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Understanding the Binomial Distribution: Probabilities and Expectations

This lesson explores the Binomial Distribution, a fundamental concept in probability theory. We define a random variable (X) that follows a Binomial distribution with parameters (n) (number of trials) and (p) (probability of success), denoted as (X sim B(n, p)). Key components such as the probability mass function (PMF), expectation (E(X)), and variance (Var(X)) are discussed. Using examples, we calculate probabilities for different scenarios, including the likelihood of a certain number of successes in trials, and cumulative probabilities for varied outcomes.

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Understanding the Binomial Distribution: Probabilities and Expectations

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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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