Chapter 5: Probability Distributions: Discrete Probability Distributions
1 1 Slide Discrete Probability Distributions (Random Variables and Discrete Probability...
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1 1 Slide Slide
Discrete Probability Distributions(Random Variables and
Discrete Probability Distributions)
Chapter 5BA 201
3 3 Slide 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. Example: 1, 2, 3, …
A continuous random variable may assume any numerical value in an interval or collection of intervals. Example: 1.0, 1.1, 1.2, …
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Let x = number of TVs sold at the store in one day, where x can take on 5 values (0, 1, 2, 3, 4)
JSL Appliances
Discrete Random Variablewith a Finite Number of Values
We can count the TVs sold, and there is a finiteupper limit on the number that might be sold (whichis the number of TVs in stock).
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Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . .
Discrete Random Variablewith an Infinite Sequence of Values
We can count the customers arriving, but there isno finite upper limit on the number that might arrive.
JSL Appliances
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Random Variables
Question Random Variable x Type
Familysize
x = Number of dependents reported on tax return
Discrete
Distance fromhome to store
x = Distance in miles from home to the store site
Continuous
Own dogor 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
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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 formula.
Discrete Probability Distributions
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The probability distribution 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
f(x) = 1
10 10 Slide Slide
Discrete Probability Distributions
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Units Sold(x)
Number of Days
f(x)
0 80 0.401 50 0.252 40 0.203 10 0.054 20 0.10
200 1.00
80/200
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.10
.20
.30
.40
.50
0 1 2 3 4Values of Random Variable x (TV sales)
Pro
babili
ty
Discrete Probability Distributions
JSL AppliancesGraphical
representationof probabilitydistribution
Graphicalrepresentationof probabilitydistribution
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Expected Value
The expected value, or mean, of a random variable is a measure of its central location.
The expected value is a weighted average of the values the random variable may assume. The weights are the probabilities.
The expected value does not have to be a value the random variable can assume.
E(x) = = x f(x)
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Variance and Standard Deviation
The variance summarizes the variability in the values of a random variable.
The variance is a weighted average of the squared deviations of a random variable from its mean. The weights are the probabilities.
Var(x) = 2 = (x - )2f(x)
The standard deviation, , is defined as the positive square root of the variance.
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Expected Value
JSL Appliances
Units Sold(x)
Number of Days
f(x) x f(x)
0 80 0.40 0.001 50 0.25 0.252 40 0.20 0.403 10 0.05 0.154 20 0.10 0.40
E(x) 1.20expected number
of TVs sold in a dayexpected number
of TVs sold in a day
0 * 0.40
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Standard deviation of daily sales = = 1.2884 TVs
Variance
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x f(x) x f(x) x - (x -)2 (x -)2
f(x)
0 0.40 0.00 -1.20 1.44 0.5761 0.25 0.25 -0.20 0.04 0.0102 0.20 0.40 0.80 0.64 0.1283 0.05 0.15 1.80 3.24 0.1624 0.10 0.40 2.80 7.84 0.784
Variance of daily sales = 2 1.660
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Discrete Uniform Probability Distribution
The discrete uniform probability distribution is the simplest example of a discrete probability distribution 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
18 18 Slide Slide
Scenario
Amount(x)
Number of
Donors
5 3510 2515 2020 1525 5
A non-profit sends donation solicitations with pre-printed amounts. Based on past years, the non-profit knows approximately how many people will donate each amount. Compute the expected value of x, variance, and standard deviation.
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Binomial Probability Distribution
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
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Binomial Probability 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.
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where: x = the number of successes p = the probability of a success on one trial n = the number of trials f(x) = the probability of x successes in n trials
( )!( ) (1 )
!( )!x n xn
f x p px n x
Binomial Probability Distribution
Binomial Probability Function
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( )!( ) (1 )
!( )!x n xn
f x p px n x
Binomial Probability Distribution
Binomial Probability Function
Probability of a particular sequence of trial outcomes with x successes in n trials
Probability of a particular sequence of trial outcomes with x successes in n trials
Number of experimental outcomes providing exactly
x successes in n trials
Number of experimental outcomes providing exactly
x successes in n trials
26 26 Slide Slide
Binomial Probability Distribution
Evans Electronics
Evans Electronics is concerned about a low
retention rate for its employees. In recent years,
management has seen a turnover of 10% of the
hourly employees annually.
Choosing 3 hourly employees at random, what is
the probability that 1 of them will leave the company
this year?
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.
27 27 Slide Slide
Binomial Probability Distribution
Evans Electronics
The probability of the first employee leaving and the
second and third employees staying, denoted (S, F, F),
is given byp(1 – p)(1 – p)With a 0.10 probability of an employee leaving
on anyone trial, the probability of an employee
leaving onthe first trial and not on the second and third
trials isgiven by
(0.10)(0.90)(0.90) = (0.10)(0.90)2 = 0.081
28 28 Slide Slide
Binomial Probability Distribution
Evans Electronics
Two other experimental outcomes also result in one
success and two failures. The probabilities for all
three experimental outcomes involving one success
follow.
ExperimentalOutcome
(S, F, F)(F, S, F)(F, F, S)
Probability ofExperimental Outcome
p(1 – p)(1 – p) = (0.1)(0.9)(0.9) = 0.081
(1 – p)p(1 – p) = (0.9)(0.1)(0.9) = 0.081
(1 – p)(1 – p)p = (0.9)(0.9)(0.1) = 0.081
Total = 0.243
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L (0.1)
S (0.9)
Binomial Probability Distribution
1st Worker 2nd Worker 3rd Worker x Prob.
L (0.1)
S (0.9)
3
2
0
2
2
0.0010
0.0090
0.0090
0.7290
0.0090
1
1
0.0810
0.0810
0.0810
S (0.9)
L (0.1)
L (0.1)
S (0.9)
S (0.9)
L (0.1)
S (0.9)
L (0.1)
1
Evans ElectronicsUsing a tree diagramUsing a tree diagram
S (0.9)
L (0.1)
30 30 Slide Slide
Binomial Probability Distribution
f xn
x n xp px n x( )
!!( )!
( )( )
1
1 23!(1) (0.1) (0.9) 3(.1)(.81) .243
1!(3 1)!f
Let: p = 0.10, n = 3, x = 1
Evans ElectronicsUsing theprobabilityfunction
Using theprobabilityfunction
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Binomial Probability Distribution
(1 )np p
E(x) = = np
Var(x) = 2 = np(1 - p)
Expected Value
Variance
Standard Deviation
32 32 Slide Slide
Binomial Probability Distribution
3(.1)(.9) .52 employees
E(x) = np = 3(.1) = 0.3 employees out of 3
Var(x) = np(1 – p) = 3(0.1)(0.9) = 0.27
• Expected Value
• Variance
• Standard Deviation
Evans Electronics
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Scenario
Kearn Manufacturing (KM) submits bids for sales a number of times each month. Based on an analysis of past bids, the sales director knows that on average 34% of bids result in sales. KM Last week the sales director submitted three bids. The sales director would like to know how likely it is that two of the bids will result in sales.
1. What are n, x (that two bids result in sales), and p.
2. What is f(x)?3. Draw a tree diagram illustrating this scenario.4. Compute the expected value of x.5. Compute the variance and standard deviation.
38 38 Slide Slide
Binomial as a Discrete Probability Distribution
x f(x) x*f(x) (x-m) (x-m)2(x-
m)2*f(x)0 0.2875 0.0000 -1.0200 1.0404 0.29911 0.4443 0.4443 -0.0200 0.0004 0.00022 0.2289 0.4578 0.9800 0.9604 0.21983 0.0393 0.1179 1.9800 3.9204 0.1541
m 1.0200 s 0.6732
)()1(**)!(!
!)( xnx pp
xnx
nxf
E(x)=m 1.0200 =n*pVar(x) 0.6732 =n*p*(1-p)
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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 Probability Distribution
41 41 Slide 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 Probability Distribution
Bell Labs used the Poisson distribution to model the arrival of phone calls.
42 42 Slide Slide
Poisson Probability Distribution
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.
43 43 Slide Slide
Poisson Probability Function
Poisson Probability Distribution
f xe
x
x( )
!
where: x = the number of occurrences in an interval f(x) = the probability of x occurrences in an interval = mean number of occurrences in an interval e = 2.71828
44 44 Slide Slide
Poisson Probability Distribution
Poisson Probability Function
In practical applications, x will eventually becomelarge enough so that f(x) is approximately zero andthe probability of any larger values of x becomesnegligible.
Since there is no stated upper limit for the numberof occurrences, the probability function f(x) isapplicable for values x = 0, 1, 2, … without limit.
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Poisson Probability Distribution
Mercy Hospital
Patients arrive at the emergency room of Mercy
Hospital at the average rate of 6 per hour onweekend evenings. What is the probability of 4 arrivals in 30
minuteson a weekend evening?
46 46 Slide Slide
Poisson Probability Distribution
4 33 (2.71828)(4) .1680
4!f
= 6/hour = 3/half-hour, x = 4
Mercy Hospital Using theprobabilityfunction
Using theprobabilityfunction
Note:y
y
nn
1
9
1
3
13
22
Example:
47 47 Slide Slide
Poisson Probability Distribution
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
Pro
bab
ilit
y
actually, the
sequencecontinues:11, 12, …
Mercy Hospital
48 48 Slide Slide
Poisson Probability Distribution
A property of the Poisson distribution is thatthe mean and variance are equal.
m = s 2
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Poisson Probability Distribution
Variance for Number of ArrivalsDuring 30-Minute Periods
m = s 2 = 3
Mercy Hospital
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Scenario
On weekdays between 11:00 a.m. and 2:00 p.m., Joe’s Pizza receives approximately one order every two minutes.
1. What is the expected number of orders per hour?
2. What is the probability three orders will arrive in a five-minute period?
3. What is the probability no orders will arrive in a five-minute period?
4. What is the variance of the number of orders arriving?
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Discrete Probability Distributions
• x takes discrete values.• Values for f(x) may be subjectively assigned,
assigned based on frequency of occurrence, or uniform.o If f(x) is uniform, f(x) = 1/n where n is the
number of values x may assume.
E(x) = μ = [x * f(x)]
Var(x) = 2 = [(x – μ)2 * f(x)]
57 57 Slide Slide
Binomial Probability Distributions
• Four properties: (1) n identical trials, (2) success or failure, (3) p does not change, and (4) trials are independent.
• x is the number of successes.
E(x) = μ = n * p
Var(x) = 2 = n * p * (1-p)
)()1(**)!(!
!)( xnx pp
xnx
nxf
58 58 Slide Slide
Poisson Probability Distributions
• Number of occurrences over a specified interval of time or in space.
• (1) Probability of occurrence e is same for any two intervals of equal length and (2) occurrences are independent.
• x is the number of occurrences in the interval.
μ = 2
!
*)(
x
exf
x