12_Homework 4 Ch. 3 Solutions

12_Homework 4 Ch. 3 Solutions - X = 1.407 b. X has...

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HOMEWORK #4 CHAPTER 3 72. a. h(x; 6,4,11) b. 18 . 2 11 4 6 = 76. The only possible values of X are 3, 4, and 5. p(3) = P(X = 3) = P(first 3 are B’s or first 3 are G’s) = 2(.5) 3 = .250 p(4) = P(two among the 1 st three are B’s and the 4th is a B) + P(two among the 1 st three are G’s and the 4th is a G) = 375 . ) 5 (. 2 3 2 4 = p(5) = 1 – p(3) – p(4) = .375 82. a. P(X = 1) = F(1;.2) – F(0;.2) = .982 - .819 = .163 b. P(X 2) = 1 – P(X 1) = 1 – F(1;.2) = 1 - .982 = .018 c. P(1 st doesn’t 2 nd doesn’t) = P(1 st doesn’t) P(2 nd doesn’t) = (.819)(.819) = .671 83. 200 1 = p ; n = 1000; λ = np = 5 a. P(5 X 8) = F(8;5) – F(4;5) = .492 b. P(X 8) = 1 – P(X 7) = 1 - .867 = .133 88. Let X = the number of diodes on a board that fail. a. E(X) = np = (200)(.01) = 2, V(X) = npq = (200)(.01)(.99) = 1.98,
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Unformatted text preview: X = 1.407 b. X has approximately a Poisson distribution with = np = 2, so P(X 4) = 1 P(X 3) = 1 F(3;2) = 1 - .857 = .143 c. P(board works properly) = P(all diodes work) = P(X = 0) = F(0;2) = .135 Let Y = the number among the five boards that work, a binomial r.v. with n = 5 and p = .135. Then P(Y 4) = P(Y = 4 ) + P(Y = 5) = 5 4 ) 865 (. ) 135 (. 5 5 ) 865 (. ) 135 (. 4 5 + = . 00144 + .00004 = .00148 92. a. P(X = 10 and no violations) = P(no violations | X = 10) P(X = 10) = (.5) 10 [F(10;10) F(9;10)] = (.000977)(.125) = .000122...
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12_Homework 4 Ch. 3 Solutions - X = 1.407 b. X has...

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