Series to Frequency Distributions in Space. 395- 



3 

 Hence r = 4, r' = 3, 6 = j-^g, 



n_2o) 2 i 



and .*. ±— — H- = - so that <7 = '6 or '4. 



We take q='Q for equation (41) shows that since \ p s0 is 

 negative q>p. 



From (21) and (22) we find c=c' = l. 

 Finally 



N 210 1 



nO-l)(n-2) . . (n-r-r' + l) ~ 10*9 . . . 5'4 " 2880' 

 .-. the given table is equivalent to the series 



2 ^(t)(.?)( 6 >^>- 



Numerical Example 2. 



The following table, based on the Registrar General's 

 report for 1910, gives the distribution according to ages of 

 the 235,252 bachelors and the same number of spinsters 

 who intermarried in 1910 and whose ages were stated. 



Taking an arbitrary origin at 22*5 years of age for each 

 sex, we obtain the following values for the raw moments : — 

 p\o= '5227798 p' 01 = -8860797 p' u =l*075566 

 p' 20 = 1-1860516 p' 02 = 1-888826 p'^ 2*7207547 

 y 30 = 2-8226710 y 03 = 5-543315 p'^ 3-1404579 

 p' 40 = 10-5105716 / , = 22-63546 



Thus the mean age at marriage of spinsters is 25*11390 y 

 and of bachelors 26*93040. "When transferring to the mean 

 we diminish p 20 by ^ and p U) by ip^o—^io *° allow for the 

 grouping of the material (cf. W. F. Sheppard, Proc. Lond. 

 Math. Soc. vol. xxix. pp. 353-380). The moments about the 

 mean are : — 



p 20 = -8294195, p 2=l'020356, /> u = '012341, o- = -910725, 



^30=1-2482896, /> 03 = 1-9137522, p 81 = 1*0295785, <r' = 1*010126, 



p 40 = 5*9016221, 7> 0i = 9-5141106, p l2 = 1-067845. 



The distribution is decidedly unsymmetrical. We will 

 first attempt a fit by moments of 3rd order. 

 The constants are as follows : — 



^ = •673016, \=-935105, £=1-505454, 77 = 1-46650, 



= 20-308, (£ = -26427; 



whence rc 3 -3-8532n+ 3*8532=0, 



giving an imaginary n. Thus it is impossible to fit this 



