Transcript document

The Poisson PD
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Are you familiar with the rock group Aerosmith? They have a
song called Toys in the Attic. If they had a song called Toys on
the Roof, then the “Toys on” part rhymes with the name
Poisson when you say “Toys on” real fast. Can you say
Poisson?
OKAY, I know this is a silly way to start this section, but I
wanted to have you be able to say Poisson correctly and “Toys
on” said real fast is all I could come up with.
The Poisson PD is useful when certain types of problems are
encountered. Some examples might be the number of people
who arrive at a Dairy Queen each hour, or the number of phone
calls that come into a bank card center each fifteen minute
interval. The number of occurrences is a random variable that
can be described by the Poisson PD.
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If, in an experiment,
1. the probability of an occurrence is the same for any two
intervals of equal length, and
2. the occurrence or nonoccurrence in any interval is independent
of the occurrence or nonoccurrence in any other interval,
then the number of occurrences can be described by the Poisson
PD. If x is the number of occurrences in the interval, then P(X) is
the probability of x occurrences.
Now P(X) = e-λλX/X!,
where λ = the expected value or mean number of occurrences, and
e = 2.71828
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Example: Say we are interested in the number of arrivals at a driveup teller window during a 1-minute period during the noon hour. Say
from past experience we know that on average 3 cars drive up to the
window in a 1 minute period.
Table E.7 has information about various mean amounts during a
interval. Since our λ = 3 we look in that column:
X
P(X)
0
.0498
1
.1494
2
.2240
And so on. On the next slide I have produced the P(X) column in
Excel and I have put in the cumulative probability column.
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Number
of Arrivals
0
1
2
3
4
5
6
7
8
9
10
11
12
Mean Number
of Occurrences
3
P(X)
0.049787068
0.149361205
0.224041808
0.224041808
0.168031356
0.100818813
0.050409407
0.021604031
0.008101512
0.002700504
0.000810151
0.00022095
5.52376E-05
Cumulative
Probability
0.049787068
0.199148273
0.423190081
0.647231889
0.815263245
0.916082058
0.966491465
0.988095496
0.996197008
0.998897512
0.999707663
0.999928613
0.999983851
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What does the Cum Prob on the previous screen do for us? It
helps us answer questions like what is the probability that 5 or
fewer arrivals occur in the interval? We might write this
P(X≤5)?
So, P(x≤5) = .9161. What about P(x>5)? This equals 1 - P(x≤5)
= .0839
Note on the previous screen that the Cum Prob does not equal 1
when x = 12. The reason is that I just listed up to 12 arrivals,
but I could have listed more.
In problems in this section they do something peculiar. They
might say something like on average 48 calls arrive per hour.
How many calls on average come in 5 minutes? Since 5
minutes is 1/12 of an hour, divide the 48 by 12, or 4. IN
OTHER WORDS, pay attention to the time interval and average
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occurrences per time interval.
In excel the Poisson function has the form
=poisson(number of arrivals, mean number of arrivals, false or
true)
The number of arrivals is in the x column, the mean number of
arrivals is at the top of the table, and use false if you want P(X)
or use true if you want the cum. prob.
You could just use the table in the book.
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Problem 22 page 170
From Table E.7 column 6.0
X
P(X)
0
.0025
a. P(X < 5) = P(0) + P(1) + P(2) + P(3)
1
.0149
+ P(4) = .2851
2
.0446
b. P(5) = .1606
3
.0892
c. P(X≥5) = 1 – P(X<5) = .7149
4
.1339
d. P(4 or 5) = P(4) + P(5) = .2945
5
.1606
6
.1377
question 37 page 201
And so on.
Discard if 3 or less – P(X≤3) =
P(0) + P(1) + P(2) + P(3) = .1512
They can expect to discard .1512(100)
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= 15.12 cookies