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Lesson #13 The Binomial Distribution. If X follows a Binomial distribution, with parameters n and p, we use the notation X ~ B(n, p) p x (1-p) (n-x) f(x)

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Presentation on theme: "Lesson #13 The Binomial Distribution. If X follows a Binomial distribution, with parameters n and p, we use the notation X ~ B(n, p) p x (1-p) (n-x) f(x)"— Presentation transcript:

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) p x (1-p) (n-x) f(x) = x = 0, 1, …, n E(X) = np Var(X) = np(1-p)

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

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 x012345x012345 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) = 1 - P(X < 3)= 1 - F(3) = P(X > 4) = 1 -.9130 =.0870

7 x012345x012345 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) = F(3) - F(1) = P(X < 3) =.9130 -.3370 =.5760 - P(X < 1) E(X) = (5)(.4) = 2

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

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

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

11 P(2 < X < 3) = P(X < 3) 0 1 2 3 4 5 - 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) 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(Y < 3)= F(3)=.9872

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

17 P(at most 4 passed) = P(X < 4) 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(Y > 6)= 1 - P(Y < 5) = 1 - F(5)= 1 -.9999 =.0001


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