Pt or f but not both pt pf 2pt and f 08 07 206 03

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P(T or F but not both) = P(T) + P(F) – 2P(T and F) = 0.8 + 0.7 – 2(0.6) = 0.3 ANS__ 0.3 ________ 38. Given this information, are having a Facebook account and sending text messages independent or dependent? Justify using probabilities (not a discussion of teenagers and their behaviors.) Independent Dependent Why: P(T) * P(F) = (0.8)(0.7) = 0.56 P(T and F ) = 0.6 0.56 ≠ 0.6 (There are several other possible explanations) 39. Select 2 teenagers at random. What’s the chance that exactly one of them has a Facebook account? P(Exactly 1 Facebook) = 2P(F)P(F C ) = 2(0.7)(1-0.7) = 0.42 ANS__ 0.42 ________ Suppose Males and Females from a certain population take a survey and are asked whether or not they eat breakfast in the mornings. Suppose 60% of the males do, and 70% of the females do. Also suppose there are 55% males and 45% females in the population. 40. Make a two-way table and fill in all the probabilities (including those in the margins). Use proper labeling telling us which variables you have, and what their possible values are. B – Eats Breakfast M – Male F – Female P(B|M) = 0.6 P(B|F) = 0.7 P(M) = 0.55 P(F) = 0.45 P(B and F) = P(B|F)P(F) = (0.7)(0.45) = 0.315 P(B and M) = P(B|M)P(M) = (0.6)(0.55) = 0.33 P(B C and F) = P(F) – P(B and F) = 0.45 – 0.315 = 0.135 P(B C and M) = P(M) – P(B and M) = 0.55 – 0.33 = 0.22 B B C F 0.315 0.135 0.45 M 0.33 0.22 0.55 0.645 0.355 1 10
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ANSWER SHEET MC AND TF NAME________________________________ Midterm 2 Stat 133 SP 10 RECITATION TIME________________ GROUP NUMBER __________________ 1. A B C D 2. A B C D 3. A B C D 4. A B C D 5. A B C D 6. A B C D 7. A B C D 8. A B C D 9. A B C D 10. A B C D 11. A B C D 12. A B C D 13. A B C D 14. A B C D 15. A B C D 16. A B C D 17. A B C D 18. A B C D 19. A B C D 20. A B C D 21. A B C D 22. A B C D 23. A B C D 24. A B C D 25. A B C D 26. A B C D 27. A B C D 28. A B C D 29. A B C D 30. A B C D 11
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