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 kpi 1)
• Then
• f(x1, x2, , xk p1, p2, , pk, n)

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Multinomial Examples

• Example 5.7 refer to page 150
• Problem 22, page 152
• Convert ratio 844 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
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Binomial vs.Hypergeometric Distribution

• Replacement and Independence
• Binomial (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.)

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Hypergeometric Example

• 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.)

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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 gt 1) 0.556 0.222 0.778

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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 52/10 1
• s2 (5/9)(52/10)(1-2/10) 0.444
• s 0.667

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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
• 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
  63/20 18/200.9
• What is the expected variance?
  s2 (14/19)(63/20)(1-3/20)
  0.5637
• What is the probability that the lot will be
  rejected?
• P(Xgt2) 1 P(0)P(1)

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Binomial Approximation

• Note, if N gtgt 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 (p0.2) 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 lt 1) 1 P(X 0)

Hypergeometric distribution Binomial distribution
1 - h(0100,5,20) 0.6807 1 - b(05,0.2) 1 - 0.3277 0.6723
(See also example 5.12, pg. 155-6)
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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?

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Negative Binomial Equation

• 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

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Example Negative Binomial Distribution

• What 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

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Geometric Distribution

• Continuing with our example in which p
  probability of success (error) 0.3
• What 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

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Additional problems

• A worn machine tool produces 1 defective parts.
  If we assume that parts produced are independent
• What is the probability that the 2nd defective
  part will be the 6th one produced?
• What is the probability that the 1st defective
  part will be seen before 3 are produced?
• How many parts can we expect to produce before we
  see the 1st defective part?
• Negative binomial or geometric? Expected value
  ?

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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 interval
• P(occurrence) during a very short interval or
  small region is proportional to the size of the
  interval and doesnt depend on number occurring
  outside the region or interval.
• P(Xgt1) in a very short interval is negligible

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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.

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Poisson Distribution Probabilities

• The probability associated with the number of
  occurrences in a given period of time is given
  by,
• Where,
• ? average number of outcomes per unit time or
  region
• t time interval or region

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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.

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Our Example ? 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 lt 2) P(X 0) P(X 1)
• The mean and variance are both ?t, so
• µ ?t ________________
• Note Table A.2, pp. 732-734, gives St p(xµ)

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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 ?t (2.7) (3) 8.1
• 8.1 is not in the table we must use basic
  equation
• Suppose ?t 8 see table with µ 8 and r 6
• P(X gt 6) 1 P(X lt 6) 1 - 0.3134
  0.6866

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Poisson Distribution