THEORY OF STATISTICS. 
2. (Figures from ref. (5).) The following are the frequencies of the 
positive classes for the girls in the same investigation :— 
N 23,713 (4B) 587 
(4) 1,618 (40) 428 
(B) 2,015 (BC) 335 
(0) 770 (ABC) 156 
Find the frequencies of the ultimate classes. 
3. (Figures from Census, England and Wales, 1891, vol. iii.) Convert the 
census statement as below into a statement in terms of (@) the positive, (b) 
the ultimate class-frequencies. 4 =blindness, B=deaf-mutism, C'=mental 
derangement. 
N 29,002,525 (4 By) 82 
(4) 23,467 (4B0) 380 
(B) 14,192 (aBC) 500 
(0) 97,383 (4B0C) 25 
4. (Of. Mill’s Logic, bk. iii, ch. xvii.,, and ref. (1).) Show that if 4 
occurs in a larger proportion of the cases where B is than where 5 is not, 
then will B occur in a larger proportion of the cases where 4 is than where 
4 is not: i.e. given (4B)/(B)>(4B)/(B), show that (4 B)/(4)>(aB)/(a). 
5. (Cf. De Morgan, Formal Logic, p. 163, and ref. (1).) Most B's are 4’s, 
most B’s are (Ps: find the least number of A4’s that are C’s, 7.e. the lowest 
possible value of (40). 
6. Given that 
(4)=(a)=(B)=(8)=3N, 
show that 
(4B)=(aB), (48) =(aB). 
7. (Cf. ref. (2), § 9, ¢“ Case of equality of contraries.”) Given that 
(A)=(a)=(B)=(B)=(C)=(y)=4D, 
and also that 
(4BC)=(aBy), 
show that 
2 (4BC)=(4B)+(4C)+(BC)-3N. 
8. Measurements are made on a thousand husbands and a thousand wives. 
If the measurements of the husbands exceed the measurements of the wives in 
800 cases for one measurement, in 700 cases for another, and in 660 cases for 
both measurements, in how many cases will both measurements on the wife 
exceed the measurements on the husband ? 
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