Math 3338.
Problems for home study
Section 2.1: 1, 4, 5, 6, 7, 8
Section 2.2: 13, 15, 18, 22, 23.
Section 2.3: 31, 32, 34, 35, 36, 37, 39, 41, 43. Solution of Problem 43, question 3
Notes: Problems 31-35 are elementary. Problem 36 uses combinations (why. It is
important to justify why we have combinations and not permutations. In Problem
37 you choose postions 1-5 for the letters A and B. This problem is not easy.
Form words of length 5 by repeating A and B. A must remain ahead of B at all
times. Can we have the first two votes be A and B with A first? Problem 39 is
easy, basic counting, but you must read the problem well. Solution for
Problem 39: If you read the problem carefully each time we perform an
experiment we are not allowed to use a configuration of
temperature-pressure-catalyst that has been used in a previous experiment for
obvious reasons. So in a series of 5 experiments we have 60 configurations for
the 1st experiment, 59 for the 2nd, 58 for the third and
so on. For the series of 5 five experiments we have (60)(59)(58)(57)(56)
outcomes in total. If you use a different catalyst for each of the five
experiments then you have (60)(48)(36)(24)(12) (favorable) outcomes in total.
The probability to use a different catalyst in each of the five trials is
[(60)(48)(36)(24)(12)]/[ (60)(59)(58)(57)(56)]=0.0456.
Problem 41 was solved in
class. Interpret the word `first' appropriately; it implies that the collection
of the 15 phones is split in two classes. You are interested in the first one.
Is the order in which the phones are repaired important?
Also check problems from Section 2.3
supplementary problems : Pg 1: 1,2, 4
Pg
2: 2, 3
Pg
3: 1,2,3
Pg
4: 1 through 7. The answers are in pg. 5.
Section 2.4: 45,47,48,49,51,59,
88(p. 90, this is not a joke);
question c of 88 requires Bayes’ theorem.
More problems from this section: 52,53,55, 61,63, 64, 89, 93, 92, 95. From page
2 of Section 2.3 supplementary problems solve problem 1.
Download solution of
Quiz 4 and to problems on specificity, sensitivity and positive predictive
value
Section 2.5: 70,71,72,73,74,75,79,81,83.
Supplementary problems
(84-109): 85, 91,
95,96,97,98a,105,107, 88, 92,93,94,101,110.
Section 3.2: 12, 13, 15, 23
Section 3.3: 28, 29, 35, 36, 37, 41
Section 3.4: 44, 45, 46, 47, 49, 50, 53, 57.
Section 3.5: 59, 61, 65,67, 68, 69, 70, 75. Remark on the solution of 69a: When calculating the probability that the claim is rejected when p=0.8, we calculate a conditional probability but without using the definition of conditional probability. To this end, we assume that the probability model governing the sample space is b(25,0.8). Now,
P(rejecting the claim: “p>=0.8” |p=0.8)=P(X<=15),
where the latter probability will be calculated using b(25,0.8). Note that the rejection of the claim “p>=0.8” is equivalent to X<=15. P(X<=15)=0.017 per table A1e. Similarly, the probability of not rejecting the claim if p=.7 is P(X>15)=1- P(X<=15)=0.811 calculated using b(25,0.7).
Section 3.6: 81, 83, 84, 87,
Section 3.7: 93, 95, 96, 97, 98, 99.
Supplementary problems (pp.
149-153): 111, 113c, 116, 117 (not for exam 1), 119, 120 (this is interesting
but challenging. Try to find what is the r.v. that applies to this problem),
123, 124, 126,
Section 4.1: 1,2,3,4,7,11
Section 4.2: 19, 24, 25, 29, 30, 31.
Section 4.3: 39, 41, 43d,e,f, 45, 47, 50, 53c, 61, 62, 63,
64.
Section 4.7: 108, 109, 110, 113, 114 (ignore
this section).
Section 3.7 (revisited): 98, 99, 100, 105, 106
Section 4.4: 69, 70a,c,f, 71. Focus on 73, 74,75.
No problems from Section 4.5
Section 5.1: 1 (all but d), 2abc, 3, 4ab, 5, 6, 7abc d, 9, 11abc,
12, 13.
Section 5.2: 18,19,20,21, 22,23,24, 25, 26.
Course completed with Section 5.2
Section 5.3: (We will
cover only pages 249 through 251 from this section): 37abcd, 39abcd, 40abcd.
Section 6.3: 27a. The objective is that you learn how to derive
the formulas 6.7, 6.8, 6.9, 6.10 in this section and the proposition in page
305.
This is the complete list of
problems you have to be able to solve to make an A in this course.