 + 84/ (;' - 1) /4 + 8/ (/ - 1) (< - 2) /S}, 
2)7, 7 = 105/ (105 + 630/2 + 504/* + 48/«), 
2)7, 9 = 945?' (105 + 840/2 _^ joOS/* + 192/«). 
{i) s or /. = 8. 
Ts, 2t = 2'2/, 8=1.3.5 (2/ - 1 ) {105 + 840/2 + 340/ (/ - 1 ) 
+ 224/ (/ - 1) (< - 2) /« + 16i (/ - 1) (/. - 2) (/ - 3) /8}, 
p^ g = 105 (105 + 3360/2 + 10080/* + 5376/* + 384/8), 
2^8,10 = 945 (105 + 4200r2 + 16800/* + 13440/^ + 1920^-8). 
