BCOR 1020 Business Statistics

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BCOR 1020Business Statistics

• Lecture 11 February 21, 2008

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Overview

• Chapter 6 Discrete Distributions
• Poisson Distribution
• Linear Transformations

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Chapter 6 Poisson Distribution

• Poisson Processes
• If the number of occurrences of interest on a
  given continuous interval (of time, length, etc.)
  are being counted, we say we have an approximate
  Poisson Process with parameter l gt 0 (occurrences
  per unit length/time) if the following conditions
  are satisfied

1. The number of occurrences in non-overlapping
   intervals are independent.

1. The probability of exactly one occurrences in a
   sufficiently short interval of length h is lh.
   (i.e. If the interval is scaled by h, we also
   scale the parameter l by h.)

1. The probability of two or more occurrences in a
   sufficiently short interval is essentially zero.
   (i.e. there are no simultaneous occurrences.)

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Chapter 6 Poisson Distribution

• Poisson Distribution
• The Poisson distribution describes the number of
  occurrences within a randomly chosen unit of time
  or space.

If X denotes the number of occurrences of
interest observed on a given interval of length 1
unit of a Poisson Process with parameter l gt 0,
then we say that X has the Poisson distribution
with parameter l.
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Chapter 6 Poisson Distribution

• Poisson Distribution
• Called the model of arrivals, most Poisson
  applications model arrivals per unit of time.

• The events occur randomly and independently over
  a continuum of time or space

One Unit One Unit
One Unit of Time of Time
of Time ?---? ?---?
?---?

Flow of Time ?

• Each dot () is an occurrence of the event of
  interest.

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Chapter 6 Poisson Distribution

• Let X the number of events per unit of time.

• X is a random variable that depends on when the
  unit of time is observed.

• For example, we could get X 3 or X 1 or X
  5 events, depending on where the randomly chosen
  unit of time happens to fall.

One Unit One Unit One Unit
of Time of Time of
Time ?---? ?---?
?---?
Flow
of Time ?
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Chapter 6 Poisson Distribution

• Arrivals (e.g., customers, defects, accidents)
  must be independent of each other.

• Some examples of Poisson models in which
  assumptions are sufficiently met are

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Chapter 6 Poisson Distribution

• Poisson Processes

• The Poisson models only parameter is l (Greek
  letter lambda).

l represents the mean number of events
(occurrences) per unit of time or space.

• The unit of time should be short enough that the
  mean arrival rate is not large (l lt 20).

• To make l smaller, convert to a smaller time unit
  (e.g., convert hours to minutes).

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Chapter 6 Poisson Distribution

• Poisson Processes

• The Poisson distribution is sometimes called the
  model of rare events.

• The number of events that can occur in a given
  unit of time is not bounded, therefore X has no
  obvious limit.

• However, Poisson probabilities taper off toward
  zero as X increases.

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Chapter 6 Poisson Distribution

• Poisson Distribution
• We can formulate the PMF, mean and variance (or
  standard deviation) of the Poisson distribution
  in terms of the parameter l

PMF of the Poisson distribution with parameter l
Mean of the Poisson distribution with parameter l
Variance and Standard Deviation of the Poisson
distribution with parameter l
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Chapter 6 Poisson Distribution
Parameters l mean arrivals per unit of time or space
PDF
Range X 0, 1, 2, ... (no obvious upper limit)
Mean l
St. Dev.
Random data Use Excels Tools Data Analysis Random Number Generation
Comments Always right-skewed, but less so for larger l.
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Chapter 6 Poisson Distribution

• Poisson Processes

Poisson distributions are always right-skewed but
become less skewed and more bell-shaped as l
increases.
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Chapter 6 Poisson Distribution

• Example Credit Union Customers
• On Thursday morning between 9 A.M. and 10 A.M.
  customers arrive and enter the queue at the
  Oxnard University Credit Union at a mean rate of
  102 customers per hour (or 1.7 customers per
  minute).
• Why would we consider this a Poisson
  distribution? Which units should we use? Why?

• Find the PDF, mean and standard deviation

Mean l 1.7 customers per minute.
Standard deviation s
1.304 cust/min
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Chapter 6 Poisson Distribution

• Example Credit Union Customers

• Here is the Poisson probability distribution for
  l 1.7 customers per minute on average.

x PDF P(X x) CDF P(X ? x)
0 .1827 .1827
1 .3106 .4932
2 .2640 .7572
3 .1496 .9068
4 .0636 .9704
5 .0216 .9920
6 .0061 .9981
7 .0015 .9996
8 .0003 .9999
9 .0001 1.0000

• Note that x represents the number of customers.

• For example, P(X4) is the probability that there
  are exactly 4 customers in the bank.

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Chapter 6 Poisson Distribution

• Using the Poisson Formula

Formula Excel function
POISSON(0,1.7,0)
POISSON(1,1.7,0)
POISSON(2,1.7,0)
POISSON(3,1.7,0)
POISSON(4,1.7,0)
These probabilities can be calculated using a
calculator or Excel
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Chapter 6 Poisson Distribution

• Here are the graphs of the distributions

• The most likely event is 1 arrival (P(1).3106 or
  31.1 chance).

• This will help the credit union schedule tellers.

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Clickers
Orders arrive at a pizza delivery franchise at an
average rate of 12 calls per hour. If we want
to model the number of calls arriving during a
randomly-selected 15 minute interval, which
distribution should we use? A Poisson
distribution with l 0.2 calls per minute B
Poisson distribution with l 0.8 calls per 15
minutes C Poisson distribution with l 3
calls per 15 min. D Poisson distribution with
l 12 calls per hour
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Clickers
Orders arrive at a pizza delivery franchise at an
average rate of 12 calls per hour. What are the
mean and standard deviation of the number of
calls arriving during a randomly-selected 15
minute interval? A m 3 and s 1.73 B
m 3 and s 3 C m 12 and s 3.46 D
m 12 and s 12
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Clickers
Orders arrive at a pizza delivery franchise at an
average rate of 12 calls per hour. What is the
probability of exactly two calls arriving during
a randomly-selected 15 minute interval? A
0.0004 B 0.1494 C 0.2240 D 0.4481
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Chapter 6 Poisson Distribution

• Compound Events
• Recall our earlier credit union example
• On Thursday morning between 9 A.M. and 10 A.M.
  customers arrive and enter the queue at the
  Oxnard University Credit Union at a mean rate of
  102 customers per hour (or 1.7 customers per
  minute).

• Cumulative probabilities can be evaluated by
  summing individual X probabilities.

• What is the probability that two or fewer
  customers will arrive in a given minute?

.1827 .3106 .2640 .7573
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Chapter 6 Poisson Distribution

• Compound Events

• What is the probability of at least three
  customers (the complimentary event)?

P(X gt 3) P(3) P(4) P(5)
Since X has no limit, this sum never ends. So,
we will use the compliment.
1 - .7573 .2427
P(X gt 3) 1 - P(X lt 2)
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Clickers
Orders arrive at a pizza delivery franchise at an
average rate of 12 calls per hour. What is the
probability that more than two calls arrive
during a randomly-selected 15 minute interval?
A 0.0498 B 0.1494 C 0.2240 D
0.4232 E 0.5768
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Chapter 6 Poisson Distribution

• Recognizing Poisson Applications

• Can you recognize a Poisson situation?

• Look for arrivals of rare independent events
  with no obvious upper limit.

• In the last week, how many credit card
  applications did you receive by mail?

• In the last week, how many checks did you write?

• In the last week, how many e-mail viruses did
  your firewall detect?

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Chapter 6 Linear Transformations

• Linear Transformations

• A linear transformation of a random variable X is
  performed by adding a constant or multiplying by
  a constant.

• For example, consider defining a random variable
  Y in terms of the random variable X as follows

Where a and b are any two constants.
Rule 1 maXb amX b (mean of a transformed

variable) Rule 2 saXb asX (standard
deviation of a
transformed variable)
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Chapter 6 Linear Transformations

• Example Total Cost
• The total cost of many goods is often modeled as
  a function of the good produced, Q (a random
  variable).

Specifically, if there is a variable cost per
unit v and a fixed cost F, then the total cost of
the good, C, is given by
where v and F are constant values.
For given values of mQ, sQ, v, and F, we can
determine the mean and standard deviation of the
total cost
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Clickers
If Q is a random variable with mean mQ 500
units and standard deviation sQ 40 units, the
variable cost is v 35 per unit, and the fixed
cost is F 24,000, the mean of the total cost
is Determine the standard deviation of the
total cost. A) sC 35 B) sC 40 C)
sC 1,400 D) sC 25,400