Transcript The Poisson Distribution

The Poisson Distribution
We can use the Poisson
distribution to estimate the
probability of arrivals at a car wash
in one hour or the number of leaks
in 100 miles of pipeline. Bell Labs
uses it to model the arrival of phone
calls.
The Poisson Distribution
The Poisson distribution is defined by:
f ( x) 
x 
 e
x!
Where f(x) is the probability of x occurrences in an
interval
 is the expected value or mean value of occurrences
within an interval
e is the natural logarithm. e = 2.71828
Properties of the Poisson Distribution
1. The probability of occurrences is the same for
any two intervals of equal length.
2. The occurrence or nonoccurrence of an event in
one interval is independent of an occurrence on
nonoccurrence of an event in any other interval
Example: Mercy Hospital

Poisson Probability Function
Patients arrive at the
emergency room of Mercy
Hospital at the average
rate of 6 per hour on
weekend evenings.
What is the
probability of 4 arrivals in
30 minutes on a weekend evening?
MERCY
Example: Mercy Hospital
 Poisson
Probability Function
 = 6/hour = 3/half-hour, x = 4
34 (2.71828)3
f (4) 
 .1680
4!
MERCY
Using Excel to Compute
Poisson Probabilities
 Formula
A
1
2
MERCY
Worksheet
B
3 = Mean No. of Occurrences ( )
Number of
3 Arrivals (x )
4
0
5
1
6
2
7
3
8
4
9
5
10
6
… and so on
Probability f (x )
=POISSON(A4,$A$1,FALSE)
=POISSON(A5,$A$1,FALSE)
=POISSON(A6,$A$1,FALSE)
=POISSON(A7,$A$1,FALSE)
=POISSON(A8,$A$1,FALSE)
=POISSON(A9,$A$1,FALSE)
=POISSON(A10,$A$1,FALSE)
… and so on
Using Excel to Compute
Poisson Probabilities
 Value
A
1
2
MERCY
Worksheet
B
3 = Mean No. of Occurrences ( )
Number of
3 Arrivals (x )
4
0
5
1
6
2
7
3
8
4
9
5
10
6
… and so on
Probability f (x )
0.0498
0.1494
0.2240
0.2240
0.1680
0.1008
0.0504
… and so on
Example: Mercy Hospital
 Poisson
MERCY
Distribution of Arrivals
Poisson Probabilities
Probability
0.25
0.20
actually,
the sequence
continues:
11, 12, …
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
Number of Arrivals in 30 Minutes
10
Problem 31, p. 229
Consider a Poisson probability distribution with an
average number of occurrences of two per period.
a. Write the appropriate Poisson distribution
b. What is the average number of occurrences in three time
periods?
c. Write the appropriate Poisson function to determine the
probability of x occurrences in three time periods.
d. Compute the probability of two occurrences in one time period.
e. Compute the probability of six occurrences in three time
periods.
f. Compute the probability of five occurrences in two time
periods.
Problem 31, p. 229
x
(a)
(b)
(c)
(d)
2
2 e
f ( x) 
X!
 6
6 x e 6
f ( x) 
X!
22 e 2 .5413
f (2) 

 .27067
2!
2
Problem 31, p. 229
6
6
(d)
6 e
f (6) 
 .16062
6!
(e)
45 e 5
f (5) 
 .15629
5!
Problem 31, p. 229
Poisson Distribution for Ex. 39, p. 229
f(x)
0.3
0.2
0.1
0
0
1
2
3
4
5
6
No. of Occurrences Per Interval
7
The Hypergeometric Distribution
This is similar to the
binominal distribution
except: (1) the trials are
NOT independent; and (2)
the probability of success
(ρ) changes from trial to
trial.
Hypergeometric Distribution
Let r denote in the population size N labeled a success.
N – r is the number of elements in the population labeled failure.
The hypergeometric
distribution is used to compute
the probability that in a random
selection of n elements,
selected without replacement,
we obtain x elements labeled
success and
N – x elements labeled failure.
Notice that the x successes
must be pulled from the r
number of successes in the
population and the n - x
failures must be drawn from a
population of N – r failures
Hypergeometric Distribution
 r  N  r 
 

x  n  x 

f ( x) 
for all 0  x  r
N
 
n 
Where
n = the number of trials.
N = number of elements in the population
r = number of elements in the population labeled a success
Hypergeometric Distribution
Number of ways a sample of
size x successes can be
selected from a population of
size r
Number of ways a sample of
size n -x failures can be
selected from a population of
size N -r
 r  N  r 
 

x  n  x 

f ( x) 
for all 0  x  r
N
 
n 
Number of ways a sample of
size n can be selected from a
population of size N
Example: Neveready

Hypergeometric Probability
Distribution
Bob Neveready has removed two
dead batteries from a flashlight and
inadvertently mingled them with
the two good batteries he intended
as replacements. The four batteries look
identical.
Bob now randomly selects two of the four
batteries. What is the probability he selects the
two good batteries?
Example: Neveready

Hypergeometric Probability Distribution
 r  N  r   2  2   2!  2! 
 x  n  x   2  0   2!0!  0!2! 
1









f (x) 


  .167
6
N
 4
 4! 
n
2
 2!2! 
 
 


where:
x = 2 = number of good batteries selected
n = 2 = number of batteries selected
N = 4 = number of batteries in total
r = 2 = number of good batteries in total
Using Excel to Compute
Hypergeometric Probabilities

Formula Worksheet
A
1
2
3
4
5
6
7
B
2
2
2
4
Number
Number
Number
Number
of Successes (x )
of Trials (n )
of Elements in the Population Labeled Success (r )
of Elements in the Population (N )
f (x ) =HYPGEOMDIST(A1,A2,A3,A4)
Using Excel to Compute
Hypergeometric Probabilities

Value Worksheet
A
1
2
3
4
5
6
7
B
2
2
2
4
Number
Number
Number
Number
f (x ) 0.1667
of Successes (x )
of Trials (n )
of Elements in the Population Labeled Success (r )
of Elements in the Population (N )