30 SUMMARIZATION OF DATA Ch. 2 



The procedures outlined in Table 2.52 can be justified as follows. 

 It is obvious that the midpoints, s«, of Table 2.51 can be rewritten 

 as follows: 



2.5 = 12.5 - 2(5), 



7.5 = 12.5 - 1(5), 



12.5 = 12.5 + 0(5), 



17.5 = 12.5 + 1(5), 



22.5 = 12.5 + 2(5). 

 Then 



2(/-z) = 2[12.5 + 2(5)] + 6[12.5 + 1(5)] + 10[12.5 + 0(5)] 

 + 5[12.5 - 1(5)] + 2[12.5 - 2(5)]; 

 = (2 + 6 + 10 + 5 + 2)(12.5) 



+ [2(2) + 6(1) + 10(0) + 5(-l) + 2(-2)](5); 



or, a bit more generally, 



S(/-z) - (S/)(12.5)+S(/-d)(I), 



if d = +2 for the top class of Tables 2.51 and 2.52, d = +1 for the 

 next class down, etc., until d — —2 for the bottom class of each of 

 those tables. (The symbol J stands for the length of the class in- 

 terval.) Therefore, 



2(/-s) S(/-d)(/) 

 (2.51) m = = 12-5 H 



1 ' 2(/) S(/) 



is the approximation to /x which can be obtained from the frequency 

 distribution table. It should be clear that some other midpoint be- 

 sides 12.5 could have been used without changing the answer ob- 

 tained. Hence if there is any best choice of a midpoint to use as a 

 base point (with d = 0), that choice must rest on its leading to 

 simpler computations. Generally, the d should be taken as zero for 

 the class with the greatest frequency. If the distribution is quite 

 non-symmetrical, it is advisable to shift the choice one way or the 

 other so that the positive and negative fd's will be more nearly bal- 

 anced. This rarely will be more than two classes from the one with 

 the greatest frequency, /. 



