Lesson #13

The BinomialDistribution

If X follows a Binomial distribution, withparameters n and p, we use the notation

X ~ B(n , p)

px(1-p)

(n-x)n

x

f(x) = x = 0, 1, … , n

E(X) = np Var(X) = np(1-p)

If X = # obese, then X ~ B(5 , .4).

P(no obese people) = P(X = 0) = f(0)

0 55= .4 .6

0

= (1)(1)(.07776) = .0778

x 5-x5f(x) = .4 .6

x

x = 0, 1, 2, 3, 4, 5

P(one obese person) = P(X = 1) = f(1)

1 45= .4 .6

1

= (5)(.4)(.1296) = .2592

P(two obese people) = P(X = 2) = f(2)

2 35= .4 .6

2

= (10)(.16)(.216) = .3456

f(3) 3 25= .4 .6

3

= (10)(.064)(.36) = .2304

f(4) 4 15= .4 .6

4

= (5)(.0256)(.6) = .0768

f(5) 5 05= .4 .6

5

= (1)(.01024)(1) = .0102

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)

= 1 - P(X < 3) = 1 - F(3)

= P(X > 4)

= 1 - .9130 = .0870

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)

= F(3) - F(1)

= P(X < 3)

= .9130 - .3370 = .5760

- P(X < 1)

E(X) = (5)(.4) = 2

P(2 < X < 3)

0 1 2 3 4 5

P(2 < X < 3)

0 1 2 3 4 5

P(2 < X < 3)

0 1 2 3 4 5

= P(X < 3)

P(2 < X < 3) = P(X < 3)

0 1 2 3 4 5

- P(X < 1)

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

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(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(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

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(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