Transcript Hypergeometric and Poisson Distributions

Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Discrete Probability Distributions
Hypergeometric & Poisson Distributions
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
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Hypergeometric Distribution
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Hypergeometric Distribution
Conditions:
• Population size = N
K members are “successes”
N - K members are “failures”
• Sample size = n
Obtained without replacement
X = the number of successes in n trials
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Probability Mass Function
 k  N  k 
 

x  n  x 

h ( x )  h ( x ; N, n , k ) 
 N
 
n
n if n  k
x = 0, 1, ...,
where
m
m!
  
 r  r!m  r !
k if n > k
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Mean & Standard Deviation
• Mean or Expected Value of
X
K
  n 
N
• Standard Deviation of
X
N n
K  K 

 n  1   
N  N 
 N 1
1
2
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Hypergeometric Distribution
Example:
Five individuals from an animal population thought
to be near extinction in a certain region have been
caught, tagged and released to mix into the
population. After they have had an opportunity to
mix in, a random sample of 10 of these animals is
selected. Let X = the number of tagged animals in
the second sample. If there are actually 25 animals
of this type in the region,
Find:
a) P( X  2)
b) P( X  2)
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Hypergeometric Distribution
Example Solution:
The parameter values are
n = 10
K = 5 (5 tagged animals in the population)
N = 25
5
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 

 

x 10  x 

hx;25,10,5 
 25 
 
 10 
X = 0, 1, 2, 3, 4, 5
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Hypergeometric Distribution
Example Solution:
 5  20 
  
2  8 

 0.385
a. P X  2   h2;10,5,25 
 25 
 
 10 
2
b.
P X  2  PX  0, 1, or 2   hx;10,5,25
x 0
 0.057  0.257  0.385
 0.699
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Hypergeometric Distribution
•
Similar to the Binomial Dist in Excel:
–
Click the Insert button on the menu bar (at the
top of the Excel page)
•
•
•
•
Go to the function option
Choose Statistical from the Function Category window
(a list of all available statistical functions will appear in
the Function Name window)
Choose the HYPGEOMDIST function
Type in parameters:
–
–
–
–
Sample_s => x
Number_sample => n
Population_s => k
Number_pop => N
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Poisson Distribution
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Properties
1. The number of outcomes occurring in one time
interval or specified region is independent of the
number that occurs in any other disjoint time
interval or region of space. In this way we say that
the Poisson process has no memory.
2. The probability that a single outcome will occur
during a very short time interval or in a small region
is proportional to the length of the time interval or
the size of the region and does not depend on the
number of outcomes occurring outside this time
interval or region.
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Properties
3. The probability that more than one outcome will
occur in such a short time interval or fall in such a
small region is negligible.
Remark: The Poisson distribution is used to describe
a number of processes, including the distribution of
telephone calls going through a switchboard system,
the demand (needs) of patients for service at a health
institution, the arrivals of trucks and cars at a
tollbooth, the number of accidents at an intersection,
etc.
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Poisson Distribution
Definition - If X is the number of outcomes occurring
during a Poisson experiment, then X has a Poisson
distribution with probability mass function

e 
p( x;  ) 
x!
x
for x  0, 1, 2, ...
where  = t and  is the average number of
outcomes per unit time, t is the time interval and
e = 2.71828...
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Poisson Distribution
• Mean or Expected Value of
X
E X   
• Variance and Standard Deviation of
X
Var X     
2
 
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Poisson Distribution
Example:
When a company tests new tires by driving them
over difficult terrain, they find that flat tires
externally caused occur on the average of once
every 2000 miles. What is the probability
that in a given 500 mile test no more than one flat
will occur?
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Poisson Distribution
Example Solution:
Here the variable t is distance, and the random
variable of interest is
X
= number of flats in 500 miles
Since E(X) is proportional to the time interval involved
in the definition of X, and since the average is given
as one flat is 2000 miles, we have
1
  t  E ( X )  flat in 500 miles
4
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Poisson Distribution
Example Solution:
The values assigned to  and t depend on the unit
of distance adopted. If we take one mile as the unit,
then t = 500,  = 0.0005, and  = t = 1/4. If we
take 1000 miles as the unit, then t = 1/2,  = 1/2,
and again t = 1/4, and so on. The important thing
is that t = 1/4, no matter what unit is chosen.
P( X  1)  p(0)  p(1)
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Poisson Distribution
Example Solution
e
1 / 4
1 / 4
0
e
0!
 0.779  0.195
 0.974
P( X  1)  0.026
1 / 4
1 / 4
1
1!
Is this acceptable?
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