SomeAnswersToHomeworkExercises

SomeAnswersToHomeworkExercises - Some answers to important...

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Some answers to important exercises 2-67. a) P(A) = 30/100 = 0.30 b) P(B) = 75/100 = 0.75 c) P(A') = 1 – 0.30 = 0.70 d) P(A B) = 20/100 = 0.2 e) P(A B) = 85/100 = 0.85 f) P(A’ B) =90/100 = 0.9 2-88. (a) 9 30 ( ) 0.39 100 P A (b) 13 9 ( ) 0.22 100 P B (c) ( ) 9/100 ( | ) 0.409 ( ) 22 /100 P A B P A B P B (d) ( ) 9/100 ( | ) 0.23 ( ) 39/100 P A B P B A P A 2-114. Let A and B denote the events that the first and second chips selected are defective, respectively. a) P(B) = P(B|A)P(A) + P(B|A')P(A') = (19/99)(20/100) + (20/99)(80/100) = 0.2 b) Let C denote the event that the third chip selected is defective. 00705 . 0 100 20 99 19 98 18 ) ( ) ( ) ( ) ( ) ( ) ( A P A B P B A C P B A P B A C P C B A P
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2-149. Denote as follows: S = signal, O = organic pollutants, V = volatile solvents, C = chlorinated compounds a) P(S) = P(S|O)P(O)+P(S|V)P(V)+P(S|C)P(C) = 0.997(0.60) + 0.9995(0.27) + 0.897(0.13) = 0.9847 b) P(C|S) = P(S|C)P(C)/P(S) = ( 0.897)(0.13)/0.9847 = 0.1184 3-86. Let X denote the number of questions answered correctly. Then, X is binomial with n = 30 and p = 0.25. a) 20 10 21 9 22 8 23 7 24 6 25 5 26 4 27 3 2 30 30 ( 20) 0.25 0.75 0.25 0.75 20 21 30 30 30 0.25 0.75 0.25 0.75 0.25 0.75 22 23 24 30 30 30 0.25 0.75 0.25 0.75 0.25 0.75 25 26 27 30 0.25 28 P X 8 2 29 1 30 0 6 30 30 0.75 0.25 0.75 0.25 0.75 29 30 1.821 10 b) 0 30 1 29 2 28 30 30 30 5 0.25 0.75 0.25 0.75 0.25 0.75 0 1 2 P X 3 27 4 26 30 30 0.25 0.75 0.25 0.75 0.0979 3 4 3-92. E(X) = 25 (0.01) = 0.25 V(X) = 25 (0.01) (0.99) = 0.248 3 0.25 3 0.248 1.74 X X a ) X is binomial with n = 25 and p = 0.01 0 25 1 24 ( 1.74) ( 2) 1 ( 1) 25 25 1 0.01 0.99 0.01 0.99 0.0258 0 1 P X P X P X
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b) X is binomial with n = 25 and p = 0.04 0 25 1 24 ( 1) 1 ( 1) 25 25 1 0.04 0.96 0.04 0.96 0.2642 0 1 P X P X c) Let Y denote the number of times X exceeds 1 in the next five samples. Then, Y is binomial with n = 5 and p = 0.190 from part b. 0 5 5 ( 1) 1 ( 0) 1 0.2642 0.7358 0.7843 0 P Y P Y The probability is 0.651 that at least one sample from the next five will contain more than one defective 3-129. a) 4 0 4 4 ( 0) 0.0183 0!
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