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- https://stats.libretexts.org/Courses/Saint_Mary's_College_Notre_Dame/BFE_1201_Statistical_Methods_for_Finance_(Kuter)/04%3A_Random_Variables/4.S%3A_Random_Variables_(Solutions-_Practice__Homework)/1.02%3A_Normal_Solution_(Practice__Homework)ounces of water in a bottle The mean becomes zero. \(z = 2\) \(z = 2.78\) \(x = 20\) \(x = 6.5\) \(x = 1\) \(x = 1.97\) \(z = –1.67\) \(z \approx –0.33\) 0.67, right 3.14, left about 68% about 4% betw...ounces of water in a bottle The mean becomes zero. \(z = 2\) \(z = 2.78\) \(x = 20\) \(x = 6.5\) \(x = 1\) \(x = 1.97\) \(z = –1.67\) \(z \approx –0.33\) 0.67, right 3.14, left about 68% about 4% between –5 and –1 about 50% about 27% The lifetime of a Sunshine CD player measured in years. \(P(x < 1)\) Yes, because they are the same in a continuous distribution: \(P(x = 1) = 0\) \(1 – P(x < 3)\) or \(P(x > 3)\) \(1 – 0.543 = 0.457\) Let \(X =\) an SAT math score and \(Y =\) an ACT math score.
- https://stats.libretexts.org/Courses/Fresno_City_College/Introduction_to_Business_Statistics_-_OER_-_Spring_2023/06%3A_The_Normal_Distribution/6.11%3A_Chapter_Solution_(Practice__Homework)ounces of water in a bottle 3. The mean becomes zero. 11. \(z = 2\) 13. \(z = 2.78\) 15. \(x = 20\) 17. \(x = 6.5\) 19. \(x = 1\) 21. \(x = 1.97\) 23. \(z = –1.67\) 25. \(z \approx –0.33\) 27. about 6...ounces of water in a bottle 3. The mean becomes zero. 11. \(z = 2\) 13. \(z = 2.78\) 15. \(x = 20\) 17. \(x = 6.5\) 19. \(x = 1\) 21. \(x = 1.97\) 23. \(z = –1.67\) 25. \(z \approx –0.33\) 27. about 68% 33. about 4% 35. about 50% 39. The lifetime of a Sunshine CD player measured in years. Yes, because they are the same in a continuous distribution: \(P(x = 1) = 0\) 47. \(1 – P(x < 3)\) or \(P(x > 3)\) 49. \(1 – 0.543 = 0.457\) 51. Let \(X =\) an SAT math score and \(Y =\) an ACT math score.
- https://stats.libretexts.org/Courses/Fresno_City_College/Book%3A_Business_Statistics_Customized_(OpenStax)/06%3A_The_Normal_Distribution/6.11%3A_Chapter_Solution_(Practice__Homework)88. \(n = 100; p = 0.1; q = 0.9\) \(\mu = np = (100)(0.10) = 10\) \(\sigma=\sqrt{n p q}=\sqrt{(100)(0.1)(0.9)}=3\) \(z=\pm 1 : x_{1}=\mu+\mathrm{z} \sigma=10+1(3)=13 \text { and } x_2=\mu-\mathrm{z} \...88. \(n = 100; p = 0.1; q = 0.9\) \(\mu = np = (100)(0.10) = 10\) \(\sigma=\sqrt{n p q}=\sqrt{(100)(0.1)(0.9)}=3\) \(z=\pm 1 : x_{1}=\mu+\mathrm{z} \sigma=10+1(3)=13 \text { and } x_2=\mu-\mathrm{z} \sigma=10-1(3)=7.68 \%\) of the defective cars will fall between seven and 13. \(z=\pm 2 : x_{1}=\mu+\mathrm{z} \sigma=10+2(3)=16 \text { and } x_2=\mu-\mathrm{z} \sigma=10-2(3)=4.95 \%\) of the defective cars will fall between four and 16 \(z=\pm 3 : x_{1}=\mu+\mathrm{z} \sigma=10+3(3)=19 \text { a…
