The Hypergeometric Distribution
The Poisson Distribution
Lecture 7: Special Probability Distributions - 2
Assist. Prof. Dr. Emel YAVUZ DUMAN
Introduction to Probability and Statistics
İstanbul Kültür University
The Hypergeometric Distribution
Outline
1
The Hypergeometric Distribution
2
The Poisson Distribution
The Poisson Distribution
The Hypergeometric Distribution
Outline
1
The Hypergeometric Distribution
2
The Poisson Distribution
The Poisson Distribution
The Hypergeometric Distribution
The Poisson Distribution
Many times we used sampling with and without replacement to
illustrate the multiplication rules for independent and dependent
events.
The Hypergeometric Distribution
The Poisson Distribution
Many times we used sampling with and without replacement to
illustrate the multiplication rules for independent and dependent
events. To obtain a formula analogous to the binomial distribution
that applies to sampling without replacement, in which case the
trials are not independent, let us consider a set of N elements for
which M are looked upon as successes and the other N − M as
failures.
The Hypergeometric Distribution
The Poisson Distribution
Many times we used sampling with and without replacement to
illustrate the multiplication rules for independent and dependent
events. To obtain a formula analogous to the binomial distribution
that applies to sampling without replacement, in which case the
trials are not independent, let us consider a set of N elements for
which M are looked upon as successes and the other N − M as
failures. In connection with the binomial distribution, we are
interested in the probability of getting x successes in n trial, but
now we are choosing, without replacement, n of the N elements
contained in the set.
The Hypergeometric Distribution
There are
The Poisson Distribution
The Hypergeometric Distribution
There are
M x way of choosing x of the M successes, and
The Poisson Distribution
The Hypergeometric Distribution
The Poisson Distribution
There are
M x way of choosing x of the M successes, and
N−M ways of choosing n − x of the N − M failure, and
n−x
The Hypergeometric Distribution
The Poisson Distribution
There are
M x way of choosing x of the M successes, and
N−M ways of choosing n − x of the N − M failure, and
n−x
M N−M hence x n−x ways of choosing x successes and n − x
failures.
The Hypergeometric Distribution
The Poisson Distribution
There are
M x way of choosing x of the M successes, and
N−M ways of choosing n − x of the N − M failure, and
n−x
M N−M hence x n−x ways of choosing x successes and n − x
failures.
Since there are Nn ways of choosing n of the N elements in
the set,
The Hypergeometric Distribution
The Poisson Distribution
There are
M x way of choosing x of the M successes, and
N−M ways of choosing n − x of the N − M failure, and
n−x
M N−M hence x n−x ways of choosing x successes and n − x
failures.
Since there are Nn ways of choosing n of the N elements in
the set, and we shall assume that they are all equally likely
(which is what we mean when we say that the selection is
random),
The Hypergeometric Distribution
The Poisson Distribution
There are
M x way of choosing x of the M successes, and
N−M ways of choosing n − x of the N − M failure, and
n−x
M N−M hence x n−x ways of choosing x successes and n − x
failures.
Since there are Nn ways of choosing n of the N elements in
the set, and we shall assume that they are all equally likely
(which is what we mean when we say that the selection is
then that the probability of x successes in n trials is
random),
M N−M N x
n−x / n .
The Hypergeometric Distribution
The Poisson Distribution
Definition 1
A random variable X has a hypergeometric distribution and it is
referred to as a hypergeometric random variable if and only if its
probability distribution is given by
M N−M h(x; n, N, M) =
x
Nn−x
n
for x = 0, 1, 2, · · · , n, x ≤ M and n − x ≤ N − M.
The Hypergeometric Distribution
The Poisson Distribution
Definition 1
A random variable X has a hypergeometric distribution and it is
referred to as a hypergeometric random variable if and only if its
probability distribution is given by
M N−M h(x; n, N, M) =
x
Nn−x
n
for x = 0, 1, 2, · · · , n, x ≤ M and n − x ≤ N − M.
Thus, for sampling without replacement, the number of successes
in n trials is a random variable having a hypergeometric
distribution with parameters n, N, and M.
The Hypergeometric Distribution
The Poisson Distribution
Example 2
As part of an air-pollution survey, an inspector decides to examine
the exhaust of six of a company’s 24 trucks. If four of the
company’s trucks emit excessive amounts of pollutants, what is the
probability that none of them will be included in the inspector’s
sample?
The Hypergeometric Distribution
The Poisson Distribution
Example 2
As part of an air-pollution survey, an inspector decides to examine
the exhaust of six of a company’s 24 trucks. If four of the
company’s trucks emit excessive amounts of pollutants, what is the
probability that none of them will be included in the inspector’s
sample?
Solution. Substituting x = 0, n = 6, N = 24, and M = 4 into the
formula for the hypergeometric distribution,
The Hypergeometric Distribution
The Poisson Distribution
Example 2
As part of an air-pollution survey, an inspector decides to examine
the exhaust of six of a company’s 24 trucks. If four of the
company’s trucks emit excessive amounts of pollutants, what is the
probability that none of them will be included in the inspector’s
sample?
Solution. Substituting x = 0, n = 6, N = 24, and M = 4 into the
formula for the hypergeometric distribution, we get
424−4
M N−M h(x; n, N, M) = h(0; 6, 24, 4) =
x
Nn−x
n
=
0
246−0
6
= 0.2880.
The Hypergeometric Distribution
The Poisson Distribution
Example 3
Draw 6 cards from a deck without replacement. What is the
probability of getting two hearts?
The Hypergeometric Distribution
The Poisson Distribution
Example 3
Draw 6 cards from a deck without replacement. What is the
probability of getting two hearts?
Solution. Substituting x = 2, n = 6, N = 52, and M = 13 into
the formula for the hypergeometric distribution,
The Hypergeometric Distribution
The Poisson Distribution
Example 3
Draw 6 cards from a deck without replacement. What is the
probability of getting two hearts?
Solution. Substituting x = 2, n = 6, N = 52, and M = 13 into
the formula for the hypergeometric distribution, we get
1352−13
M N−M h(x; n, N, M) = h(2; 6, 52, 13) =
x
Nn−x
n
=
2
526−2
6
= 0.31513.
The Hypergeometric Distribution
The Poisson Distribution
Example 4
49 balls are numbered 1 - 49. You select six numbers between 1
and 49. The ones you write on your lotto card. What is the
probability that they contain
The Hypergeometric Distribution
The Poisson Distribution
Example 4
49 balls are numbered 1 - 49. You select six numbers between 1
and 49. The ones you write on your lotto card. What is the
probability that they contain (a) match 4?
The Hypergeometric Distribution
The Poisson Distribution
Example 4
49 balls are numbered 1 - 49. You select six numbers between 1
and 49. The ones you write on your lotto card. What is the
probability that they contain (a) match 4? (b) match 6?
The Hypergeometric Distribution
The Poisson Distribution
Example 4
49 balls are numbered 1 - 49. You select six numbers between 1
and 49. The ones you write on your lotto card. What is the
probability that they contain (a) match 4? (b) match 6?
Solution. (a) Substituting x = 4, n = 6, N = 49, and M = 6 into
the formula for the hypergeometric distribution,
The Hypergeometric Distribution
The Poisson Distribution
Example 4
49 balls are numbered 1 - 49. You select six numbers between 1
and 49. The ones you write on your lotto card. What is the
probability that they contain (a) match 4? (b) match 6?
Solution. (a) Substituting x = 4, n = 6, N = 49, and M = 6 into
the formula for the hypergeometric distribution, we get
649−6
M N−M h(x; n, N, M) = h(4; 6, 49, 6) =
x
n−x
N n
=
4
496−4
6
= 2.3062×10−5 .
The Hypergeometric Distribution
The Poisson Distribution
Example 4
49 balls are numbered 1 - 49. You select six numbers between 1
and 49. The ones you write on your lotto card. What is the
probability that they contain (a) match 4? (b) match 6?
Solution. (a) Substituting x = 4, n = 6, N = 49, and M = 6 into
the formula for the hypergeometric distribution, we get
649−6
M N−M h(x; n, N, M) = h(4; 6, 49, 6) =
x
n−x
N n
=
4
496−4
= 2.3062×10−5 .
6
(b) Substituting x = 6, n = 6, N = 49, and M = 6 into the
formula for the hypergeometric distribution,
The Hypergeometric Distribution
The Poisson Distribution
Example 4
49 balls are numbered 1 - 49. You select six numbers between 1
and 49. The ones you write on your lotto card. What is the
probability that they contain (a) match 4? (b) match 6?
Solution. (a) Substituting x = 4, n = 6, N = 49, and M = 6 into
the formula for the hypergeometric distribution, we get
649−6
M N−M h(x; n, N, M) = h(4; 6, 49, 6) =
x
n−x
N =
4
n
496−4
= 2.3062×10−5 .
6
(b) Substituting x = 6, n = 6, N = 49, and M = 6 into the
formula for the hypergeometric distribution, we get
649−6
M N−M h(x; n, N, M) = h(6; 6, 49, 6) =
x
Nn−x
n
=
6
496−6
6
= 7.1511×10−8 .
The Hypergeometric Distribution
The Poisson Distribution
Theorem 5
The mean and the variance of the hypergeometric distribution are
μ=
nM(N − M)(N − n)
nM
and σ 2 =
.
N
N 2 (N − 1)
The Hypergeometric Distribution
The Poisson Distribution
Example 6
Suppose that a researcher goes to a small college of 200 faculty, 12
of which have blood type O-negative. She obtains a simple random
sample of 20 of the faculty. Determine the mean and standard
deviation of the number of randomly selected faculty that will have
blood type O-negative.
The Hypergeometric Distribution
The Poisson Distribution
Example 6
Suppose that a researcher goes to a small college of 200 faculty, 12
of which have blood type O-negative. She obtains a simple random
sample of 20 of the faculty. Determine the mean and standard
deviation of the number of randomly selected faculty that will have
blood type O-negative.
Solution. Substituting n = 20, N = 200, and M = 12 into the
formula for the hypergeometric distribution’s mean and variance
we obtain
The Hypergeometric Distribution
The Poisson Distribution
Example 6
Suppose that a researcher goes to a small college of 200 faculty, 12
of which have blood type O-negative. She obtains a simple random
sample of 20 of the faculty. Determine the mean and standard
deviation of the number of randomly selected faculty that will have
blood type O-negative.
Solution. Substituting n = 20, N = 200, and M = 12 into the
formula for the hypergeometric distribution’s mean and variance
we obtain
20 · 12
nM
=
= 1.2
μ=
N
200
The Hypergeometric Distribution
The Poisson Distribution
Example 6
Suppose that a researcher goes to a small college of 200 faculty, 12
of which have blood type O-negative. She obtains a simple random
sample of 20 of the faculty. Determine the mean and standard
deviation of the number of randomly selected faculty that will have
blood type O-negative.
Solution. Substituting n = 20, N = 200, and M = 12 into the
formula for the hypergeometric distribution’s mean and variance
we obtain
20 · 12
nM
=
= 1.2
μ=
N
200
and
nM(N − M)(N − n)
20 · 12(200 − 12)(200 − 20)
=
= 1.0101
σ=
2
N (N − 1)
2002 · (200 − 1)
The Hypergeometric Distribution
The Poisson Distribution
Example 6
Suppose that a researcher goes to a small college of 200 faculty, 12
of which have blood type O-negative. She obtains a simple random
sample of 20 of the faculty. Determine the mean and standard
deviation of the number of randomly selected faculty that will have
blood type O-negative.
Solution. Substituting n = 20, N = 200, and M = 12 into the
formula for the hypergeometric distribution’s mean and variance
we obtain
20 · 12
nM
=
= 1.2
μ=
N
200
and
nM(N − M)(N − n)
20 · 12(200 − 12)(200 − 20)
=
= 1.0101
σ=
2
N (N − 1)
2002 · (200 − 1)
We expect that, in a random sample of 20 faculty members, 1.2
will have blood type O-negative.
The Hypergeometric Distribution
The Poisson Distribution
Example 7
A case of wine has 12 bottles, 3 of which contains spoiled wine. A
sample of 4 bottles is randomly selected from the case.
The Hypergeometric Distribution
The Poisson Distribution
Example 7
A case of wine has 12 bottles, 3 of which contains spoiled wine. A
sample of 4 bottles is randomly selected from the case.
(a) Find the probability distribution for X , the number of spoiled
wine in the sample
The Hypergeometric Distribution
The Poisson Distribution
Example 7
A case of wine has 12 bottles, 3 of which contains spoiled wine. A
sample of 4 bottles is randomly selected from the case.
(a) Find the probability distribution for X , the number of spoiled
wine in the sample
(b) What are the mean and variance of X ?
The Hypergeometric Distribution
The Poisson Distribution
Example 7
A case of wine has 12 bottles, 3 of which contains spoiled wine. A
sample of 4 bottles is randomly selected from the case.
(a) Find the probability distribution for X , the number of spoiled
wine in the sample
(b) What are the mean and variance of X ?
Solution. For this example n = 4, N = 12, and M = 3. Then
3 9 h(x; 4, 12, 3) =
x
124−x
.
4
The Hypergeometric Distribution
h(x; 4, 12, 3) =
The Poisson Distribution
9
(x3)(4−x
)
.
12
(4)
The Hypergeometric Distribution
h(x; 4, 12, 3) =
The Poisson Distribution
9
(x3)(4−x
)
.
12
(4)
(a) The possible values for X are 0, 1, 2 and 3,
The Hypergeometric Distribution
h(x; 4, 12, 3) =
The Poisson Distribution
9
(x3)(4−x
)
.
12
(4)
(a) The possible values for X are 0, 1, 2 and 3, with probabilities
39
39
h(0; 4, 12, 3) = 0124 = 0.25, h(1; 4, 12, 3) = 1123 = 0.51,
4
4
The Hypergeometric Distribution
h(x; 4, 12, 3) =
The Poisson Distribution
9
(x3)(4−x
)
.
12
(4)
(a) The possible values for X are 0, 1, 2 and 3, with probabilities
39
39
h(0; 4, 12, 3) = 0124 = 0.25, h(1; 4, 12, 3) = 1123 = 0.51,
4
4
39
39
h(2; 4, 12, 3) = 2122 = 0.22, h(3; 4, 12, 3) = 3121 = 0.02.
4
4
The Hypergeometric Distribution
h(x; 4, 12, 3) =
The Poisson Distribution
9
(x3)(4−x
)
.
12
(4)
(a) The possible values for X are 0, 1, 2 and 3, with probabilities
39
39
h(0; 4, 12, 3) = 0124 = 0.25, h(1; 4, 12, 3) = 1123 = 0.51,
4
4
39
39
h(2; 4, 12, 3) = 2122 = 0.22, h(3; 4, 12, 3) = 3121 = 0.02.
4
4
(b) The mean is given by
μ=
4·3
nM
=
=1
N
12
The Hypergeometric Distribution
h(x; 4, 12, 3) =
The Poisson Distribution
9
(x3)(4−x
)
.
12
(4)
(a) The possible values for X are 0, 1, 2 and 3, with probabilities
39
39
h(0; 4, 12, 3) = 0124 = 0.25, h(1; 4, 12, 3) = 1123 = 0.51,
4
4
39
39
h(2; 4, 12, 3) = 2122 = 0.22, h(3; 4, 12, 3) = 3121 = 0.02.
4
4
(b) The mean is given by
μ=
4·3
nM
=
=1
N
12
and the variance is
σ2 =
4 · 3(12 − 3)(12 − 4)
nM(N − M)(N − n)
=
= 0.5455.
2
N (N − 1)
122 (12 − 1)
The Hypergeometric Distribution
The Poisson Distribution
Binomial Approximation to Hypergeometric Distribution
When N is large and n is relatively small compared to N
The Hypergeometric Distribution
The Poisson Distribution
Binomial Approximation to Hypergeometric Distribution
When N is large and n is relatively small compared to N (the usual
rule of thumb is that n should not exceed 5 percent of N),
The Hypergeometric Distribution
The Poisson Distribution
Binomial Approximation to Hypergeometric Distribution
When N is large and n is relatively small compared to N (the usual
rule of thumb is that n should not exceed 5 percent of N), there is
not much difference between sampling with replacement and
sampling without replacement,
The Hypergeometric Distribution
The Poisson Distribution
Binomial Approximation to Hypergeometric Distribution
When N is large and n is relatively small compared to N (the usual
rule of thumb is that n should not exceed 5 percent of N), there is
not much difference between sampling with replacement and
sampling without replacement, and the formula for the binomial
distribution with the parameters n and θ = M
N may be used to
approximate hypergeometric probabilities.
The Hypergeometric Distribution
The Poisson Distribution
Example 8
Among the 120 applicants for a job, only 80 are actually qualified.
If five of the applicants are randomly selected for an in-depth
interview, find the probability that only two of the five will be
qualified for the job
The Hypergeometric Distribution
The Poisson Distribution
Example 8
Among the 120 applicants for a job, only 80 are actually qualified.
If five of the applicants are randomly selected for an in-depth
interview, find the probability that only two of the five will be
qualified for the job by using
(a) the formula for the hypergeometric distribution;
The Hypergeometric Distribution
The Poisson Distribution
Example 8
Among the 120 applicants for a job, only 80 are actually qualified.
If five of the applicants are randomly selected for an in-depth
interview, find the probability that only two of the five will be
qualified for the job by using
(a) the formula for the hypergeometric distribution;
(b) the formula for the binomial distribution with θ = 80/120 as
an approximation.
The Hypergeometric Distribution
The Poisson Distribution
Solution. (a) Substituting x = 2, n = 5, N = 120, and M = 80
into the formula for the hypergeometric distribution,
The Hypergeometric Distribution
The Poisson Distribution
Solution. (a) Substituting x = 2, n = 5, N = 120, and M = 80
into the formula for the hypergeometric distribution, we get
8040
h(x; n, N, M) = h(2; 5, 120, 80) =
1203
2
5
rounded to three decimals;
= 0.164.
The Hypergeometric Distribution
The Poisson Distribution
Solution. (a) Substituting x = 2, n = 5, N = 120, and M = 80
into the formula for the hypergeometric distribution, we get
8040
h(x; n, N, M) = h(2; 5, 120, 80) =
1203
2
= 0.164.
5
rounded to three decimals;
(b) substituting x = 2, n = 5, N = 120, and θ =
formula for the binomial distribution,
80
120
=
2
3
into the
The Hypergeometric Distribution
The Poisson Distribution
Solution. (a) Substituting x = 2, n = 5, N = 120, and M = 80
into the formula for the hypergeometric distribution, we get
8040
h(x; n, N, M) = h(2; 5, 120, 80) =
1203
2
= 0.164.
5
rounded to three decimals;
(b) substituting x = 2, n = 5, N = 120, and θ =
formula for the binomial distribution, we get
2
b 2; 5,
3
80
120
=
2 5
2
2 3
=
= 0.165
1−
2
3
3
rounded to three decimals.
2
3
into the
The Hypergeometric Distribution
The Poisson Distribution
Solution. (a) Substituting x = 2, n = 5, N = 120, and M = 80
into the formula for the hypergeometric distribution, we get
8040
h(x; n, N, M) = h(2; 5, 120, 80) =
1203
2
= 0.164.
5
rounded to three decimals;
(b) substituting x = 2, n = 5, N = 120, and θ =
formula for the binomial distribution, we get
2
b 2; 5,
3
80
120
=
2
3
into the
2 5
2
2 3
=
= 0.165
1−
2
3
3
rounded to three decimals. As can be seen from these results, the
approximation is very close.
The Hypergeometric Distribution
The Poisson Distribution
Example 9
Boxes contain 2000 items of which 10% are defective. Find the
probability that no more than 2 defectives will be obtained in a
sample of size 10.
The Hypergeometric Distribution
The Poisson Distribution
Example 9
Boxes contain 2000 items of which 10% are defective. Find the
probability that no more than 2 defectives will be obtained in a
sample of size 10.
Solution. For this question x is equal to 0, 1 or 2, n = 10,
N = 2000 and M = 2000 · 0.10 = 200. Since
n = 10 ≤ 100 = 2000 · 0.05 = N · 0.05
this means n is not exceed 5 percent of N we may use the method
of binomial approximation to the hypergeometric distribution also.
The Hypergeometric Distribution
The Poisson Distribution
(a) The hypergeometric distribution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
2001800 2001800 2001800
=
200010
0
10
+
20009
1
10
+
20008
2
10
= 0.3476 + 0.3881 + 0.1939 = 0.9296.
The Hypergeometric Distribution
The Poisson Distribution
(a) The hypergeometric distribution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
2001800 2001800 2001800
=
200010
0
10
+
20009
1
10
+
20008
2
10
= 0.3476 + 0.3881 + 0.1939 = 0.9296.
(b) Binomial approximation to the hypergeometric distribution
with θ = M/N = 200/2000 = 0.1:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
10
10
10
0
1
1
9
0.1 0.9 +
0.12 0.98
=
0.1 0.9 0 +
1
2
0
= 0.3487 + 0.3874 + 0.1937 = 0.9298.
The Hypergeometric Distribution
Outline
1
The Hypergeometric Distribution
2
The Poisson Distribution
The Poisson Distribution
The Hypergeometric Distribution
The Poisson Distribution
When n, the number of trial, is large the calculation of binomial
probabilities with the formula of binomial distribution will usually
involve a prohibitive amount of work.
The Hypergeometric Distribution
The Poisson Distribution
When n, the number of trial, is large the calculation of binomial
probabilities with the formula of binomial distribution will usually
involve a prohibitive amount of work. In this section we shall
present a probability distribution that can be used to approximate
binomial probabilities of this kind.
The Hypergeometric Distribution
The Poisson Distribution
When n, the number of trial, is large the calculation of binomial
probabilities with the formula of binomial distribution will usually
involve a prohibitive amount of work. In this section we shall
present a probability distribution that can be used to approximate
binomial probabilities of this kind. Specifically, we shall investigate
the limiting form of the binomial distribution when n → ∞, θ → 0,
while nθ remains constant.
The Hypergeometric Distribution
The Poisson Distribution
Definition 10
A random variable X has a Poisson distribution and it is referred
to as Poisson random variable if and only if its probability
distribution is given by
p(x; λ) =
λx e −λ
for x = 0, 1, 2 · · ·
x!
where λ, the mean number of successes.
The Hypergeometric Distribution
The Poisson Distribution
Definition 10
A random variable X has a Poisson distribution and it is referred
to as Poisson random variable if and only if its probability
distribution is given by
p(x; λ) =
λx e −λ
for x = 0, 1, 2 · · ·
x!
where λ, the mean number of successes.
In general, Poisson distribution will provide a good approximation
to binomial probabilities when n ≥ 20 and θ ≤ 0.05. When
n ≥ 100 and nθ < 10, the approximation will generally be excellent.
The Hypergeometric Distribution
The Poisson Distribution
Example 11
If 2 percent of books bound at a certain bindery have defective
bindings, use the Poisson approximation to the binomial
distribution to determine the probability that five of 400 books
bound by this bindery will have defective bindings.
The Hypergeometric Distribution
The Poisson Distribution
Example 11
If 2 percent of books bound at a certain bindery have defective
bindings, use the Poisson approximation to the binomial
distribution to determine the probability that five of 400 books
bound by this bindery will have defective bindings.
Solution. Substituting x = 5, λ = nθ = 400 · 0.02 = 8 into the
formula for Poisson distribution, we get
p(5; 8) =
85 e −8
= 0.09160.
5!
The Hypergeometric Distribution
The Poisson Distribution
Example 12
Records show that the probability is 0.00005 that a car will have a
flat tire while crossing a certain bridge. Use the Poisson
distribution to approximate the binomial probabilities that, among
10,000 cars crossing the bridge
(a) exactly two will have a flat tire;
(b) at most two will have a flat tire.
The Hypergeometric Distribution
The Poisson Distribution
Example 12
Records show that the probability is 0.00005 that a car will have a
flat tire while crossing a certain bridge. Use the Poisson
distribution to approximate the binomial probabilities that, among
10,000 cars crossing the bridge
(a) exactly two will have a flat tire;
(b) at most two will have a flat tire.
Solution. (a) Substituting x = 2,
λ = nθ = 10, 000 · 0.00005 = 0.5 into the formula for Poisson
distribution, we get
p(2; 0.5) =
0.52 e −0.5
= 0.07582.
2!
The Hypergeometric Distribution
The Poisson Distribution
Example 12
Records show that the probability is 0.00005 that a car will have a
flat tire while crossing a certain bridge. Use the Poisson
distribution to approximate the binomial probabilities that, among
10,000 cars crossing the bridge
(a) exactly two will have a flat tire;
(b) at most two will have a flat tire.
Solution. (a) Substituting x = 2,
λ = nθ = 10, 000 · 0.00005 = 0.5 into the formula for Poisson
distribution, we get
p(2; 0.5) =
0.52 e −0.5
= 0.07582.
2!
(b)
0.52 e −0.5 0.51 e −0.5 0.50 e −0.5
+
+
2!
1!
0!
= 0.07582 + 0.30327 + 0.60653 = 0.98562.
p(2; 0.5) + p(1; 0.5)+p(0; 0.5) =
The Hypergeometric Distribution
The Poisson Distribution
Having derived the Poisson distribution as a limiting form of the
binomial distribution, we can obtain formulas for its mean and its
variance by applying the same limiting conditions (n → ∞, θ → 0
and nθ = λ remains constant) to mean and the variance of the
binomial distribution.
The Hypergeometric Distribution
The Poisson Distribution
Having derived the Poisson distribution as a limiting form of the
binomial distribution, we can obtain formulas for its mean and its
variance by applying the same limiting conditions (n → ∞, θ → 0
and nθ = λ remains constant) to mean and the variance of the
binomial distribution. For the mean we get μ = nθ = λ
The Hypergeometric Distribution
The Poisson Distribution
Having derived the Poisson distribution as a limiting form of the
binomial distribution, we can obtain formulas for its mean and its
variance by applying the same limiting conditions (n → ∞, θ → 0
and nθ = λ remains constant) to mean and the variance of the
binomial distribution. For the mean we get μ = nθ = λ and for the
variance we get σ 2 = nθ(1 − θ) = λ(1 − θ) which approaches λ
when θ → 0.
The Hypergeometric Distribution
The Poisson Distribution
Having derived the Poisson distribution as a limiting form of the
binomial distribution, we can obtain formulas for its mean and its
variance by applying the same limiting conditions (n → ∞, θ → 0
and nθ = λ remains constant) to mean and the variance of the
binomial distribution. For the mean we get μ = nθ = λ and for the
variance we get σ 2 = nθ(1 − θ) = λ(1 − θ) which approaches λ
when θ → 0.
Theorem 13
The mean and the variance of the Poisson distribution are given by
μ = λ and σ 2 = λ.
The Hypergeometric Distribution
The Poisson Distribution
Having derived the Poisson distribution as a limiting form of the
binomial distribution, we can obtain formulas for its mean and its
variance by applying the same limiting conditions (n → ∞, θ → 0
and nθ = λ remains constant) to mean and the variance of the
binomial distribution. For the mean we get μ = nθ = λ and for the
variance we get σ 2 = nθ(1 − θ) = λ(1 − θ) which approaches λ
when θ → 0.
Theorem 13
The mean and the variance of the Poisson distribution are given by
μ = λ and σ 2 = λ.
Theorem 14
The moment generating function of the Poisson distribution is
given by
t
MX (t) = e λ(e −1) .
The Hypergeometric Distribution
The Poisson Distribution
Although the Poisson distribution has been derived as a limiting
form of the binomial distribution, it has many applications that
have no direct connection with binomial distribution.
The Hypergeometric Distribution
The Poisson Distribution
Although the Poisson distribution has been derived as a limiting
form of the binomial distribution, it has many applications that
have no direct connection with binomial distribution.
In many practical situations we are interested in measuring how
many times a certain event occurs in a specific time interval or in a
specific length or area.
The Hypergeometric Distribution
The Poisson Distribution
Although the Poisson distribution has been derived as a limiting
form of the binomial distribution, it has many applications that
have no direct connection with binomial distribution.
In many practical situations we are interested in measuring how
many times a certain event occurs in a specific time interval or in a
specific length or area. For instance:
The Hypergeometric Distribution
The Poisson Distribution
Although the Poisson distribution has been derived as a limiting
form of the binomial distribution, it has many applications that
have no direct connection with binomial distribution.
In many practical situations we are interested in measuring how
many times a certain event occurs in a specific time interval or in a
specific length or area. For instance:
1
the number of phone calls received at an exchange or call
center in an hour;
The Hypergeometric Distribution
The Poisson Distribution
Although the Poisson distribution has been derived as a limiting
form of the binomial distribution, it has many applications that
have no direct connection with binomial distribution.
In many practical situations we are interested in measuring how
many times a certain event occurs in a specific time interval or in a
specific length or area. For instance:
1
the number of phone calls received at an exchange or call
center in an hour;
2
the number of customers arriving at a toll booth per day;
The Hypergeometric Distribution
The Poisson Distribution
Although the Poisson distribution has been derived as a limiting
form of the binomial distribution, it has many applications that
have no direct connection with binomial distribution.
In many practical situations we are interested in measuring how
many times a certain event occurs in a specific time interval or in a
specific length or area. For instance:
1
the number of phone calls received at an exchange or call
center in an hour;
2
the number of customers arriving at a toll booth per day;
3
the number of flaws on a length of cable;
The Hypergeometric Distribution
The Poisson Distribution
Although the Poisson distribution has been derived as a limiting
form of the binomial distribution, it has many applications that
have no direct connection with binomial distribution.
In many practical situations we are interested in measuring how
many times a certain event occurs in a specific time interval or in a
specific length or area. For instance:
1
the number of phone calls received at an exchange or call
center in an hour;
2
the number of customers arriving at a toll booth per day;
3
the number of flaws on a length of cable;
4
the number of cars passing using a stretch of road during a
day.
The Hypergeometric Distribution
The Poisson Distribution
Although the Poisson distribution has been derived as a limiting
form of the binomial distribution, it has many applications that
have no direct connection with binomial distribution.
In many practical situations we are interested in measuring how
many times a certain event occurs in a specific time interval or in a
specific length or area. For instance:
1
the number of phone calls received at an exchange or call
center in an hour;
2
the number of customers arriving at a toll booth per day;
3
the number of flaws on a length of cable;
4
the number of cars passing using a stretch of road during a
day.
The Poisson distribution plays a key role in modeling such
problems.
The Hypergeometric Distribution
The Poisson Distribution
Suppose we are given an interval (this could be time, length, area
or volume) and we are interested in the number of successes in
that interval.
The Hypergeometric Distribution
The Poisson Distribution
Suppose we are given an interval (this could be time, length, area
or volume) and we are interested in the number of successes in
that interval. Assume that the interval can be divided into very
small subintervals such that:
The Hypergeometric Distribution
The Poisson Distribution
Suppose we are given an interval (this could be time, length, area
or volume) and we are interested in the number of successes in
that interval. Assume that the interval can be divided into very
small subintervals such that:
1
the probability of more than one success in any subinterval is
zero;
The Hypergeometric Distribution
The Poisson Distribution
Suppose we are given an interval (this could be time, length, area
or volume) and we are interested in the number of successes in
that interval. Assume that the interval can be divided into very
small subintervals such that:
1
the probability of more than one success in any subinterval is
zero;
2
the probability of one success in a subinterval is constant for
all subintervals and is proportional to its length;
The Hypergeometric Distribution
The Poisson Distribution
Suppose we are given an interval (this could be time, length, area
or volume) and we are interested in the number of successes in
that interval. Assume that the interval can be divided into very
small subintervals such that:
1
the probability of more than one success in any subinterval is
zero;
2
the probability of one success in a subinterval is constant for
all subintervals and is proportional to its length;
3
subintervals are independent of each other.
The Hypergeometric Distribution
We assume the following.
The Poisson Distribution
The Hypergeometric Distribution
The Poisson Distribution
We assume the following.
1
The random variable X denotes the number of successes in
the whole interval.
The Hypergeometric Distribution
The Poisson Distribution
We assume the following.
1
2
The random variable X denotes the number of successes in
the whole interval.
λ is the mean number of successes in the interval.
The Hypergeometric Distribution
The Poisson Distribution
We assume the following.
1
2
The random variable X denotes the number of successes in
the whole interval.
λ is the mean number of successes in the interval.
X has a Poisson Distribution with parameter λ and
P(X = x) = p(x; λ) =
λx e −λ
, x = 0, 1, 2, · · · .
x!
The Hypergeometric Distribution
The Poisson Distribution
Example 15
The average number of trucks on any one day at a truck depot in
a certain city is known to be 12. What is the probability that on a
given day fewer than nine trucks will arrive at this depot?
The Hypergeometric Distribution
The Poisson Distribution
Example 15
The average number of trucks on any one day at a truck depot in
a certain city is known to be 12. What is the probability that on a
given day fewer than nine trucks will arrive at this depot?
Solution. Let X be the number of trucks arriving on a given day.
Then, using Poisson distribution with λ = 12, we get
P(X < 9) =
8
p(x; 12) =
x=0
8
12x e −12
x=0
121
x!
122
123 124
120
+
+
+
+
0!
1!
2!
3!
4!
5
6
7
8
12
12
12
12
+
+
+
+
5!
6!
7!
8!
= 0.1550.
= e −12
The Hypergeometric Distribution
The Poisson Distribution
Example 16
The number of flaws in a fiber optic cable follows a Poisson
distribution. The average number of flaws in 50m of cable is 1.2.
The Hypergeometric Distribution
The Poisson Distribution
Example 16
The number of flaws in a fiber optic cable follows a Poisson
distribution. The average number of flaws in 50m of cable is 1.2.
(a) What is the probability of exactly three flaws in 150m of
cable?
The Hypergeometric Distribution
The Poisson Distribution
Example 16
The number of flaws in a fiber optic cable follows a Poisson
distribution. The average number of flaws in 50m of cable is 1.2.
(a) What is the probability of exactly three flaws in 150m of
cable?
(b) What is the probability of at least two flaws in 100m of cable?
The Hypergeometric Distribution
The Poisson Distribution
Example 16
The number of flaws in a fiber optic cable follows a Poisson
distribution. The average number of flaws in 50m of cable is 1.2.
(a) What is the probability of exactly three flaws in 150m of
cable?
(b) What is the probability of at least two flaws in 100m of cable?
(c) What is the probability of exactly one flaw in the first 50m of
cable and exactly one flaw in the second 50m of cable?
The Hypergeometric Distribution
The Poisson Distribution
Example 16
The number of flaws in a fiber optic cable follows a Poisson
distribution. The average number of flaws in 50m of cable is 1.2.
(a) What is the probability of exactly three flaws in 150m of
cable?
(b) What is the probability of at least two flaws in 100m of cable?
(c) What is the probability of exactly one flaw in the first 50m of
cable and exactly one flaw in the second 50m of cable?
Solution. (a) Mean number of flaws in 150m of cable is
1.2 · 3 = 3.6. So the probability of exactly three flaws in 150m of
cable is
3.63 e −3.6
= 0.21247
p(3; 3.6) =
3!
The Hypergeometric Distribution
The Poisson Distribution
(b) Mean number of flaws in 100m of cable is 1.2 · 2 = 2.4. Let X
be the number of flaws in 100m of cable.
P(X ≥ 2) = 1 − P(X < 2) = 1 − (P(X = 0) + P(X = 1))
= 1 − p(0; 2.4) − p(1; 2.4)
2.40 e −2.4 2.41 e −2.4
−
0!
1!
= 0.69156
=1−
The Hypergeometric Distribution
The Poisson Distribution
(b) Mean number of flaws in 100m of cable is 1.2 · 2 = 2.4. Let X
be the number of flaws in 100m of cable.
P(X ≥ 2) = 1 − P(X < 2) = 1 − (P(X = 0) + P(X = 1))
= 1 − p(0; 2.4) − p(1; 2.4)
2.40 e −2.4 2.41 e −2.4
−
0!
1!
= 0.69156
=1−
(c) Now let X denote the number of flaws in a 50m section of
cable. Then we know that
1.21 e −1.2
= 0.36143.
P(X = 1) = p(1; 1.2) =
1!
The Hypergeometric Distribution
The Poisson Distribution
(b) Mean number of flaws in 100m of cable is 1.2 · 2 = 2.4. Let X
be the number of flaws in 100m of cable.
P(X ≥ 2) = 1 − P(X < 2) = 1 − (P(X = 0) + P(X = 1))
= 1 − p(0; 2.4) − p(1; 2.4)
2.40 e −2.4 2.41 e −2.4
−
0!
1!
= 0.69156
=1−
(c) Now let X denote the number of flaws in a 50m section of
cable. Then we know that
1.21 e −1.2
= 0.36143.
P(X = 1) = p(1; 1.2) =
1!
As X follows a Poisson distribution, the occurrence of flaws in the
first and second 50m of cable are independent. Thus the
probability of exactly one flaw in the first 50m and exactly one flaw
in the second 50m is
(0.36143)(0.36143) = 0.13063.
The Hypergeometric Distribution
The Poisson Distribution
Example 17
Births in a hospital occur randomly at an average rate of 1.8 births
per hour. What is the probability of observing 4 births in a given
hour at the hospital?
The Hypergeometric Distribution
The Poisson Distribution
Example 17
Births in a hospital occur randomly at an average rate of 1.8 births
per hour. What is the probability of observing 4 births in a given
hour at the hospital?
Solution. If we let X be the number of births in an hour, then X
has a Poisson distribution:
P(X = 4) = p(4; 1.8) =
1.84 e −1.8
= 0.072302.
4!
The Hypergeometric Distribution
The Poisson Distribution
Example 18
Consider a telephone operator who, on the average, handles five
calls every 3 minutes.
The Hypergeometric Distribution
The Poisson Distribution
Example 18
Consider a telephone operator who, on the average, handles five
calls every 3 minutes. (a) What is the probability that there will
be no calls in the next minute?
The Hypergeometric Distribution
The Poisson Distribution
Example 18
Consider a telephone operator who, on the average, handles five
calls every 3 minutes. (a) What is the probability that there will
be no calls in the next minute? (b) At least one call?
The Hypergeometric Distribution
The Poisson Distribution
Example 18
Consider a telephone operator who, on the average, handles five
calls every 3 minutes. (a) What is the probability that there will
be no calls in the next minute? (b) At least one call?
Solution. If we let X be the number of calls in a minute, then X
has a Poisson distribution with λ = 53 .
The Hypergeometric Distribution
The Poisson Distribution
Example 18
Consider a telephone operator who, on the average, handles five
calls every 3 minutes. (a) What is the probability that there will
be no calls in the next minute? (b) At least one call?
Solution. If we let X be the number of calls in a minute, then X
has a Poisson distribution with λ = 53 . So
(a) P(no calls in the next minute) = P(X = 0) = p(0; 5/3) =
(5/3)0 e −5/3
= 0.1889
0!
The Hypergeometric Distribution
The Poisson Distribution
Example 18
Consider a telephone operator who, on the average, handles five
calls every 3 minutes. (a) What is the probability that there will
be no calls in the next minute? (b) At least one call?
Solution. If we let X be the number of calls in a minute, then X
has a Poisson distribution with λ = 53 . So
(a) P(no calls in the next minute) = P(X = 0) = p(0; 5/3) =
(5/3)0 e −5/3
= 0.1889
0!
(b) P(At least one call) = P(X ≥ 1) = 1 − P(X = 0) =
1 − p(0; 5/3) = 1 − 0.1889 = 0.8111.
The Hypergeometric Distribution
The Poisson Distribution
Example 19
A certain kind of sheet metal has on the average, five defects per
10-square-feet. If we assume a Poisson distribution, what is the
probability that a 15-square-feet sheet of the metal will have at
least six defects?
The Hypergeometric Distribution
The Poisson Distribution
Example 19
A certain kind of sheet metal has on the average, five defects per
10-square-feet. If we assume a Poisson distribution, what is the
probability that a 15-square-feet sheet of the metal will have at
least six defects?
Solution. Let X denote the number of defects in a 15-square-foot
sheet of the metal.
The Hypergeometric Distribution
The Poisson Distribution
Example 19
A certain kind of sheet metal has on the average, five defects per
10-square-feet. If we assume a Poisson distribution, what is the
probability that a 15-square-feet sheet of the metal will have at
least six defects?
Solution. Let X denote the number of defects in a 15-square-foot
sheet of the metal. Then, since the unit area is 10-square-feet, we
have
λ = 5 · 1.5 = 7.5
The Hypergeometric Distribution
The Poisson Distribution
Example 19
A certain kind of sheet metal has on the average, five defects per
10-square-feet. If we assume a Poisson distribution, what is the
probability that a 15-square-feet sheet of the metal will have at
least six defects?
Solution. Let X denote the number of defects in a 15-square-foot
sheet of the metal. Then, since the unit area is 10-square-feet, we
have
λ = 5 · 1.5 = 7.5
and
P(X ≥ 6) = 1 − P(X ≤ 5) = 1 − (P(X = 0) + P(X = 1) + P(X = 2)
+P(X = 3) + P(X = 4) + P(X = 5))
0
7.51
7.52 7.53
7.54
7.55
7.5
+
+
+
+
+
= 1 − e −7.5
0!
1!
2!
3!
4!
5!
= 1 − (0.2414)
= 0.7586.
The Hypergeometric Distribution
The Poisson Distribution
Thank You!!!