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Transcript
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
Binomial, Negative Binomial,
Geometric Distributions
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Binomial Distribution
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The Binomial Model:
Let X be the number of successes in n trials.
If
1. The trials are identical and independent
2. Each trial results in one of two possible
outcomes success or failure
3. The probability of success on a single trial is p,
and is constant from trial to trial,
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Binomial Model:
then X has the Binomial Distribution with Probability
Mass Function given by:
 n  x n -x
b( x)  b( x; n, p)    p q
 x
x  0,1,..., n
where
n
n!
  
 x  x!n  x !
and
q  1 p
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Binomial Distribution
Rule:
 n  x  p 
b( x  1; n, p)  
 b( x; n, p)
 x  1  q 
for x = 0, …, n - 1
Rule:
 n  x n x
1  p  q     p q
x 0  x 
n
n
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Binomial Distribution
• Mean or Expected Value
 = np
• Standard Deviation
σ  npq 
1
2
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Binomial Model in Excel
•
Excel provides an easier way of
finding probabilities:
–
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 BINOMDIST function
Type in parameters:
–
–
–
–
number_s => X
Trials => N
Probability_s => p
Cumulative (logical) => TRUE for cumulative
function, FALSE for mass function.
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
7
Example - Binomial Distribution
When circuit boards used in the manufacture of
compact disc players are tested, the long-run
percentage of defects is 5%. Let X=the number of
defective boards in a random sample size n=25, so
X~B(25,0.05).
a) Determine P(X  2).
b) Determine P(X  5).
c) Determine P(1  X  4).
d) What is the probability that none of the 25
boards are defective?
e) Calculate the expected value and standard
deviation of X.
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Solution – Binomial Distribution
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Negative Binomial Distribution
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Negative Binomial Model:
The negative binomial distribution is based on an
experiment satisfying the following conditions:
1. The experiment consists of a sequence of
independent trials
2. Each trial can result in either a success S, or a
failure F
3. The probability of success is constant from trial
to trial, so P(S on trial i) = p for i = 1, 2, 3, …
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Negative Binomial Model:
4. The experiment continues (trials are performed)
until a total of r successes have been observed,
where r is a specified positive integer.
The associated random variable is:
X = number of failures that precede the rth success
X is called the negative binomial random variable
because, in contrast to the binomial random
variable, the number of successes is fixed and the
number of trials is random.
Possible values of X are x = 0, 1, 2, ...
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
12
The Negative Binomial Model:
The probability mass function of the negative
binomial random variable X with parameters
r = number of successes and
p = probability of success on a single trial
 x  r  1 r
x
 p 1  p  , x  0,1,2,...
NBx; r , p   
 r 1 
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Negative Binomial Model:
The negative binomial model can also be
expressed as:
X = total number of trials to get k successes
k = number of successes and
p = probability of success on a single trial
 x  1 k
x k
 p 1  p  , x  0,1,2,...
NBx; k , p   
 k  1
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Negative Binomial Probability Mass Function Derivation
th
• Consider the probability of a success on the x
trial preceded by k  1 successes and x  k failures
in some specified order.
• Since the trials are independent, we can multiply
all the probabilities corresponding to each desired
outcome.
• Each success occurs with probability
failure with probability q  1  p
pand each
• Therefore, the probability for the specified order,
ending in a success, is p k 1q xk p  p k q xk
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• The total number of sample points in the
experiment ending in a success, after the
occurrence of k  1successes and x  k failures in
any order, is equal to the number of partitions of
x  1 trials into two groups with k  1 successes
corresponding to one group and x  k failures
corresponding to the other group.
• The sample space consists of  x  1 points, each


 k  1
mutually exclusive and occurring with equal
k xk
p
q
probability
k xk
p
• We obtain the general formula by q multiplying
by  x  1


 k  1
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Stracener_EMIS 7370/STAT 5340_Sum 07_06.05.07
The Negative Binomial Model: Example
Find the probability that a person tossing three coins
will get either all heads or all tails for the second time
on the fifth toss.
Solution
p  Psuccess 
 P[( HHH )or (TTT )]
1 1
 
8 8
1

4
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Since a success must occur on the fifth toss of the 3
coins, the following outcomes are possible:
probability
Toss
1
S
F
F
F
2
F
S
F
F
3
F
F
S
F
4
F
F
F
S
5
S
S
S
S
27
1024
27
1024
27
1024
27
1024
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Therefore, the probability of obtaining all heads or all
tails for the second time on the fifth toss is
27
27
4.

1024 256
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The Negative Binomial Model:
Example Solution continued
Or, using the Negative Binomial Distribution with
r = 2, p = 0.25, and x = 3 gives
 4
27
3
2
NB3;2,0.25   0.25 0.75 
 0.105
256
1
20
Stracener_EMIS 7370/STAT 5340_Sum 07_06.05.07
The Negative Binomial Model
If X is a negative binomial random variable with
probability mass function nb(x;r,p) then
r 1  p 
EX  
p
and
r 1  p 
Var  X  
2
p
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Negative Binomial Model
Note: By expanding the binomial coefficient in
front of pr(1 - p)x and doing some cancellation, it
can be seen that NB(x;r,p) is well defined even
when r is not an integer. This generalized negative
binomial distribution has been found to fit the
observed data quite well in a wide variety of
applications.
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Negative Binomial Model in Excel
•
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 NEGBINOMDIST function
Type in parameters:
– number_f => X
– Number_s => r
– Probability_s => p
23
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Geometric Distribution
If repeated independent trials can result in a
success with probability p and a failure with
probability q = 1 - p, then the probability mass
function of the random variable x, the number of
the trial on which the first success occurs, is:
g(x;p) = pqx-1,
x = 1,2,3…
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Geometric Distribution
The mean and variance of a random variable
following the geometric distribution are
1
μ ,
p
1 p
σ  2
p
2
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Geometric Distribution - Example
At “busy time” a telephone exchange is very near
capacity, so callers have difficulty placing their
calls. It may be on interest to know the number of
attempts necessary in order to gain a connection.
Suppose that we let p = 0.05 be the probability of
a connection during a busy time. We are
interested in knowing the probability that 5
attempts are necessary for a successful call.
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Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
The Geometric Distribution - Example Solution
The random variable X is the number of attempts
for a successful call. Then
X~G(0.05),
So that for with x = 5 and p = 0.05 yields:
P(X=x) = g(5;0.05)
= (0.05)(0.95)4
= 0.041
1
And the expected number of attempts is μ 
 20
0.05
27
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08