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The Binomial

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If X = # obese, then X ~ B(5 , .4) P(no obese people) = P(X = 0) = f ... P(two obese people) = P(X = 2) = f(2) = (10)(.16)(.216) = .3456. f(3) = (10)(.064)(.36) ... – PowerPoint PPT presentation

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Title: The Binomial


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
(No Transcript)
5
(No Transcript)
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 lt 2)
F(2) .6826
P(at least 4 obese)
P(X gt 4)
1 - P(X lt 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 lt X lt 3)
P(X lt 3)
- P(X lt 1)
F(3) - F(1)
.9130 - .3370 .5760
E(X) (5)(.4) 2
8
P(2 lt X lt 3)
9
P(2 lt X lt 3)
10
P(2 lt X lt 3)
P(X lt 3)
11
P(2 lt X lt 3) P(X lt 3)
- P(X lt 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 gt 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 gt 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 gt 7)
P(Y lt 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
P(at most 4 passed)
P(X lt 4)
17
P(at most 4 passed)
P(X lt 4)
P(Y gt 6)
1 - P(Y lt 5)
1 - F(5)
1 - .9999 .0001
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