Solutions To Mathematics Textbooks/Introduction to Mathematical Statistics (5th edition) (ISBN 8178086301)/Chapter 1/Section 2

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Contents

Section 2.1Edit

1.1Edit

Find the union   and the intersection   of the two sets   and   where

(a)  

(b)  

(c)  

Solution (c):

 

 

Rest are similar.


1.3Edit

List all the possible arrangements of the four letters m,a,r and y. Let   be the collection of the arrangements in which y is in the last position. Let   be the collection of the arrangements in which m is in the first position. Find the union and intersection of   and  .

Solution: The arrangements are: mary, mayr, mray, mrya, myra, myar, amry, amyr, aymr, ayrm, arym, army, ryma, ryam, ramy, raym, rmya, rmay, yrma, yram, yamr, yarm, ymra, ymar.

 

 . The rest is obvious.


1.5Edit

If a sequence of sets   is such that   the sequence is said to be a nondecreasing sequence. Give an example of this kind of sequence of sets.

Solution: Let   where n = 1, 2, 3...


1.6Edit

If a sequence of sets   is such that   the sequence is said to be a nonincreasing sequence. Give an example of this kind of sequence of sets.

Solution: Let   where n=1,2,3...


1.7Edit

If   are sets such that   is defined as the union  . Find   if

(a)  

(b)  

Solution: (b)  


1.8Edit

If   are sets such that   is defined as the intersection  . Find   if

(a)  

(b)  

(c)  

Solution: (c)  


1.9Edit

For every one dimensional set A let   where  . If   and  , find   and  .

Solution:   and   using the formulae of sum of a geometric series.


1.10Edit

For every one dimensional set A for which the integral exists, let   where  , otherwise let   be undefined. If   and  , find   and  .

Solution:   can be found by integrating f(x) from   to  ,   as integral over a set of measure zero is 0 and   can be found by integrating f(x) from 0 to 1.


1.11Edit

Let   for every two dimensional set A for which the integral exists; otherwise let   be undefined. If   and  , find   and  .

Solution:  ,   is zero since it represents volume of a sheet and   if we introduce polar coordinates. This evaluates to  .


1.15Edit

To join a certain club, a person must be either a statistician or a mathematician or both. Of the 25 members in this club, 19 are statisticians and 16 are mathematicians. How many persons in the club are both a statistician and a mathematician?

Soluion: The number of persons which are statisticians or mathematicians = number of statisticians + number of mathematicians - number of persons which are both statisticians and mathematicians. This can be proved directly using properties of the counting measure. The general proof is as follows: Let A and B be two finite sets. Endow the natural numbers with the counting measure which we shall denote by n. We shall denote union by + and intersection by . and the context will make it clear whether we mean actual addition, multiplication or union, intersection. Now A-B (Here - represents the relative complement, AB and B-A are disjoint and as n is sigma additive so

  1. n(A+B)=n(A-B)+n(B-A)+n(AB)
  2. n(A)=n(A-B)+n(AB) which implies n(A-B)=n(A)-n(AB)
  3. n(B)=n(B-A)+n(AB) which implies n(B-A)=n(B)-n(AB)

Substituting n(A-B) and n(B-A) from (2) and (3) into (1) gives us n(A+B)=n(A)+n(B)-n(AB). So the answer is 10.


1.16Edit

After a hard fought football game, it was reported that of the 11 starting players, 8 hurt a hip, 6 hurt an arm, 5 hurt a knee, 3 hurt both a hip and an arm, 2 hurt both a hip and a knee, 1 hurt both an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.

Solution: We shall first establish that in the language of the previous problem:


 .


Now,


       


Now if we let A, B and C to be the set of players with injured hips, knees and arms respectively, we would find the contradiction on substituting the appropriate terms in the above formula. Hence the report is not accurate.