VARIABILITY OF A SINGLE CHARACTER 119 
Circumference of ear? Number of rows on cob # 
rm S F; F i f yo Tip see — 
425 I a 10 I 
4.8 2 5 I rio 6 2 7 
g4 15 13 I T4 3 25 sr 
5-4 28 19 9 16 85 77 70) 
5-7 40 20 a 18 103 115 79 
6.0 70 4s 50 20 59 SI g 
6.3 58 So 67 se at 26 18 
6.6 50 67 89 2 8 I I 
6.9 34 58 85 26 : 4 
2 5 18 23 28 I 
75 I 5 13 
75 I 
8.1 — on 
310 344 307 313 335 253 
M=6.121 M= 6.304 M= 6.505 M 
g=0.530 ¢=0.579 6 =0.499 o 
C=866 C= 9.18 C=6.90 iC 
17-911 M=18.107 M= 
2.501 C= S417 oS. 2.377 
= 13.96 C=13.35 C=14.09 
Original problems. Besides the solving of distributions given 
in the text, the student should have practice in devising and 
solving problems of his own. I know of no better method of 
teaching variability, and at the same time insuring rational 
conceptions of heredity, than by this methodical and accurate 
study of characters taken singly. 
For this purpose the student may use not only dimensions 
like length and circumference, but he may use weights and num- 
bers. He may take the heights of pupils in the school, the 
grades they make in classes, or he may take the yield of milk 
of many cows, or the weights of milk at the creamery, Anything 
1 Experience shows that it is better to take values by 0.25 instead of 0.30; 
thus, 4.50, 4.75, 5.00, 5.25, etc. These distributions were made smooth only 
by careful assignment of alternate measurements. This scheme of grouping 
has been discarded for this reason. 
2 Below each f column will be found the corresponding values. Thus of the 
first frequency, footing 310, the M = 6.121, the « = 0.530, and the C = 8.66. 
In this way the answers can be identified for each problem contained in these 
tables. 
