CHAPTER 6 Discrete probability distributions to accompany Introduction to business statistics fourth edition by ronald M. Weiers Presentation by priscilla Chaffe-Stengel Donald N. stengel C 2002 The Wadsworth Group
CHAPTER 6 Discrete Probability Distributions to accompany Introduction to Business Statistics fourth edition, by Ronald M. Weiers Presentation by Priscilla Chaffe-Stengel Donald N. Stengel © 2002 The Wadsworth Group
l Chapter 6- learning objectives Distinguish between discrete and continuous random variables Differentiate between the binomial and the Poisson discrete probability distributions and heir applications Construct a probability distribution for a discrete random variable, determine its mean and variance and specify the probability that a discrete random variable will have a given value or value in a given range o 2002 The Wadsworth Group
Chapter 6 - Learning Objectives • Distinguish between discrete and continuous random variables. • Differentiate between the binomial and the Poisson discrete probability distributions and their applications. • Construct a probability distribution for a discrete random variable, determine its mean and variance, and specify the probability that a discrete random variable will have a given value or value in a given range. © 2002 The Wadsworth Group
ll Chapter 6-Key terms Randomⅴ ariables Discrete Continuous Bernoulli process Probability distributions Binomial distribution Poisson distribution o 2002 The Wadsworth Group
Chapter 6 - Key Terms • Random variables – Discrete – Continuous • Bernoulli process • Probability distributions – Binomial distribution – Poisson distribution © 2002 The Wadsworth Group
l Discrete us continuous variables Discrete Variables: Continuous variables: Can take on only Can take on any value certain values along at any point along an an interval interval the number of sales the depth at which a made in a week drilling team strikes oil the volume of milk the volume of milk bought at a store produced by a cow the number of the proportion of defective parts defective parts o 2002 The Wadsworth Group
Discrete vs Continuous Variables • Discrete Variables: Can take on only certain values along an interval – the number of sales made in a week – the volume of milk bought at a store – the number of defective parts • Continuous Variables: Can take on any value at any point along an interval – the depth at which a drilling team strikes oil – the volume of milk produced by a cow – the proportion of defective parts © 2002 The Wadsworth Group
l Describing the distribution for a Discrete random variable The probability distribution for a discrete random variable defines the probability of a discrete value x -Mean:=E(x)=∑xP(x) Variance: 02=El(x-u)2I ∑(x1-+)2P(x) o 2002 The Wadsworth Group
Describing the Distribution for a Discrete Random Variable • The probability distribution for a discrete random variable defines the probability of a discrete value x. – Mean: µ= E(x) = – Variance: s 2 = E[(x – µ)2 ] = © 2002 The Wadsworth Group ( ) i i x P x ( − ) ( ) 2 i i x P x
l The Bernoulli Process Characteristics 1. There are two or more consecutive trials 2. In each trial, there are just two possible outcomes 3. The trials are statistically independent 4. The probability of success remains constant trial-to-trial o 2002 The Wadsworth Group
The Bernoulli Process, Characteristics 1. There are two or more consecutive trials. 2. In each trial, there are just two possible outcomes. 3. The trials are statistically independent. 4. The probability of success remains constant trial-to-trial. © 2002 The Wadsworth Group
l The Binomial distribution The binomial probability distribution defines the e proba ability of exactly x successes in n trials of the Bernoulli process P(x) x!(n-X 可x(-=xy-x for each value of x Mean =E(x)=1π Variance:o2=F[(x-1)2]=nπ(1-π o 2002 The Wadsworth Group
The Binomial Distribution • The binomial probability distribution defines the probability of exactly x successes in n trials of the Bernoulli process. – for each value of x. – Mean: µ= E(x) = n p – Variance: s2 = E[(x – µ)2 ] = n p (1 – p) P(x) = n! x!(n– x)!px(1–p)n– x © 2002 The Wadsworth Group
l The Binomial Distribution An example worked by equation Problem 6.23: a study by the International Coffee Association found that 52% of the U.S. population aged 10 and over drink coffee For a randomly selected group of 4 individuals, what is the probability that 3 of them are coffee drinkers? Number Proportion Coffee drinkers(x) 3 52 Noncoffee drinkers 1 48 Totals 4 1.00 So,p=0.52,(1-p)=0.48,x=3,(m-x) o 2002 The Wadsworth Group
The Binomial Distribution, An Example Worked by Equation • Problem 6.23: A study by the International Coffee Association found that 52% of the U.S. population aged 10 and over drink coffee. For a randomly selected group of 4 individuals, what is the probability that 3 of them are coffee drinkers? Number Proportion Coffee drinkers (x) 3 .52 Noncoffee drinkers 1 .48 Totals 4 1.00 So, p = 0.52, (1 – p) = 0.48, x = 3, (n – x) = 1 . © 2002 The Wadsworth Group
l The Binomial distribution Working with the equation To solve the problem we substitute x(n-x) Tx(1-丌)n-x P(3)=3(4-3) (.52)3(8)1 4(140608)(48)=269967≈0.2700 o 2002 The Wadsworth Group
The Binomial Distribution, Working with the Equation • To solve the problem, we substitute: P(x) = n! x!(n– x)!px(1–p)n– x = P(3) = 4! 3!(4–3)! (.52)3(.48)1 = = 4(.140608)(.48) = .269967 0.2700 © 2002 The Wadsworth Group
l The Binomial distribution An example worked with tables Problem: According to a corporate association, 50.0% of the population of Vermont were boating participants during the most recent year. For a randomly selected sample of 20 Vermont residents, with x=the number sampled who were boating participants that year, determine a.E(x)=nT=20×0.50=10 b P(a<8)Go to Appendix, Table A 2, n=20. For T=0.5 andk=8,P(x≤8)=0.2517 C P(=10) Go to Appendix, Table A 1, n=20. For T=0.5 andk=10,P(x=10)=0.1762 o 2002 The Wadsworth Group
The Binomial Distribution, An Example Worked with Tables • Problem: According to a corporate association, 50.0% of the population of Vermont were boating participants during the most recent year. For a randomly selected sample of 20 Vermont residents, with x = the number sampled who were boating participants that year, determine: a. E(x) = n p = 20 x 0.50 = 10 b. P(x 8) Go to Appendix, Table A.2, n = 20. For p = 0.5 and k = 8, P(x 8) = 0.2517 c. P(x = 10) Go to Appendix, Table A.1, n = 20. For p = 0.5 and k = 10, P(x = 10) = 0.1762 © 2002 The Wadsworth Group
