Download Chapter 5

Discrete Probability
Distributions
Chapter 5
MSIS 111
Prof. Nick Dedeke
Learning Objectives
Distinguish between discrete random
variables and continuous random variables.
Learn how to determine the mean and
variance of a discrete distribution.
Identify the type of statistical experiments
that can be described by the binomial
distribution, and learn how to work such
problems.
Learning Objectives -- Continued
Decide when to use the Poisson
distribution in analyzing statistical
experiments, and know how to work
such problems.
Decide when binomial distribution
problems can be approximated by the
Poisson distribution, and know how to
work such problems.
Random Variable
Random Variable -- a variable which
associates a numerical value to the
outcome of a chance experiment
Example: Random Variable  Result of baseball game
Outcomes Values  Win, Lose, Tie
Discrete vs. Continuous Distributions
Discrete Random Variable – arise from
counting experiments. It has a finite number
of possible values or an infinite number of
possible values that can be arranged in sequence



Number of new subscribers to a magazine
Number of bad checks received by a restaurant
Number of absent employees on a given day
Continuous Random Variable – arise from
measuring experiments. It takes on values at
every point over a given interval



Current annual income of motorcycle distributorships
Elapsed time between arrivals of bank customers
Percent of the labor force that is unemployed
Experiment
Experiment: We want to flip a coin
twice. What are all the possible
outcomes of the experiment?
Experiment
Experiment: We want to flip a coin twice.
What are all the possibilities of the
experiment?
Outcomes possible: HH, HT, TH, TT
To assign numerical variables to the
outcomes, we have to define the random
variables
Let X be number of heads in the outcome
Let Y be number of tails in the outcome
Deriving Discrete Distributions for
Random Variable Experiments
Experiment: We want to flip a coin twice. What are all the
possibilities of the experiment?
Outcomes possible: HH, HT, TH, TT
Let X be number of heads in outcome
Let Y be number of tails in outcome
Outcome
HH
HT
TH
TT
values of X
2
1
1
0
values of Y
0
1
1
2
Exercise:
Deriving Discrete Distributions for
Random Variable Experiments
Experiment: We want to flip a coin thrice. What are all
the possibilities of the experiment? What are the
random variables?
Outcomes possible?:
What are the values for the distributions if random
variables are defined as follows?
Let X be number of heads in outcome
Let Y be number of tails in outcome
Tree diagram for Sample Spaces
H
H
T
H
HHH
T
H
HHT
HTH
HTT
T
H
T
H
T
T
H
THH
THT
TTH
TTT
T
Number of elements in sample space 2n = 23
The number 2 came from the possible outcomes per toss (experiment)
Response: Deriving Random Variables
Experiment: We want to flip a coin thrice. What are all
the possibilities of the experiment?
Let X be number of heads in outcome
Let Y be number of tails in outcome
Outcomes values of X values of Y
HHH
3
0
HHT
2
1
HTH
2
1
HTT
1
2
THH
2
1
THT
1
2
TTH
1
2
TTT
0
3
Exercise: Deriving Discrete Probability
Distribution
Experiment: We flip a coin thrice. What is the probability
distribution for X random variable the experiment? Let X be the
number of heads in outcome. What is the probability distribution
for Y random variable?
Outcome values of X
HHH
3
HHT
2
HTH
2
HTT
1
THH
2
THT
1
TTH
1
TTT
0
values of Y
0
1
1
2
1
2
2
3
X
0
1
2
3
Fi
1
3
3
1
8
P(x)
0.125
0.375
0.375
0.125
1.00
Exercise: Deriving Discrete Probability
Distribution
Experiment: Before we registered at a gym club, we investigated samples
of members of the gym that had injured themselves The data is provided
below. Let X be the number of muscle injuries in outcome. What is the
probability distribution for X random variable for the experiment? Let Y be
the number of bone injuries in outcome. What is the probability
distribution for Y random variable?
Samples
1st pool
2nd pool
3rd pool
4th pool
5th pool
6th pool
7th pool
8th pool
values of X
4
2
4
4
3
3
1
0
values of Y
0
1
1
2
1
2
2
0
Response: Deriving Discrete Probability
Distribution
Experiment: Before we registered at a gym club, we investigated samples
of members of the gym that had injured themselves The data is provided
below. Let X be the number of muscle injuries in outcome. What is the
probability distribution for X random variable for the experiment? Let Y be
the number of bone injuries in outcome. What is the probability
distribution for Y random variable?
Samples
1st pool
2nd pool
3rd pool
4th pool
5th pool
6th pool
7th pool
8th pool
values of X
4
2
4
4
3
3
1
0
values of Y
0
1
1
2
1
2
2
0
X
0
1
2
3
4
Fi
1
1
1
2
3
8
P(X)
0.125
0.125
0.125
0.250
0.375
1.00
Y
0
1
2
Fi
2
3
3
8
P(Y)
0.25
0.375
0.375
1.00
Example: Discrete Distributions & Graphs
Distribution of Daily
Crises
Number of
Probability
Crises
0
1
2
3
4
5
0.37
0.31
0.18
0.09
0.04
0.01
P
r
o
b
a
b
i
l
i
t
y
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Number of Crises
4
5
Requirements for a
Discrete Probability Functions
Probabilities are between 0 and 1,
inclusively
0  P( X )  1 for all X
Total of all probabilities equals 1
 P( X )  1
over all x
Do Any of these Distributions meet the
Requirements for Discrete Probability
Functions?
X
P(X)
X
P(X)
X
P(X)
-1
0
1
2
3
.1
.2
.4
.2
.1
1.0
-1
0
1
2
3
-.1
.3
.4
.3
.1
1.0
-1
0
1
2
3
.1
.3
.4
.3
.1
1.2
PROBABILITY
DISTRIBUTION
Do Any of these Distributions meet the
Requirements for Discrete
Probability Functions?
X
P(X)
X
P(X)
X
P(X)
-1
0
1
2
3
.1
.2
.4
.2
.1
1.0
-1
0
1
2
3
-.1
.3
.4
.3
.1
1.0
-1
0
1
2
3
.1
.3
.4
.3
.1
1.2
PROBABILITY
DISTRIBUTION
: YES
NO
NO
Example: Mean of a Discrete Distribution
  E X    X  P( X )
X
-1
0
1
2
3
P(X) X  P( X)
.1
.2
.4
.2
.1
-.1
.0
.4
.4
.3
1.0
 = 1.0
Variance and Standard Deviation
of a Discrete Distribution
  E  X    X  P( X )  1


2
2


P
(
X
)

1
.
2

  X 

  1.2  1.10
2
X
P(X)
X 
-1
0
1
2
3
.1
.2
.4
.2
.1
-2
-1
0
1
2
( X   ) ( X   )  P( X )
2
4
1
0
1
4
2
.4
.2
.0
.2
.4
1.2
Example: Mean of the Crises Data
  E X    X  P( X )  115
.
X
P(X)
X P(X)
0
.37
.00
1
.31
.31
2
.18
.36
3
.09
.27
4
.04
.16
5
.01
.05
P
r
o
b
a
b
i
l
i
t
y
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Number of Crises
1.15
4
5
Example: Variance and Standard
Deviation of Crises Data
    X     P( X )  1.41
2
2
X
P(X)
(X- )
(X- )
0
.37
-1.15
1.32
.49
1
.31
-0.15
0.02
.01
2
.18
0.85
0.72
.13
3
.09
1.85
3.42
.31
4
.04
2.85
8.12
.32
5
.01
3.85
14.82
.15
2
(X- ) 2  P(X)
1.41



2
 141
.  119
.
Exercise: Deriving Discrete Probability
Distribution
Experiment: We investigated the number of people in two gyms
who have injured themselves. The data is provided below. What is
the probability distribution for X1 and X2 random variables for the
experiment? What is the mean and standard deviations for each
gym? What is the probability that equal to or greater than 3
injuries occur in each gym? Let X1 and X2 be the random variables
for the number of injuries in outcome.
X1
4
2
4
4
3
3
1
0
X2
4
2
3
4
3
2
2
1
Exercise: Deriving Discrete Probability
Distribution
Experiment: We investigate the number of people in two gyms that
injured themselves. The data is provided below. What is the
probability distribution for X random variable for each experiment?
What is the mean and standard deviations for each gym? What is
the probability that equal to or greater than 3 injuries occur in
each gym? Let X be the number of injuries in outcome.
X1
4
2
4
4
3
3
1
0
X2
4
2
3
4
3
2
2
1
X2
1
2
3
4
Fi
1
3
2
2
8
P(x2)
0.125
0.375
0.250
0.250
1.00
X1
0
1
2
3
4
Fi
1
1
1
2
3
8
P(x1)
0.125
0.125
0.125
0.250
0.375
1.00
Response: Deriving Discrete Probability
Distribution (Gym I)
What is the mean and standard deviations for each gym? What is
the probability that equal to or greater than 3 injuries occur in
each gym? Let X1 be the number of injuries in outcome.
X1
0
1
2
3
4
Fi
1
1
1
2
3
8
P(x1)
0.125
0.125
0.125
0.250
0.375
1.00

X*P(x1) 
0
-2.625
0.125 -1.625
0.250 -0.625
0.750 0.375
1.50
1.375
2.625
X 1 

X 1 

2

P ( x1) X 1 
6.8906
2.6406
0.3906
0.1406
1.8906
11.953

2
0.86125
0.330
0.0488
0.0351
0.07089
1.9842
Standard deviation = σ = 1.984 = 1.408 injuries
P(X1<=2) = P(X1=0)+ P(X1=1)+ P(X1=2)=0.125+0.125+0.125= 0.375
P(X1 >=3) = 1 - P(X1<=2) = 1 –0.375 = 0.625
  E  X    X  P( X )
    X     P( X )
2
2
Response: Deriving Discrete Probability
Distribution (Gym II)
What is the mean and standard deviations for each gym? What is
the probability that equal to or greater than 3 injuries occur in
each gym? Let X be the number of injuries in outcome.
X2
1
2
3
4
Fi
1
3
2
2
8
P(x2)
0.125
0.375
0.250
0.250
1.00
X*P(x2) (X2-μ) )(X2-μ)2
0.125 -1.625 2.6406
0.750 -0.625 0.3906
0.750 0.375 0.1406
1.00
1.375 1.8906
2.625
11.953
P(x2)(X2-μ)2
0.330
0.1464
0.0351
0.4726
0.9842
Standard deviation = σ = 0.9842 = 0.9921 injuries
P(X1<=2) = P(X1=1)+ P(X1=2)= 0.125+0.375
P(X1 >=3) = 1 - P(X1<=2) = 1 –0.5 = 0.5
  E  X    X  P( X )
    X     P( X )
2
2
Response: Choice between Two Gyms – Use of
Discrete Probability Distribution
We should select Gym II because it has lower probability of
yielding high injuries.
Gym I
Mean = σ = 2.625 injuries
Standard deviation = σ = 1.408 injuries
Probability of 3 or higher injuries = 1 - P(X1<=2) = 1 –0.5 = 0.625
Gym II
Mean = σ = 2.625 injuries
Standard deviation = σ = 0.9921 injuries
Probability of 3 or higher injuries = 1 - P(X1<=2) = 1 –0.5 = 0.5
Exercise: Deriving Discrete Probability
For the data below, find the following probabilities:
What is the probability that greater than 3 injuries occur in the gym?
What is the probability that less than 2 injuries occur in the gym?
What is the probability that greater than or equal to 2 but less than 4
injuries occur in the gym? Let X1 be the number of injuries in
outcome.
X1
0
1
2
3
4
Fi
1
1
1
2
3
8
P(x1)
0.125
0.125
0.125
0.250
0.375
1.00

X*P(x1) 
0
-2.625
0.125 -1.625
0.250 -0.625
0.750 0.375
1.50
1.375
2.625
X 1 

X 1 

2
6.8906
2.6406
0.3906
0.1406
1.8906
11.953

P ( x1) X 1 

2
0.86125
0.330
0.0488
0.0351
0.07089
1.9842
Response: Deriving Discrete Probability
For the data below, find the following probabilities:
What is the probability that greater than 3 injuries occur in the gym?
What is the probability that less than 2 injuries occur in the gym?
What is the probability that greater than or equal to 2 but less than 4
injuries occur in the gym? Let X1 be the number of injuries in
outcome.
X1
0
1
2
3
4
Fi
1
1
1
2
3
8
P(x1)
0.125
0.125
0.125
0.250
0.375
1.00

X*P(x1) 
0
-2.625
0.125 -1.625
0.250 -0.625
0.750 0.375
1.50
1.375
2.625
X 1 

X 1 

2
6.8906
2.6406
0.3906
0.1406
1.8906
11.953

P ( x1) X 1 

2
0.86125
0.330
0.0488
0.0351
0.07089
1.9842
P(X1>3) = P(X1=4) = 0.375
P(X1<2) = P(X1=0)+ P(X1=1) = 0.125+0.125 = 0.25
P(2=<X1<4) = P(X1=2)+ P(X1=3)=0.125+0.25 = 0.375
Casino Case
There are two casinos that we could play at
in the coming months. Design an experiment
that should be used to compare the two
locations. We want to be able to make the
most money. Create a sample table using
fictitious data to show how you will make a
decision. We would choose one location.
Work in groups and prepare a submission.
Illustration of Casino problem
Daily wins
Daily wins
Day 1
Day 2
Day 3
Day 1
Day 2
Day 3
300,000
Day
Day
Day
Day
Day
Day
Day
Day
200,000
4
5
6
7
1,000,000
4
5
6
7
Day 8
Day 8
Day 9
Day 10
Day 9
Day 10
1,000,000
Casino 1
150,000
200,000
150,000
1,000,000
Casino 2
Illustration of Casino problem
X1
0
1
Fi
8
2
P(x1)
0.8
0.2
X2
0
1
Fi
8
2
P(x1)
0.8
0.2
Sum
10
1
Sum
10
1
Casino 1
Casino 2
Calculate arithmetical average for each casino.
X the random variable, is defined as wins greater than
$50,000 in the table shown above.
Which casino should we choose?
What happens if random variable X is defined as wins greater
than $200,000?
Use of Tables
In the exercises that we have done, we
assumed that the person doing the
decision making had to conduct
experiments on gym members. This is
because, she does not know ahead of
hand, the probability distribution of the
experiment. If she knew this, she would
not need to do experiments, she would
just pick the appropriate probability
distributions and do the calculations.
Examples of standard: Discrete and
Continuous Probability Distributions
Discrete


Binomial distributions
Poisson distributions
Continuous






Uniform distributions
normal distributions
exponential distributions
T distributions
chi-square distributions
F distributions