multinomial experiments
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Multinomial Experiments. What if there are more than 2 possible outcomes? (e.g., acceptable, scrap, rework) That is, suppose we have: n independent trials k outcomes that are mutually exclusive (e.g., ♠, ♣, ♥, ♦) exhaustive (i.e., ∑ all k p i = 1) Then - PowerPoint PPT PresentationTRANSCRIPT
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 1
Multinomial ExperimentsWhat if there are more than 2 possible outcomes?
(e.g., acceptable, scrap, rework)That is, suppose we have:
n independent trialsk outcomes that are
mutually exclusive (e.g., ♠, ♣, ♥, ♦) exhaustive (i.e., ∑all kpi = 1)
Then f(x1, x2, …, xk; p1, p2, …, pk, n) =
kxk
xx
k
pppxxx
n...
,...,,21
2121
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 2
Multinomial Examples Example 5.7 refer to page 150
Problem 22, page 152 Convert ratio 8:4:4 to probabilities (8/16, 4/16)
f( __, __, __; ___, ___, ___, __) =___.25, 8) = (8 choose 5,2,1)(0.5)5(0.25)2(0.25)1 = 8!/(5!2!1!)* )(0.5)5(0.25)2(0.25)1 = 21/256 or 0.082031
x1 = _______ p1 = 0.50
x2 = _______ p2 = 0.25
x3 = _______ p3 = 0.25
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 3
Binomial vs.Hypergeometric Distribution
Replacement and IndependenceBinomial (assumes sampling “with replacement”)
and hypergeometric (sampling “without replacement”)
Binomial assumes independence, while hypergeometric does not.
Hypergeometric: The probability associated with getting x successes in the sample (given k successes in the lot.)
n
N
xn
kN
x
k
knNxhxXP ),,;()(
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 4
Example from Complete Business Statistics, 4th ed (McGraw-Hill) Automobiles arrive in a dealership in lots of 10. Five out of
each 10 are inspected. For one lot, it is known that 2 out of 10 do not meet prescribed safety standards. What is probability that at least 1 out of the 5 tested from that lot will be found not meeting safety standards?
This example follows a hypergeometric distribution: A random sample of size n is selected without replacement
from N items. k of the N items may be classified as “successes” and N-k are
“failures.” The probability associated with getting x successes in the sample
(given k successes in the lot.)
n
N
xn
kN
x
k
knNxhxXP ),,;()(
Hypergeometric Example
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 5
Solution: Hypergeometric Example In our example,
k = number of “successes” = 2 n = number in sample = 5
N = the lot size = 10 x = number found = 1 or 2
P(X > 1) = 0.556 + 0.222 = 0.778
5
10
25
210
2
2
5
10
15
210
1
2
)2,5,10;2()2,5,10;1(
)2()1()(
hh
XPXPxXP
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 6
Expectations: Hypergeometric Distribution
The mean and variance of the hypergeometric distribution are given by
What are the expected number of cars that fail inspection in our example? What is the standard deviation?
μ = nk/N = 5*2/10 = 1
σ2 = (5/9)(5*2/10)(1-2/10) = 0.444
σ = 0.667
N
k
N
kn
N
nNN
nk
1**1
2
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 7
Additional problems …A worn machine tool produced defective parts for a period of time before the problem was discovered. Normal sampling of each lot of 20 parts involves testing 6 parts and rejecting the lot if 2 or more are defective. If a lot from the worn tool contains 3 defective parts:
1. What is the expected number of defective parts in a sample of six from the lot? N = 20 n = 6 k = 3 μ = nk/N = 6*3/20 =18/20=0.9
2. What is the expected variance? σ2
= (14/19)(6*3/20)(1-3/20) = 0.5637
3. What is the probability that the lot will be rejected?
P(X>2) = 1 – [P(0)+P(1)]
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 8
Binomial Approximation Note, if N >> n, then we can approximate the
hypergeometric with the binomial distribution. Example: Automobiles arrive in a dealership in lots of 100. 5
out of each 100 are inspected. 2 /10 (p=0.2) are indeed below safety standards. What is probability that at least 1 out of 5 inspected will not meet safety standards?
Recall: P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0)
Hypergeometric distribution Binomial distribution
1 - h(0;100,5,20)
= 0.6807
1 - b(0;5,0.2)
1 - 0.3277 = 0.6723
(See also example 5.12, pg. 155-6)
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 9
Negative Binomial Distribution b* A binomial experiment in which trials are
repeated until a fixed number of successes occur.
Example:Historical data indicates that 30% of all bits transmitted through a digital transmission channel are received in error. An engineer is running an experiment to try to classify these errors, and will start by gathering data on the first 10 errors encountered.
What is the probability that the 10th error will occur on the 25th trial?
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 10
This example follows a negative binomial distribution: Repeated independent trials. Probability of success = p and probability of failure = q = 1-p. Random variable, X, is the number of the trial on which the kth
success occurs. The probability associated with the kth success occurring
on trial x is given by,
Where,k = “success number”x = trial number on which k occurs p = probability of success (error) q = 1 – p
,...2,1,,1
1),;(*
kkkxqpk
xpkxb kxk
Negative Binomial Equation
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 11
Example: Negative Binomial DistributionWhat is the probability that the 10th error will
occur on the 25th trial? k = “success number” = 10
x = trial number on which k occurs = 25
p = probability of success (error) = 0.3
q = 1 – p = 0.7
102510 )7.0()3.0(110
125)3.0,10;25(*
b
037.0)7.0()3.0(9
24)3.0,10;25(* 1510
b
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 12
Geometric DistributionContinuing with our example in which p =
probability of success (error) = 0.3What is the probability that the 1st bit received in
error will occur on the 5th trial?This is an example of the geometric distribution,
which is a special case of the negative binomial in which k = 1.The probability associated with the 1st success
occurring on trial x is
P = (0.3)(0.7)4 = 0.072
...,3,2,1);( 1 xpqpxg x
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 13
Additional problems …A worn machine tool produces 1% defective parts. If we assume that parts produced are independent:
1.What is the probability that the 2nd defective part will be the 6th one produced?
2.What is the probability that the 1st defective part will be seen before 3 are produced?
3.How many parts can we expect to produce before we see the 1st defective part?
Negative binomial or geometric? Expected value = ?
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 14
Poisson Process
The number of occurrences in a given interval or region with the following properties: “memoryless” ie number in one interval is independent of
the number in a different intervalP(occurrence) during a very short interval or small region
is proportional to the size of the interval and doesn’t depend on number occurring outside the region or interval.
P(X>1) in a very short interval is negligible
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 15
Poisson Process Situations
Number of bits transmitted per minute.Number of calls to customer service in an hour.Number of bacteria present in a given sample.Number of hurricanes per year in a given region.
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 16
Poisson Distribution ProbabilitiesThe probability associated with the number of
occurrences in a given period of time is given by,
Where,λ = average number of outcomes per unit time or region
t = time interval or region
...,2,1,0,!
)();(
xx
tetxp
xt
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 17
Service Call Example - Poisson Process
Example
An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. To determine personnel and equipment needs to maintain a desired level of service, the plant manager needs to be able to determine the probabilities associated with numbers of service calls.
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 18
Our Example: λ = 2.7 and t = 1 minute
What is the probability that fewer than 2 calls will be received in any given minute?
The probability that fewer than 2 calls will be received in any given minute is
P(X < 2) = P(X = 0) + P(X = 1)
The mean and variance are both λt, so
μ = λt =________________
Note: Table A.2, pp. 732-734, gives Σt p(x;μ)
!1
)7.2(
!0
)7.2()2(
17.207.2
ee
XP
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 19
Service Call Example (Part 2) If more than 6 calls are received in a 3-minute period,
an extra service technician will be needed to maintain the desired level of service. What is the probability of that happening? μ = λt = (2.7) (3)= 8.1 8.1 is not in the table; we must use basic equation
Suppose λt = 8; see table with μ = 8 and r = 6
P(X > 6) = 1 – P(X < 6) = 1 - 0.3134 = 0.6866
!6
)1.8(
!1
)1.8(
!0
)1.8(1)6(1)6(
61.811.801.8 eeeXpXp
JMB Ch. 5 Part 2 rev 2012b EGR 252.001 2012 9th ed. Slide 20
Poisson Distribution
05
101520253035404550
Fre
qu
ency
Calls per minute