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
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
Chapter 6 - Key Terms
? Random variables
– Discrete
– Continuous
? Bernoulli process
? Probability distributions
– Binomial distribution
– Poisson distribution
? 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
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,s2 = E[(x – μ) 2]
=
? 2002 The Wadsworth Group
? ? )( ii xPx
? ?? )()( 2 ii xPx ?
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
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 )! ?p x ? ( 1 – p ) n – x
? 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
The Binomial Distribution,
Working with the Equation
? To solve the problem,we substitute:
P ( x ) ? n !
x !?( n – x )!
?p x ? ( 1 – p ) n – x ?
P ( 3 ) ? 4 !
3 !?( 4 – 3 )!
(, 52 ) 3 ?(,48 ) 1 ?
? 4 ?(,14 060 8 ) ? (,48 ) ?, 26 996 7 ? 0, 27 00
? 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
Example,Binomial 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:
d,P(x = 12) Go to Appendix,Table A.1,n = 20,For p = 0.5
and k = 12,P(x = 12) = 0.1201
e,P(7 ? x ? 13) Go to Appendix,Table A.2,n = 20,For p = 0.5
and k = 13,P(x ? 13) = 0.9423
For p = 0.5 and k = 6,P(x ? 6) = 0.0577
P(7 ? x ? 13) = 0.9423 – 0.0577 = 0.8846
? 2002 The Wadsworth Group
Example,Using Microsoft Excel
? 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:
b,P(x ? 8) In a cell on an Excel worksheet,type
=BINOMDIST(8,20,0.5,true)
and you will see the answer,0.2517
c,P(x = 10) In a cell on an Excel worksheet,type
=BINOMDIST(10,20,0.5,false)
and you will see the answer,= 0.1762
? 2002 The Wadsworth Group
Example,Using Microsoft Excel
? 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:
d,P(x = 12) In a cell on an Excel worksheet,type
=BINOMDIST(12,20,0.5,false)
and you will see the answer,= 0.1201
e,P(7 ? x ? 13) In a cell on an Excel worksheet,type
=BINOMDIST(13,20,0.5,true)- BINOMDIST(6,20,0.5,true)
and you will see the answer,= 0.8846
? 2002 The Wadsworth Group
The Poisson Distribution
? The Poisson distribution defines the probability that an
event will occur exactly x times over a given span of time,
space,or distance.
– where l = the mean number of
occurrences over the span
e = 2.71828,a constant
– Mean = Variance = l
– Example,Problem 6.36,During the 12 p.m,– 1 p.m,noon
hour,arrivals at a curbside banking machine have been
found to be Poisson distributed with a mean of 1.3
persons per minute,If x = number of arrivals per minute,
determine:
a,E(x) = l = 1.3
P ( x ) ? l x ?e – lx !
? 2002 The Wadsworth Group
The Poisson Distribution,
Working with the Equation
? Example,Problem 6.36,During the 12 p.m,– 1 p.m,
noon hour,arrivals at a curbside banking machine have
been found to be Poisson distributed with a mean of 1.3
persons per minute,If x = number of arrivals per minute,
determine:
– b.
– c,
– d,P(x ? 2) = 0.8571 from Appendix Table A.4,l = 1.3,
k = 2
P ( x ? 0 ) ? ( 1, 3 ) 0 ( 2, 7 18 28 ) – 1, 30 ! ? 0, 2 72 51 ? 0, 2 72 5
P ( x ? 1 ) ? ( 1, 3 ) 1 ( 2, 7 18 28 ) – 1, 31 ! ? 0, 3 54 31 ? 0, 3 54 3
? 2002 The Wadsworth Group
The Poisson Distribution,
Using Microsoft Excel
? Example,Problem 6.36,
– b,P(x = 0) In a cell in an Excel spreadsheet,type
=POISSON(0,1.3,false)
and you will see the answer,= 0.2725
– c,P(x = 1) In a cell in an Excel spreadsheet,type
=POISSON(1,1.3,false)
and you will see the answer,= 0.3543
– d,P(x 2) In a cell in an Excel spreadsheet,type
=POISSON(2,1.3,true)
and you will see the answer,= 0.8571
? 2002 The Wadsworth Group
?
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
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
Chapter 6 - Key Terms
? Random variables
– Discrete
– Continuous
? Bernoulli process
? Probability distributions
– Binomial distribution
– Poisson distribution
? 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
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,s2 = E[(x – μ) 2]
=
? 2002 The Wadsworth Group
? ? )( ii xPx
? ?? )()( 2 ii xPx ?
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
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 )! ?p x ? ( 1 – p ) n – x
? 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
The Binomial Distribution,
Working with the Equation
? To solve the problem,we substitute:
P ( x ) ? n !
x !?( n – x )!
?p x ? ( 1 – p ) n – x ?
P ( 3 ) ? 4 !
3 !?( 4 – 3 )!
(, 52 ) 3 ?(,48 ) 1 ?
? 4 ?(,14 060 8 ) ? (,48 ) ?, 26 996 7 ? 0, 27 00
? 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
Example,Binomial 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:
d,P(x = 12) Go to Appendix,Table A.1,n = 20,For p = 0.5
and k = 12,P(x = 12) = 0.1201
e,P(7 ? x ? 13) Go to Appendix,Table A.2,n = 20,For p = 0.5
and k = 13,P(x ? 13) = 0.9423
For p = 0.5 and k = 6,P(x ? 6) = 0.0577
P(7 ? x ? 13) = 0.9423 – 0.0577 = 0.8846
? 2002 The Wadsworth Group
Example,Using Microsoft Excel
? 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:
b,P(x ? 8) In a cell on an Excel worksheet,type
=BINOMDIST(8,20,0.5,true)
and you will see the answer,0.2517
c,P(x = 10) In a cell on an Excel worksheet,type
=BINOMDIST(10,20,0.5,false)
and you will see the answer,= 0.1762
? 2002 The Wadsworth Group
Example,Using Microsoft Excel
? 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:
d,P(x = 12) In a cell on an Excel worksheet,type
=BINOMDIST(12,20,0.5,false)
and you will see the answer,= 0.1201
e,P(7 ? x ? 13) In a cell on an Excel worksheet,type
=BINOMDIST(13,20,0.5,true)- BINOMDIST(6,20,0.5,true)
and you will see the answer,= 0.8846
? 2002 The Wadsworth Group
The Poisson Distribution
? The Poisson distribution defines the probability that an
event will occur exactly x times over a given span of time,
space,or distance.
– where l = the mean number of
occurrences over the span
e = 2.71828,a constant
– Mean = Variance = l
– Example,Problem 6.36,During the 12 p.m,– 1 p.m,noon
hour,arrivals at a curbside banking machine have been
found to be Poisson distributed with a mean of 1.3
persons per minute,If x = number of arrivals per minute,
determine:
a,E(x) = l = 1.3
P ( x ) ? l x ?e – lx !
? 2002 The Wadsworth Group
The Poisson Distribution,
Working with the Equation
? Example,Problem 6.36,During the 12 p.m,– 1 p.m,
noon hour,arrivals at a curbside banking machine have
been found to be Poisson distributed with a mean of 1.3
persons per minute,If x = number of arrivals per minute,
determine:
– b.
– c,
– d,P(x ? 2) = 0.8571 from Appendix Table A.4,l = 1.3,
k = 2
P ( x ? 0 ) ? ( 1, 3 ) 0 ( 2, 7 18 28 ) – 1, 30 ! ? 0, 2 72 51 ? 0, 2 72 5
P ( x ? 1 ) ? ( 1, 3 ) 1 ( 2, 7 18 28 ) – 1, 31 ! ? 0, 3 54 31 ? 0, 3 54 3
? 2002 The Wadsworth Group
The Poisson Distribution,
Using Microsoft Excel
? Example,Problem 6.36,
– b,P(x = 0) In a cell in an Excel spreadsheet,type
=POISSON(0,1.3,false)
and you will see the answer,= 0.2725
– c,P(x = 1) In a cell in an Excel spreadsheet,type
=POISSON(1,1.3,false)
and you will see the answer,= 0.3543
– d,P(x 2) In a cell in an Excel spreadsheet,type
=POISSON(2,1.3,true)
and you will see the answer,= 0.8571
? 2002 The Wadsworth Group
?
