1 Slide
Chapter 5
Discrete Probability Distribution
Learning objectives
1. Understand random variables and probability distributions.
1.1. Distinguish discrete and continuous random variables.
2. Able to compute Expected value and Variance of discrete random variable.
3. Understand:
3.1. Discrete uniform distribution 3.2. Binomial distribution
3.3. Poisson distribution
3 Slide
A random variable is a numerical description of the outcome of an experiment.
Random Variables
A discrete random variable may assume either a finite number of values or an infinite sequence of values.
A continuous random variable may assume any numerical value in an interval or collection of intervals.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Let x = number of TVs sold at the store in one day, where x can take on 5 values (0, 1, 2, 3, 4)
Example: JSL Appliances
n Discrete random variable with a finite number of values
n Discrete random variable with an infinite sequence of values
• We can count the customers arriving, but there is no finite upper limit on the number that might arrive.
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . .
5 Slide
Random Variables
Question Random Variable x Type Family
size
x= Number of dependents reported on tax return
Discrete
Distance from home to store
x= Distance in miles from home to the store site
Continuous
Own dog
or cat x= 1 if own no pet;
= 2 if own dog(s) only;
= 3 if own cat(s) only;
= 4 if own dog(s) and cat(s)
Discrete
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable.
We can describe a discrete probability distribution with a table, graph, or equation.
Discrete Probability Distributions
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
7 Slide
The probability distribution for discrete random variable is defined by a probability function, denoted by f(x), which provides the probability for each value of the random variable.
The required conditions for a discrete probability function are:
Discrete Probability Distributions
f(x) > 0 Σf(x) = 1
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
n a tabular representation of the probability distribution for TV sales was developed.
n Using past data on TV sales, …
Number Units Sold of Days
0 80
1 50
2 40
3 10
4 20
200
x f(x)
0 .40
1 .25
2 .20
3 .05
4 .10
1.00
80/200
Discrete Probability Distributions
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
9 Slide
.10 .20 .30
.40 .50
0 1 2 3 4
Values of Random Variable x (TV sales)
Probability
Discrete Probability Distributions
n Graphical Representation of Probability Distribution
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Expected Value and Variance
The expected value, or mean, of a random variable is a measure of its central location.
The variance summarizes the variability in the values of a random variable.
The standard deviation, σ, is defined as the positive square root of the variance.
Var(x) = σ2= Σ(x -µ)2f(x) E(x) = µ = Σxf(x)
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
11 Slide n Expected Value
expected number of TVs sold in a day
x f(x) xf(x)
0 .40 .00
1 .25 .25
2 .20 .40
3 .05 .15
4 .10 .40
E(x) = 1.20
Expected Value and Variance
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
n Variance and Standard Deviation
0 1 2 3 4
-1.2 -0.2 0.8 1.8 2.8
1.44 0.04 0.64 3.24 7.84
.40 .25 .20 .05 .10
.576 .010 .128 .162 .784 x -µ (x -µ)2 f(x) (x -µ)2f(x)
Variance of daily sales = σ 2= 1.660 x
squaredTVs
Standard deviation of daily sales = 1.2884 TVs
Expected Value and Variance
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
13 Slide
In-class Exercise
n Random variables
•
# 2 (page 188)•
# 5 (page 188)n Expected value and variance
•
#16 (page 196)•
#17 (page 197)Discrete Uniform Probability Distribution
The discrete uniform probability distribution is the simplest example of a discrete probabilitydistribution given by a formula.
The discrete uniform probability function is f(x) = 1/n
where:
n= the number of values the random variable may assume
the values of the random variable are equally likely
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
15 Slide
Binomial Distribution
n Four Properties of a Binomial Experiment
3. The probability of a success, denoted by p, does not change from trial to trial.
4. The trials are independent.
2. Two outcomes, success and failure, are possible on each trial.
1. The experiment consists of a sequence of n identical trials.
stationarity assumption
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Binomial Distribution
Our interest is in the number of successes occurring in the n trials.We let x denote the number of successes occurring in the n trials.
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
17 Slide
where:
f(x) = the probability of x successes in n trials n= the number of trials
p= the probability of success on any one trial
−
−
= −
= −
−
( )
( )
( ) (1 )
! (1 )
!( )!
n x x
n x x
f x n p p
x
n p p
x n x
−
−
= −
= −
−
( )
( )
( ) (1 )
! (1 )
!( )!
n x x
x n x
f x n p p
x
n p p
x n x
Binomial Distribution
n Binomial Probability Function
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
( )
( ) ! (1 )
!( )!
x n x
f x n p p
x n x
= − −
−
( )
( ) ! (1 )
!( )!
x n x
f x n p p
x n x
= − −
−
Binomial Distribution
!
!( )!
n x n x−
!
!( )!
n
x n x− ppxx(1(1−−pp))((n xn x−− ))
n Binomial Probability Function
Probability of a particular sequence of trial outcomes with x successes in n trials Number of experimental
outcomes providing exactly xsuccesses in n trials
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
19 Slide
Example: Evans Electronics
Evans is concerned about a low retention rate for employees. In recent years, management has seen a turnover of 10% of the hourly employees annually.
Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
n Using the Binomial Probability Function
Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year?
f x n
x n x px p n x
( ) !
!( )! ( )( )
= − 1− −
f x n
x n x px p n x
( ) !
!( )! ( )( )
= − 1− −
1 2
(1) 3! (0.1) (0.9) 3(.1)(.81) .243 1!(3 1)!
f = = =
−
1 2
(1) 3! (0.1) (0.9) 3(.1)(.81) .243 1!(3 1)!
f = = =
−
Let: p = .10, n = 3, x = 1
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Example: Evans Electronics
21 Slide n Tree Diagram
1stWorker 2ndWorker 3rdWorker x Prob.
Leaves (.1)
Stays (.9)
3
2
0 2 2
Leaves (.1) Leaves (.1)
S (.9) Stays (.9)
Stays (.9)
S (.9) S (.9)
S (.9) L (.1)
L (.1) L (.1)
L (.1) .0010
.0090
.0090
.7290 .0090
1 1
.0810
.0810 .0810 1
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Example: Evans Electronics
n Using Tables of Binomial Probabilities
n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50 3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250
1 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750 2 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750 3 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250
p
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Example: Evans Electronics
23 Slide
Binomial Distribution
(1 )
np p
σ = np(1−p)
σ = −
E(x) = µ = np
Var(x) = σ2= np(1 −p)
n Expected Value
n Variance
n Standard Deviation
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
3(.1)(.9) .52 employees σ = 3(.1)(.9) .52 employees=
σ= =
E(x) = µ = 3(.1) = .3 employees out of 3
Var(x) = σ2= 3(.1)(.9) = .27
n Expected Value
n Variance
n Standard Deviation
Example: Evans Electronics
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
25 Slide
In-class Exercise
n #26 (page 207)
n #37 (page 208)
A Poisson distributed random variable is often useful in estimating the number of occurrences over a specified interval of time or space
It is a discrete random variable that may assume an infinite sequence of values (x = 0, 1, 2, . . . ).
Poisson Distribution
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
27 Slide
Examples of a Poisson distributed random variable:
the number of knotholes in 14 linear feet of pine board
the number of vehicles arriving at a toll booth in one hour
Poisson Distribution
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Poisson Distribution
n Two Properties of a Poisson Experiment
2. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.
1. The probability of an occurrence is the same for any two intervals of equal length.
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
29 Slide n Poisson Probability Function
Poisson Distribution
f x e
x
x
( )= µ !−µ
f x e
x
x
( )= µ !−µ where:
f(x) = probability of x occurrences in an interval µ = mean number of occurrences in an interval
e= 2.71828
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Example: Mercy Hospital
MERCY
Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings.
What is the
probability of 4 arrivals in
30 minutes on a weekend evening?
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
31 Slide n Using the Poisson Probability Function
4 3
3 (2.71828)
(4) .1680
f =3 (2.71828)4 4! −3 =
(4) .1680
f 4!
= − =
MERCY
µ = 6/hour = 3/half-hour, x = 4
Example: Mercy Hospital
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
n Using Poisson Probability Tables
µ
x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 0 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .0498 1 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .1494 2 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .2240 3 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .2240 4 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .1680 5 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .1008 6 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .0504 7 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .0216 8 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081
MERCY
Example: Mercy Hospital
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
33 Slide n Poisson Distribution of Arrivals
MERCY
Poisson Probabilities
0.00 0.05 0.10 0.15 0.20 0.25
0 1 2 3 4 5 6 7 8 9 10
Number of Arrivals in 30 Minutes
Probability
actually, the sequence
continues:
11, 12, …
Example: Mercy Hospital
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Poisson Distribution
A property of the Poisson distribution is that the mean and variance are equal.
µ = σ 2
•L.O.
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
35 Slide
MERCY
n Variance for Number of Arrivals During 30-Minute Periods
µ = σ2= 3
•Random variables
•EV and Variance
•Uniform discrete dist.
•Binomial dist.
•Poisson dist.
Example: Mercy Hospital
In-class Exercise
n #38 (page 211)
n #41 (page 212)
37 Slide