VARIATION 23 



The average length of all the nuts is very nearly 44. 

 In order to determine the average deviation we should 

 have to find the value of the series 2 (44 — 30) +7 

 (44-34)+28, {44-38)+59, (44-42)+49, (46-44)+33 

 (50— 44)+6, (54-44)+l, (58-44), and divide the total 

 by 185. 



Xow, in practice, the average deviation is not found 

 to be the most convenient expression, a somewhat 

 different function, knowii as the standard deviation, being 

 more useful. The method of determining this differs 

 from that just given, in that one squares the deviation 

 in each case, and after dividing the total of the products 

 by the number of individuals, one extracts the square 

 root of the answer. 



Our case would be worked out as follows : — 



(44-30)2 = 14- = 196 X 2= 392 

 (44 - 34)2 = 102 = 100 X 7 _ 70O 

 (44 - 38)2 _ g2 _ 36 X 28 = 1008 

 (44-42)2 = 22 = 4 X 59 = 236 

 (46 - 44)2 3= 22 = 4 X 49 = 196 

 (50-44)2= 62= 36x33 = 1188 

 (54-44)2 _ 1Q2 = 100 X 6 = 600 

 (58-44)2 = 142 = 196 X 1 = 196 



Total, 4516 



4516 , 



-jg5- = 24-4; v24-4 = about 4-9. 



4*9 millimetres, then, is the standard deviation of the 

 lot of nuts measured, with regard to length. And this 

 is a convenient measure of the amount of variation 

 with regard to this character. 



In order to obtain comparative figures mth regard to 

 different characters, it is usual to express the standard 

 deviation as a percentage of the mean or average. This 

 gives what is kno^^■n as the coeificient of variation. In 

 our ."■'se, 4*9 expressed as a percentage of the average, 

 44, givet> a coefficient of variation of about 11 per cent. 



If, then, we are in a position to give the mean value 

 of a character, to state the type of curve which the 

 variation follows, and to give the standard deviation 

 or the coefficient of variation we have described the 



