sololdmidterm2

# sololdmidterm2 - ST 351 Midterm 2 solutions 70 points Fall...

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ST 351 Midterm 2 solutions – 70 points Fall 2005 Form 2 Part I : 1. a 6. d 9. a 11. c (2.6 x 0.25) 13. a 15. b 16. a 2. a 7. d 10. a 12. c 14. d (0.3 x 0.4 x 0.3) 3. b 8. d (three ways one child could get the disease. Each has the same probability: (.2)(.8)(.8) = 4. a 0.128. Multiply this probability by 3 to take into account the 3 different ways of one child 5. b getting the disease: 3(0.128) = 0.384) Part II: 1. a. No. Multiple ways to show this – here’s one: Let A = person snorkeled and let B = person is an adult. P(A) = 33/94 = .3511. P(B) = 61/94 = .6489. P(A and B) = 25/94 = .26596. P(A) x P(B) = (.3511)(.6489) = .2278 T .26596. Therefore, “activity at the beach” and “age” are NOT independent. b. P(surf | youth) = P(surf and youth) 7/94 = P(youth) 33/94 = .2121 2. First three do NOT hit the gold spot and the fourth does. Therefore, find P(1 st no gold AND 2 nd no gold AND 3 rd no gold AND 4 th gold) = P(no gold) x P(no gold) x P(no gold) x P(gold) (since independent)

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## This note was uploaded on 04/07/2008 for the course ST 351 taught by Professor Kollath during the Spring '05 term at Oregon State.

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sololdmidterm2 - ST 351 Midterm 2 solutions 70 points Fall...

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