Presentation is loading. Please wait.

Presentation is loading. Please wait.

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)

Published by Modified over 9 years ago

Similar presentations

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

Download ppt "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)"

Similar presentations

Acknowledgement: Thanks to Professor Pagano

Acknowledgement: Thanks to Professor Pagano
Unit 18 Section 18C The Binomial Distribution. Example 1: If a coin is tossed 3 times, what is the probability of obtaining exactly 2 heads Solution:

Unit 18 Section 18C The Binomial Distribution. Example 1: If a coin is tossed 3 times, what is the probability of obtaining exactly 2 heads Solution:
Continuous Probability Distributions

Continuous Probability Distributions
Section 2.4 For any random variable X, the cumulative distribution function (c.d.f.) of X is defined to be F(x) = P(X  x).

Section 2.4 For any random variable X, the cumulative distribution function (c.d.f.) of X is defined to be F(x) = P(X  x).
Section 5.7 Suppose X 1, X 2, …, X n is a random sample from a Bernoulli distribution with success probability p. Then Y = X 1 + X 2 + … + X n has a distribution.

Section 5.7 Suppose X 1, X 2, …, X n is a random sample from a Bernoulli distribution with success probability p. Then Y = X 1 + X 2 + … + X n has a distribution.
P robability 1 07. Midterm Practice Condition Independence 郭俊利 2009/04/13.

P robability Midterm Practice Condition Independence 郭俊利 2009/04/13.
Statistics 270 - Lecture 15. Percentile for Normal Distributions The 100p th percentile of the N( ,  2 ) distribution is  +  (p)  Where  (p) is.

Statistics Lecture 15. Percentile for Normal Distributions The 100p th percentile of the N( ,  2 ) distribution is  +  (p)  Where  (p) is.
Lesson #15 The Normal Distribution. For a truly continuous random variable, P(X = c) = 0 for any value, c. Thus, we define probabilities only on intervals.

Lesson #15 The Normal Distribution. For a truly continuous random variable, P(X = c) = 0 for any value, c. Thus, we define probabilities only on intervals.
Section 10.6 Recall from calculus: lim= lim= lim= x  y  1 1 + — x x 1 1 + — x kx k 1 + — y y eekek (Let y = kx in the previous limit.) ekek If derivatives.

Section 10.6 Recall from calculus: lim= lim= lim= x  y  — x x — x kx k 1 + — y y eekek (Let y = kx in the previous limit.) ekek If derivatives.
The Erik Jonsson School of Engineering and Computer Science Chapter 2 pp. 49-100 William J. Pervin The University of Texas at Dallas Richardson, Texas.

The Erik Jonsson School of Engineering and Computer Science Chapter 2 pp William J. Pervin The University of Texas at Dallas Richardson, Texas.
STARTER The probability that a particular page in a maths book has a misprint is 0.2. Find the probability that of 12 pages in the book: 4 of them have.

STARTER The probability that a particular page in a maths book has a misprint is 0.2. Find the probability that of 12 pages in the book: 4 of them have.
Mutually Exclusive: P(not A) = 1- P(A) Complement Rule: P(A and B) = 0 P(A or B) = P(A) + P(B) - P(A and B) General Addition Rule: Conditional Probability:

Mutually Exclusive: P(not A) = 1- P(A) Complement Rule: P(A and B) = 0 P(A or B) = P(A) + P(B) - P(A and B) General Addition Rule: Conditional Probability:
Write each expression in standard form. Then name each polynomial by its degree and number of terms. 1.(2x 4 + 3x – 4) + (–3x + 4 + x 4 ) 2.(–3r + 4r 2.

Write each expression in standard form. Then name each polynomial by its degree and number of terms. 1.(2x 4 + 3x – 4) + (–3x x 4 ) 2.(–3r + 4r 2.
Chapter 5 Discrete Probability Distribution I. Basic Definitions II. Summary Measures for Discrete Random Variable Expected Value (Mean) Variance and Standard.

Chapter 5 Discrete Probability Distribution I. Basic Definitions II. Summary Measures for Discrete Random Variable Expected Value (Mean) Variance and Standard.
Binomial test (a) p = 0.07 n = 17 X ~ B(17, 0.07) (i) P(X = 2) = 0.22437 = 0.224 using Bpd on calculator Or 17 C 2 x 0.07 2 x 0.93 15 (ii) n = 50 X ~ B(50,

Binomial test (a) p = 0.07 n = 17 X ~ B(17, 0.07) (i) P(X = 2) = = using Bpd on calculator Or 17 C 2 x x (ii) n = 50 X ~ B(50,
Chapter 5.4: Bernoulli and Binomial Distribution

Chapter 5.4: Bernoulli and Binomial Distribution
20/6/1435 h Sunday Lecture 11 Jan 2009 1. Mathematical Expectation مثا ل قيمة Y 13 المجموع P(y)3/41/41 Y p(y)3/4 6/4.

20/6/1435 h Sunday Lecture 11 Jan Mathematical Expectation مثا ل قيمة Y 13 المجموع P(y)3/41/41 Y p(y)3/4 6/4.
The Binomial Distribution Permutations: How many different pairs of two items are possible from these four letters: L, M. N, P. L,M L,N L,P M,L M,N M,P.

The Binomial Distribution Permutations: How many different pairs of two items are possible from these four letters: L, M. N, P. L,M L,N L,P M,L M,N M,P.
The Binomial Distribution

The Binomial Distribution
Lecture 4 1 Discrete distributions Four important discrete distributions: 1.The Uniform distribution (discrete) 2.The Binomial distribution 3.The Hyper-geometric.

Lecture 4 1 Discrete distributions Four important discrete distributions: 1.The Uniform distribution (discrete) 2.The Binomial distribution 3.The Hyper-geometric.

Similar presentations

Ads by Google