Hypergeometric, Poisson & Joint Distributionscathy/Math3339/Lecture/lecture6_3339_sum19c.pdfOutline 1 Hypergeometric Distribution 2 Poisson Distribution 3 Joint Distribution Cathy

Hypergeometric, Poisson & Joint DistributionsSec 4.7 - 4.9

Cathy Poliak, [email protected] in Fleming 11c

Department of MathematicsUniversity of Houston

Lecture 6 - 3339

Cathy Poliak, Ph.D. [email protected] Office in Fleming 11c (Department of Mathematics University of Houston )Sec 4.7 - 4.9 Lecture 6 - 3339 1 / 30

Outline

1 Hypergeometric Distribution

2 Poisson Distribution

3 Joint Distribution


Beginning Example

Each of 12 refrigerators of a certain type has been returned to adistributor because of an audible, high-pitched, oscillating noise whenthe refrigerator is running. Suppose that 7 of these refrigerators have adefective compressor and the other 5 have less serious problems. Thetechnition looks at 6 refrigerators, what is the probability that exactly 5have a defective compressor?


Conditions for a Hypergeometric Distribution

1. The population or set to be sampled consists of N individuals,objects or elements (a finite population).

2. Each individual can be characterized as a "success" or "failure."There are m successes in the population, and n failures in thepopulation. Notice: m + n = N.

3. A sample size of k individuals is selected without replacement insuch a way that each subset of size k is equally likely to bechosen.

The parameters of a hypergeometric distribution is m,n, k . We writeX ∼ Hyper(m,n, k). The probability mass function for ahypergeometric is:

fX (x) = P(X = x) =

(mx

)( nk−x

)(m+nk

) =C(m, x)C(n, k − x)

C(m + n, k)


Using R

R commands:P(X = x) = dhyper(x,m,n,k) andP(X ≤ x) = phyper(x ,m,n, k)

Example going back to the refrigerator example, m = 7, n = 5,k = 6. P(X = 5)> dhyper(5,7,5,6)[1] 0.1136364

P(X ≤ 4)> phyper(4,7,5,6)[1] 0.8787879


Mean and Variance of a Hypergeometric Distribution

Let Y have a hypergeometric distribution with parameter, m,n, and k .The mean of Y is:

µY = E(Y ) = k(

mm + n

)= kp.

The variance of Y is:

σ2Y = var(Y ) = kp(1− p)

(1− k − 1

m + n − 1

).

1− k−1m+n−1 is called the finite population correction factor. As,

the population increases, this factor will get closer to 1.


Voters

A small voting district has 101 female voters and 95 male voters. Arandom sample of 10 voters is drawn.

1. What is the probability exactly 7 of the voters will be female?

2. What is the probability that at most 7 of the voters will be female?

3. What is is the probability that at least 7 of the voters will befemale?


Example

Suppose that the average number of emails received by aparticular student at UH is five emails per hour.

We want to know what is the probability that a particular studentwill get X number of emails in any given hour.

Suppose we want to know P(X = 3), the probability of gettingexactly 3 emails in one hour.

What about in two hours?


The Poisson Distribution

The Poisson distribution is appropriate under the following conditions.1. Let X be the number of successes occurring per unit of measure.

X = 0,1,2,3, . . ..

2. Let µ be the mean number of successes occurring per unit ofmeasure.

3. The number of successes that occur in two non-overlapping unitsof measure are independent.

4. The probability that success will occur in a unit of measure is thesame for all unites of equal size and is proportional to the size ofthe unit.

5. The probability that more than one event occurs in a unit ofmeasure is negligible for very small-sized units. In other words,the events occur one at a time.


The Probability Function of the Poisson Distribution

A random variable X with non-negative integer values has a Poissondistribution if its frequency function is:

f (x) = P(X = x) = e−µµx

x!

for x = 0,1,2, . . ., where µ > 0 is a constant. If X has a Poissondistribution with parameter µ, we can write X ∼ Pois(µ).


The Mean and Variance of the Poisson Distribution

Let X ∼ Pois(µ)The mean of X is µ per unit of measure. By the conditions of thePoisson distribution.

The variance of X is also µ per unit of measure.

The standard deviation of X is√µ.


Example

The number of people arriving for treatment at an emergency room canbe modeled by a Poisson process with a mean of five people per hour.

1. What is the probability that exactly four arrivals occur at aparticular hour?

2. What is the probability that at least four people arrive during aparticular hour?

3. How many people do you expect to arrive during a 45-min period?


Using R to Compute Probabilities of the PoissonDistribution

R commands:P(X = x) = dpois(x,µ) and P(X ≤ x) = ppois(x , µ)P(X =4)> dpois(4,5)[1] 0.1754674

P(X ≥ 4) = 1− P(X ≤ 3)> 1-ppois(3,5)[1] 0.7349741


Review Questions

For the following information indicate the type of distribution.

a. Binomial b. Poisson c.Hypergoemetric d. Neither

1. A company makes sports bikes. 90% pass final inspection (and 10% failand need to be fixed). We randomly select 10 bikes and want to knowwhat is the probability that 7 will pass.

2. A company makes sports bikes. 90% pass final inspection (and 10% failand need to be fixed). What is the probability that the 10th bike will notpass final inspection?

3. A small company made 100 bikes in a month. 10 of these bikes willneed to be fixed. Suppose we select 4 bikes, what is the probability thatat least one needs to be fixed?

4. A small company makes bicycles. The average amount that thiscompany makes in a month is 100 bikes. What is the probability that thecompany will make less than 30 bikes in a week?


Beginning Example

Suppose two random variables X and Y have a joint functionf (x , y) = 2x−y

5 for X = 0.75 and 1 and Y = 0 and 1. What are thepossible values of f (x , y)?


Joint Probability Distribution For Discrete RandomVariables

The function f (x , y) is a joint probability distribution of the discreterandom variables X and Y if

1. f (x , y) ≥ 0 for all (x , y).

2.∑

x∑

y f (x , y) = 1.

3. P(X = x ,Y = y) = f (x , y).For any region in the xy plane, P[(X ,Y ) ∈ A] =

∑A∑

f (x , y).


Joint Probability Distribution

Let X and Y be discrete random variables that have the jointprobability distribution f (x , y). Then

1. fY (y) =∑

x f (x , y) for all y is the marginal probability massfunction of Y .

2. fX (x) =∑

y f (x , y) for all x is the marginal probability massfunction of X .

3. fY |X (y |x) =f (x ,y)f (x) if fX (x) > 0. This is the conditional frequency

function.

4. X and Y are independent if and only if f (x , y) = fX (x)fY (y) for allx , y .

5.∑

X∑

Y f (x , y) = 1.


Example 2

Suppose that the following table represents the joint pmf of X and Y.

Y/X 1 21 2/k 3/k2 3/k 4/k3 4/k 5/k

1. What should be the value of k?

2. Determine P(X = 2,Y ≥ 2)

3. Determine P(X = 2|Y = 2)


Rule for Means

If X and Y are two different random variables, then the mean ofthe sums of the pairs of the random variable is the same as thesum of their means:

E [X + Y ] = E [X ] + E [Y ].

This is called the addition rule for means.The mean of the difference of the pairs of the random variable isthe same as the difference of their means:

E [X − Y ] = E [X ]− E [Y ].


Rule for Variances

If X and Y are independent random variables

σ2X+Y = Var [X + Y ] = Var [X ] + Var [Y ].

andσ2

X−Y = Var [X − Y ] = Var [X ] + Var [Y ].


Example of Rules

Tamara and Derek are sales associates in a large electronics andappliance store. The following table shows their mean and standarddeviation of daily sales. Assume that daily sales among the salesassociates are independent.

Sales associate Mean Standard deviationTamara X E [X ] = $1100 σX = $100Derek Y E [Y ] = $1000 σY = $80

1. Determine the mean of the total of Tamara’s and Derek’s dailysales, E [X + Y ].

2. Determine the variance of the total of Tamara’s and Derek’s dailysales, σ2

(X+Y ).

3. Determine the standard deviation of the total of Tamara’s andDerek’s daily sales, σ(X+Y ).


Expected Values

Let X and Y be two random variables, then:E(X + Y ) = E(X ) + E(Y )

E(XY ) = x1y1p11 + x1y1p12 + . . .+ xnynpnn

For more than two random variables X1,X2,X3, · · · ,Xn,E(X1 + X2 + · · ·+ Xn) = E(X1) + E(X2) + · · ·+ E(Xn)


Covariance of X and Y

Let X and Y be jointly distributed random variables with respectivemeans µx and µy and standard deviation σx , σy . The covariance of Xand Y is

cov(X ,Y ) = E((x − µx)(Y − µy )).

or an easier calculation:

cov(X ,Y ) = E(XY )− E(X )E(Y ).

The correlation of X and Y is;

cor(X ,Y ) =cov(X ,Y )

σxσy.


Properties of Covariance

1. cov(X ,Y ) = cov(Y ,X )

2. cov(X ,X ) = var(X )

3. If X , Y , and Z are jointly distributed and a and b are constants

cov(X ,aY + bZ ) = a[cov(X ,Y )] + b[cov(X ,Z )].

4. If X and Y are jointly distributed,

var(X + Y ) = var(X ) + var(Y ) + 2cor(X ,Y )sd(X )sd(Y )

var(X − Y ) = var(X ) + var(Y )− 2cor(X ,Y )sd(X )sd(Y )

5. If X and Y are independent, cov(X ,Y ) = 0.

6. If jointly distributed random variables X1,X2, · · · ,Xn are pairwiseuncorrelated, then

var(X1 + X2 + · · ·+ Xn) = var(X1) + var(X2) + · · ·+ var(Xn)


Example of a Joint Probability Distribution

The following table is a joint probability table of X = number of sportsinvolved and Y = age of high school students.

X = Number of sports involvedY = Age 0 1 2 3

14 0.03 0.09 0.06 0.0215 0.015 0.045 0.03 0.0116 0.045 0.135 0.09 0.0317 0.03 0.09 0.06 0.0218 0.03 0.09 0.06 0.02

1. Determine E(X ).2. Determine E(Y ).3. Determine E(XY ).4. Determine cov(X ,Y ).


R code

I put the values in an Excel .csv file named "jointprob."

> library(readr)> jointprob <- read_csv("F:/Lectures uh/MATH 3339/jointprob.csv")> View(jointprob)> sum(jointprob$x*jointprob$y*jointprob$pxy)[1] 21.735> 1.35*16.1[1] 21.735> sum(jointprob$x*jointprob$pxy)[1] 1.35> sum(jointprob$y*jointprob$pxy)[1] 16.1> sum(jointprob$x*jointprob$y*jointprob$pxy)[1] 21.735


You Try Question

Suppose we have two random variables, X and Y with frequencyfunctions, fX (x), fy (y) and joint frequency function fX ,Y (x , y). If X andY are independent, which statement is false.a) fX ,Y (x , y) = 0b) fX ,Y (x , y) = fX (x)fY (y)c) fX |Y (x |y) = fX (x)d) cov(X ,Y ) = 0e) E(XY ) = E(X )E(Y )


Example of Joint Probability Distribution

Suppose that a fair, 6 sided die is rolled. Let X indicate the event thatan even number is rolled (in other words, X = 1 if an even number isrolled and X = 0 otherwise). Let Y indicate the event that 3, 4, or 5 isrolled (in other words, Y = 1 if 3, 4, or 5 is rolled and Y = 0 otherwise).Find P (X = 0, Y = 1).


Example

A class has 10 mathematics majors, 6 computer science majors and 4statistics majors. A committee of two is selected at random to work ana problem. Let X be the number of mathematics majors and let Y bethe number of computer science majors chosen.

1. Determine all possible (X ,Y ) pairs for randomly selecting twostudents.

2. Determine fX ,Y (0,0), that is the probability that the two pick areneither a mathematics major nor a computer science major.

3. Determine fX (0), that is the probability that we do not pick anymathematics majors.


Joint Probability Table

Y/X 0 1 2 Total0

1

2

Total

1. Determine E(X ).2. Determine E(Y ).3. Determine E(XY ).4. Determine cov(X ,Y ).5. Determine cor(X ,Y ).


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Hypergeometric, Poisson & Joint Distributionscathy/Math3339/Lecture/lecture6_3339_sum19c.pdfOutline 1 Hypergeometric Distribution 2 Poisson Distribution 3 Joint Distribution Cathy