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hw3solu_427

# hw3solu_427 - HW 3 Solutions All are from Chapter 3 12 b c...

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HW 3: Solutions All are from Chapter 3 12. a. In order for the flight to accommodate all the ticketed passengers who show up, no more than 50 can show up. We need Y ≤ 50. P(Y ≤ 50) = .05 + .10 + .12 + .14 + .25 + .17 = .83 b. Using the information in a. above, P(Y > 50) = 1 - P(Y ≤ 50) = 1 - .83 = .17 c. For you to get on the flight, at most 49 of the ticketed passengers must show up. P(Y ≤ 49) = .05 + .10 + .12 + .14 + .25 = .66. For the 3 rd person on the standby list, at most 47 of the ticketed passengers must show up. P(Y ≤ 47) = .05 + .10 + .12 = .27 13. a. P(X 3) = p(0) + p(1) + p(2) + p(3) = .10+.15+.20+.25 = .70 b. P(X < 3) = P(X 2) = p(0) + p(1) + p(2) = .45 c. P(3 X) = p(3) + p(4) + p(5) + p(6) = .55 d. P( 2 X 5) = p(2) + p(3) + p(4) + p(5) = .71 22. The jumps in F(x) occur at x = 0, 1, 2, 3, 4, 5, and 6, so we first calculate F( ) at each of these values: F(0) = P(X 0) = P(X = 0) = .10 F(1) = P(X 1) = p(0) + p(1) = .25 F(2) = P(X 2) = p(0) + p(1) + p(2) = .45 F(3) = .70, F(4) = .90, F(5) = .96, and F(6) = 1. The c.d.f. is F(x) = 00 . 1 96 . 90 . 70 . 45 . 25 . 10 . 00 . x x x x x x x x 6 6 5 5 4 4 3 3 2 2 1 1 0 0 Then P(X 3) = F(3) = .70, P(X < 3) = P(X 2) = F(2) = .45, P(3 X) = 1

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hw3solu_427 - HW 3 Solutions All are from Chapter 3 12 b c...

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