Discrete Probability Distributions

Chapter 5MSIS 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, TTTo assign numerical variables to the outcomes, we have to define the random variablesLet 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 values of X values of Y HH 2 0

HT 1 1TH 1 1TT 0 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

T

HT

H

T

HT

H

T

H

T

HT

H

HHH

HHTHTHHTT

THHTHT

TTHTTT

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 outcomeOutcomes values of X values of Y

HHH 3 0HHT 2 1HTH 2 1HTT 1 2THH 2 1THT 1 2TTH 1 2TTT 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 values of YHHH 3 0HHT 2 1HTH 2 1HTT 1 2THH 2 1THT 1 2TTH 1 2TTT 0 3

X Fi P(x)0 1 0.1251 3 0.3752 3 0.3753 1 0.125

8 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 values of X values of Y1st pool 4 02nd pool 2 13rd pool 4 14th pool 4 25th pool 3 16th pool 3 27th pool 1 28th pool 0 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 values of X values of Y1st pool 4 02nd pool 2 13rd pool 4 14th pool 4 25th pool 3 16th pool 3 27th pool 1 28th pool 0 0

Y Fi P(Y)

0 20.25

1 30.375

2 30.3758 1.00

X Fi P(X)

0 10.125

1 10.125

2 10.125

3 20.250

4 30.3758 1.00

Example: Discrete Distributions & Graphs

012345

0.370.310.180.090.040.01

Number of Crises Probability

Distribution of Daily Crises

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5

Probability

Number of Crises

Requirements for a Discrete Probability Functions

Probabilities are between 0 and 1, inclusively

Total of all probabilities equals 1

0 1 P X( ) for all X

P X( )over all x 1

Do Any of these Distributions meet the Requirements for Discrete Probability Functions?

X P(X)

-10123

.1

.2

.4

.2

.11.0

X P(X)

-10123

-.1.3.4.3.1

1.0

X P(X)

-10123

.1

.3

.4

.3

.11.2

PROBABILITYDISTRIBUTION

Do Any of these Distributions meet the Requirements for Discrete Probability Functions?

X P(X)

-10123

.1

.2

.4

.2

.11.0

X P(X)

-10123

-.1.3.4.3.1

1.0

X P(X)

-10123

.1

.3

.4

.3

.11.2

PROBABILITYDISTRIBUTION

:: YESYES NONO NONO

Example: Mean of a Discrete Distribution

E X X P X( )

X-10123

P(X).1.2.4.2.1

-.1.0.4.4.3

1.0

X P X ( )

= 1.0= 1.0

Variance and Standard Deviation

of a Discrete Distribution

2.1)(22 XPX 10.12.12

X-10123

P(X).1.2.4.2.1

-2-1012

X 41014

.4

.2

.0

.2

.41.2

)(2X

2( ) ( )X P X

( ) 1E X X P X

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

1.15

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5

Probability

Number of Crises

Example: Variance and Standard Deviation of Crises Data

41.1)(22 XPX

2141 119. .

X P(X) (X- ) (X-) 2 (X- ) 2 P(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

1.41

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. X142443310

X242343221

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 Fi

P(x1)0 1

0.1251 1

0.1252 1

0.1253 2

0.2504 3

0.3758 1.00

X142443310

X2 42343221

X2 Fi P(x2)

1 10.125

2 30.375

3 20.250

4 20.2508 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 Fi P(x1) X*P(x1) 0 1 0.125 0 -2.625 6.8906 0.861251 1 0.125 0.125 -1.625 2.6406 0.3302 1 0.125 0.250 -0.625 0.3906 0.04883 2 0.250 0.750 0.375 0.1406 0.03514 3 0.375 1.50 1.375 1.8906 0.07089

8 1.00 2.625 11.953 1.9842

22 ( )P XX ( )E X X P X

1X 2

1X 2

( 1) 1P x X

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

Standard deviation = σ = 1.984 = 1.408 injuries

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 Fi P(x2) X*P(x2) (X2-μ) )(X2-μ)2

1 1 0.125 0.125 -1.625 2.6406 0.330

2 3 0.375 0.750 -0.625 0.3906 0.1464

3 2 0.250 0.750 0.375 0.1406 0.0351

4 2 0.250 1.00 1.375 1.8906 0.47268 1.00 2.625 11.953 0.9842

22 ( )P XX ( )E X X P X

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

Standard deviation = σ = 0.9842 = 0.9921 injuries

P(x2)(X2-μ)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.

Probability of 3 or higher injuries = 1 - P(X1<=2) = 1 –0.5 = 0.5

Standard deviation = σ = 0.9921 injuries

Mean = σ = 2.625 injuries

Gym II

Probability of 3 or higher injuries = 1 - P(X1<=2) = 1 –0.5 = 0.625

Standard deviation = σ = 1.408 injuries

Mean = σ = 2.625 injuries

Gym I

Exercise: Deriving Discrete ProbabilityFor 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 Fi P(x1) X*P(x1) 0 1 0.125 0 -2.625 6.8906 0.861251 1 0.125 0.125 -1.625 2.6406 0.3302 1 0.125 0.250 -0.625 0.3906 0.04883 2 0.250 0.750 0.375 0.1406 0.03514 3 0.375 1.50 1.375 1.8906 0.07089

8 1.00 2.625 11.953 1.9842

1X 2

1X 2

( 1) 1P x X

Response: Deriving Discrete ProbabilityFor 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 Fi P(x1) X*P(x1) 0 1 0.125 0 -2.625 6.8906 0.861251 1 0.125 0.125 -1.625 2.6406 0.3302 1 0.125 0.250 -0.625 0.3906 0.04883 2 0.250 0.750 0.375 0.1406 0.03514 3 0.375 1.50 1.375 1.8906 0.07089

8 1.00 2.625 11.953 1.9842

1X 2

1X 2

( 1) 1P x X

P(X1>3) = P(X1=4) = 0.375

P(2=<X1<4) = P(X1=2)+ P(X1=3)=0.125+0.25 = 0.375

P(X1<2) = P(X1=0)+ P(X1=1) = 0.125+0.125 = 0.25

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 problemDaily wins

Daily wins

Day 1 Day 1 300,000

Day 2 Day 2

Day 3 Day 3

Day 4 Day 4 200,000

Day 5 1,000,000

Day 5

Day 6 Day 6

Day 7 Day 7 150,000

Day 8 Day 8 200,000

Day 9 Day 9 150,000

Day 10 1,000,000

Day 10 1,000,000

Casino 1 Casino 2

Illustration of Casino problem

X1 Fi P(x1) X2 Fi P(x1)

0 8 0.8 0 8 0.8

1 2 0.2 1 2 0.2

Sum 10 1 Sum 10 1Casino

1Casino 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 TablesIn 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 distributionsContinuous Uniform distributions normal distributions exponential distributions T distributions chi-square distributions F distributions