Negative Hypergeometric Distribution - Lecture Notes | STAT 341, Study notes of Probability and Statistics

Material Type: Notes; Class: INTR THY PROBAB&S I; Subject: STATISTICS; University: Iowa State University; Term: Unknown 1989;

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Negative Hypergeometric Distribution
A random variable Yhas a negative hypergeometric distribution if
Objects are selected without replacement from a population of size N.
Each member of the population belongs to one of two groups; success or failure. The number
of successes in the population is denoted as rand therefore the number of failures in the
population is Nr.
Objects are selected from the population until the kth success occurs.
The random variable Yis defined as the number of the trial on which the kth success occurs.
The parameters for the hypergeometric random variable Yare value of k, the population size
Nand the number of successes in the population r.
The probability distribution function of Y is
P(Y=y) = p(y) = y1
k1Ny
rk
N
ry=k, k + 1, . . . , N r+k
The theoretical mean of the negative hypergeometric random variable Yis
µ=E(Y) = kN+ 1
r+ 1
The theoretical variance of the negative hypergeometric random variable Yis
σ2=V(Y) = k(N+ 1)(Nr)(r+ 1 k)
(r+ 1)2(r+ 2)
Working with negative hypergeometric random variables in R.
R does not include the negative hypergeometric random variable as a built-in function. In order to
work with this random variable, you will need to copy and paste the following function into R.
dneghyper<- function(y,r,N,k){choose(y-1,k-1)*choose(N-y,r-k)/choose(N,r)}
To find a probability P(Y=y) = p(y) for a single value y, the command in R is
dneghyper(y,r,N,k)
To find the probability P(Yy), use the sum command to add up all p(y) values for ybetween
and including kand y.
sum(dneghyper(k:y,r,N,k))
To find the probability P(y1Yy2), use the sum command to add up all p(y) values for y
between and including y1and y2.
sum(dneghyper(y1:y2,r,N,k))
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Negative Hypergeometric Distribution

A random variable Y has a negative hypergeometric distribution if

  • Objects are selected without replacement from a population of size N.
  • Each member of the population belongs to one of two groups; success or failure. The number of successes in the population is denoted as r and therefore the number of failures in the population is N − r.
  • Objects are selected from the population until the kth success occurs.
  • The random variable Y is defined as the number of the trial on which the kth success occurs.
  • The parameters for the hypergeometric random variable Y are value of k, the population size N and the number of successes in the population r.
  • The probability distribution function of Y is

P (Y = y) = p(y) =

(y− 1 k− 1

)(N −y r−k

) (N r

) y^ =^ k, k^ + 1,... , N^ −^ r^ +^ k

  • The theoretical mean of the negative hypergeometric random variable Y is

μ = E(Y ) = k

( N + 1 r + 1

)

  • The theoretical variance of the negative hypergeometric random variable Y is

σ^2 = V (Y ) = k(N + 1)(N − r)(r + 1 − k) (r + 1)^2 (r + 2)

Working with negative hypergeometric random variables in R.

R does not include the negative hypergeometric random variable as a built-in function. In order to work with this random variable, you will need to copy and paste the following function into R.

dneghyper<- function(y,r,N,k){choose(y-1,k-1)*choose(N-y,r-k)/choose(N,r)}

To find a probability P (Y = y) = p(y) for a single value y, the command in R is

dneghyper(y,r,N,k)

To find the probability P (Y ≤ y), use the sum command to add up all p(y) values for y between and including k and y.

sum(dneghyper(k:y,r,N,k))

To find the probability P (y 1 ≤ Y ≤ y 2 ), use the sum command to add up all p(y) values for y between and including y 1 and y 2.

sum(dneghyper(y1:y2,r,N,k))

To find the probability P (Y ≥ y) = 1 − P (Y < y) = 1 − P (Y ≤ y − 1), use the sum command to find P (Y ≤ y − 1) and subtract this value from 1.

1 - sum(dneghyper(k:y-1,r,N,k))

Problems.

  1. How is the probability distribution function p(y) derived?
  2. A box contains 40 balls: 10 red, 15 yellow, and 15 green. Balls are selected from this box without replacement.

(a) What is the probability that the third yellow ball will be the seventh ball chosen? (b) What is the probability that the third yellow ball will be the 10th ball chosen? (c) Find the expected number of draws needed to obtain the third yellow ball. (d) Find the variance of the number of draws needed to obtain the third yellow ball.

  1. A company receives a shipment of 50 condensors, of which 5 are defective.

(a) When sampling without replacement, what is the probability the second defective con- densor will be 10th condensors chosen? (b) When sampling without replacement, what is the probability the third defective con- densor will be chosen within the first 25 draws. (c) Find the expected number of draws needed to obtain the second defective condensor. (d) Find the variance of the number of draws needed to obtain the second defective conden- sor.