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Lesson #13 The Binomial Distribution

Lesson #13 The Binomial Distribution. If X follows a Binomial distribution, with parameters n and p, we use the notation X ~ B(n , p). x. (n-x). p. (1-p). f(x) =. x = 0, 1, … , n. E(X) = np. Var(X) = np(1-p). If X = # obese, then X ~ B(5 , .4). x = 0, 1, 2, 3, 4, 5.

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Lesson #13 The Binomial Distribution

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  1. Lesson #13 The Binomial Distribution

  2. If X follows a Binomial distribution, with parameters n and p, we use the notation X ~ B(n , p) x (n-x) p (1-p) f(x) = x = 0, 1, … , n E(X) = np Var(X) = np(1-p)

  3. . If X = # obese, then X ~ B(5 , .4) x = 0, 1, 2, 3, 4, 5 P(no obese people) = P(X = 0) = f(0) = (1)(1)(.07776) = .0778

  4. P(one obese person) = P(X = 1) = f(1) = (5)(.4)(.1296) = .2592 P(two obese people) = P(X = 2) = f(2) = (10)(.16)(.216) = .3456

  5. f(3) = (10)(.064)(.36) = .2304 f(4) = (5)(.0256)(.6) = .0768 f(5) = (1)(.01024)(1) = .0102

  6. x 0 1 2 3 4 5 f(x) .0778 .2592 .3456 .2304 .0768 .0102 F(x) .0778 .3370 .6826 .9130 .9898 1.0000 P(no more than 2 obese) = P(X < 2) = F(2) = .6826 P(at least 4 obese) = P(X > 4) = 1 - P(X < 3) = 1 - F(3) = 1 - .9130 = .0870

  7. x 0 1 2 3 4 5 f(x) .0778 .2592 .3456 .2304 .0768 .0102 F(x) .0778 .3370 .6826 .9130 .9898 1.0000 P( 2 to 3, inclusive, obese) = P(2 < X < 3) = P(X < 3) - P(X < 1) = F(3) - F(1) = .9130 - .3370 = .5760 E(X) = (5)(.4) = 2

  8. 0 1 2 3 4 5 P(2 < X < 3)

  9. 0 1 2 3 4 5 P(2 < X < 3)

  10. 0 1 2 3 4 5 P(2 < X < 3) = P(X < 3)

  11. 0 1 2 3 4 5 P(2 < X < 3) = P(X < 3) - P(X < 1)

  12. If X = # who passed, X ~ B(10 , .9) Let Y = # who did not pass, Y ~ B(10 , .1) X + Y = 10, so Y = 10 - X E(X) = (10)(.9) = 9

  13. P(at least 7 passed) = P(X > 7) X 0 1 2 3 4 5 6 7 8 9 10 Y 10 9 8 7 6 5 4 3 2 1 0

  14. P(at least 7 passed) = P(X > 7) X 0 1 2 3 4 5 6 7 8 9 10 Y 10 9 8 7 6 5 4 3 2 1 0

  15. P(at least 7 passed) = P(X > 7) = P(Y < 3) = F(3) = .9872 X 0 1 2 3 4 5 6 7 8 9 10 Y 10 9 8 7 6 5 4 3 2 1 0

  16. X 0 1 2 3 4 5 6 7 8 9 10 Y 10 9 8 7 6 5 4 3 2 1 0 P(at most 4 passed) = P(X < 4)

  17. X 0 1 2 3 4 5 6 7 8 9 10 Y 10 9 8 7 6 5 4 3 2 1 0 P(at most 4 passed) = P(X < 4) = P(Y > 6) = 1 - P(Y < 5) = 1 - F(5) = 1 - .9999 = .0001

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