Poisson Distribution

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Example A student attempts a multiple choice
exam (options A to F for each question), but
having done no work, selects his answers to each
question by rolling a fair die (A 1, B 2,
etc.). If the exam contains 100 questions, what
is the probability of obtaining a mark below 20?
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Simulation
Now, let us simulate a large number of
realisations of students using this random method
of answering multiple choice questions. We still
require the same Binomial distribution with n100
and a This can be done on R using the command
rbinom.
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For example, lets simulate 1000 students.
gt xsimrbinom(1000,100,1/6) gt xsim 1 18 22
9 17 18 20 21 16 8 18 11 16 16 13 16 14 25 15 16
17 21 13 25 11 24 17 16 13 21 10 17 18 10 17
18 19 17 19 15 13 12 41 15 11 21 23 19 14 19
25 23 19 20 17 17 15 16 14 13 16 17 14 61 24
21 19 8 18 20 22 16 15 20 19 17 13 15 13 21 22
12 12 12 81 11 14 11 12 16 16 17 21 17 16 17
14 9 17 16 17 12 20 16 17 101 18 13 15 16 12
15 17 16 17 26 18 14 21 15 10 23 12 16 16 12
121 17 18 22 17 18 14 19 22 13 17 21 15 21 16
17 16 16 28 16 17 141 18 19 16 11 14 18 16 18
18 14 20 13 19 19 22 22 13 17 19 17 161 18 20
11 22 19 25 15 15 17 18 5 15 14 13 18 15 17 15
20 17 181 16 14 23 17 16 10 12 16 21 30 16 13
22 14 15 16 17 14 16 18 201 14 20 16 19 25 14
15 24 22 19 15 17 22 10 20 13 10 15 14 22 221
17 12 16 19 20 17 15 21 14 13 21 11 19 9 21 22
16 13 13 12 241 14 13 18 8 14 18 10 16 10 12
21 18 15 17 16 8 19 17 11 18 261 23 17 20 16
12 20 11 16 22 17 16 13 22 20 15 15 20 17 22 14
281 18 23 18 20 20 16 19 16 15 19 18 17 14 22
15 24 17 15 17 22
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301 18 22 10 19 24 21 16 14 11 14 20 15 21 11
17 16 20 19 13 14 321 17 17 19 15 17 13 18 23
16 12 25 13 13 21 19 16 20 27 19 18 341 18 24
15 23 13 13 14 15 23 13 19 15 11 19 17 12 15 15
17 14 361 18 20 17 13 16 14 13 20 18 15 18 16
17 20 14 19 21 12 13 17 381 22 17 19 16 14 18
16 18 12 16 13 15 16 9 15 16 18 22 14 16 401
14 17 12 16 21 16 21 13 14 19 18 18 16 19 17 17
17 13 17 11 421 16 16 13 10 26 12 20 17 11 19
18 12 15 14 14 20 15 15 15 11 441 18 23 20 23
13 12 18 22 12 16 13 21 22 14 18 21 17 12 19 16
461 17 18 15 22 22 20 15 16 13 12 19 22 16 20
19 19 16 8 15 12 481 29 26 19 16 20 15 11 22
15 20 21 14 16 13 17 15 10 13 17 12 501 18 20
17 14 13 19 23 11 27 19 17 16 17 20 21 15 20 20
21 19 521 21 16 13 21 16 19 13 9 10 20 12 18
14 13 18 19 22 19 21 18 541 6 17 17 19 19 22
23 18 13 12 17 16 21 16 18 21 19 13 22 19 561
20 17 18 15 17 15 15 10 18 13 23 17 14 23 22 10
18 11 11 18 581 16 17 14 13 9 12 14 14 21 23
24 19 12 15 17 18 11 14 19 19 601 19 16 17 13
13 15 17 18 17 13 9 19 18 22 17 13 14 22 13 23
621 23 19 19 16 24 14 17 18 17 13 16 12 7 15
17 16 18 22 19 15 641 16 18 18 13 20 18 12 6
15 11 16 19 12 13 11 17 11 15 11 19
6
661 17 16 16 21 12 18 20 19 16 14 18 17 16 14
11 17 17 16 17 17 681 17 18 16 18 12 18 18 20
19 13 12 16 14 13 13 6 15 12 19 14 701 20 17
16 14 21 19 15 26 17 20 12 24 13 11 19 21 18 13
9 16 721 9 16 17 16 15 12 11 21 21 13 19 13
13 16 11 17 15 19 22 19 741 11 13 14 16 20 15
16 12 18 14 12 14 21 12 23 21 19 10 24 17 761
17 19 19 15 18 12 14 14 14 20 12 20 12 21 19 20
21 20 17 18 781 15 12 16 23 16 16 19 15 12 14
21 25 12 19 20 22 17 16 21 20 801 23 24 17 20
17 19 14 22 20 25 10 12 15 16 7 14 14 18 22 10
821 15 22 23 18 12 10 14 18 15 15 18 10 21 11
20 15 20 10 13 16 841 16 17 22 19 19 16 8 20
17 13 21 16 25 16 13 17 14 17 19 21 861 17 19
14 22 20 18 14 19 17 23 20 18 14 11 16 18 26 24
24 18 881 21 16 23 20 14 16 15 13 14 11 12 13
14 16 18 17 16 17 13 20 901 22 8 17 17 16 16
14 22 17 18 18 21 15 11 20 21 18 15 19 21 921
16 22 14 12 16 20 16 21 11 13 19 14 23 12 12 17
14 15 26 17 941 18 14 21 17 14 24 21 12 21 13
20 22 11 20 10 16 16 15 19 13 961 16 15 16 17
9 14 11 12 19 17 16 15 21 14 15 14 15 17 15 16
981 19 11 15 17 17 17 11 18 21 14 15 17 18 16
11 22 19 16 14 15
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It makes sense now to look at properties of these
1000 simulations which have been placed in the
vector xsim.
gt mean(xsim) 1 16.624 gt median(xsim) 1 17 gt
sd(xsim) 1 3.778479 gt var(xsim) 1 14.2769 gt
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Now compare the actual values from the
simulations, with the theoretical values from the
probability distribution. SIMULATION
THEORETICAL MEAN 16.624 16.66667 VARIANCE 14.276
9 13.88889
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A full summary of the results of the simulation
is given with
gt table(xsim) xsim 5 6 7 8 9 10 11
12 13 14 15 16 17 1 3 2 7 10 21
40 57 72 80 82 118 118 18 19 20 21 22
23 24 25 26 27 28 29 30 85 83 61 55
46 25 14 9 6 2 1 1 1 gt
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A Histogram can also be plotted of this
gt hist(xsim)
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Notice that a BARPLOT of xsim does NOT produce a
useful graph!
gt barplot(xsim)
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A barplot of the TABLE of xsim does work,though.
gt barplot(table(xsim))
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Poisson Distribution
The Poisson distribution is used to model the
number of events occurring within a given time
interval. The formula for the Poisson
probability density (mass) function is ? is
the shape parameter which indicates the average
number of events in the given time interval.
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Some events are rather rare - they don't happen
that often. For instance, car accidents are the
exception rather than the rule. Still, over a
period of time, we can say something about the
nature of rare events. An example is the
improvement of traffic safety, where the
government wants to know whether seat belts
reduce the number of death in car accidents.
Here, the Poisson distribution can be a useful
tool to answer questions about benefits of seat
belt use.
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Other phenomena that often follow a Poisson
distribution are death of infants, the number of
misprints in a book, the number of customers
arriving, and the number of activations of a
Geiger counter.
The distribution was derived by the French
mathematician Siméon Poisson in 1837, and the
first application was the description of the
number of deaths by horse kicking in the Prussian
army.
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Example Arrivals at a bus-stop follow a Poisson
distribution with an average of 4.5 every quarter
of an hour. Obtain a barplot of the distribution
(assume a maximum of 20 arrivals in a quarter of
an hour) and calculate the probability of fewer
than 3 arrivals in a quarter of an hour.
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The probabilities of 0 up to 2 arrivals can be
calculated directly from the formula
with ? 4.5
So p(0) 0.01111
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Similarly p(1)0.04999 and p(2)0.11248 So the
probability of fewer than 3 arrivals is 0.01111
0.04999 0.11248 0.17358
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R Code
As with the Binomial distribution, the
codes dpois and ppois will do the
calculations for you.
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gt xdpois(020,4.5) gt x 1 1.110900e-02
4.999048e-02 1.124786e-01 1.687179e-01
1.898076e-01 6 1.708269e-01 1.281201e-01
8.236295e-02 4.632916e-02 2.316458e-02 11
1.042406e-02 4.264389e-03 1.599146e-03
5.535504e-04 1.779269e-04 16 5.337808e-05
1.501258e-05 3.973919e-06 9.934798e-07
2.352979e-07 21 5.294202e-08 gt
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gt barplot(x,names020)
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Now check that ppois gives the same answer (ppois
is a cumulative distribution).
gt ppois(2,4.5) 1 0.1735781 gt
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Consider a collection of graphs for different
values of ?
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?3
28
?4
29
?5
30
?6
31
?10
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In the last case, the probability of 20 arrivals
is no longer negligible, so values up to, say, 30
would have to be considered.
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Properties of Poisson
The mean and variance are both equal to ?. The
sum of independent Poisson variables is a further
Poisson variable with mean equal to the sum of
the individual means. As well as cropping up in
the situations already mentioned, the Poisson
distribution provides an approximation for the
Binomial distribution.
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Approximation If n is large and p is small,
then the Binomial distribution with parameters n
and p, ( B(np) ), is well approximated by the
Poisson distribution with parameter np, i.e. by
the Poisson distribution with the same mean
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Example Binomial situation, n 100,
p0.075 Calculate the probability of fewer than
10 successes.
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gt pbinom(9,100,0.075) 1 0.7832687 gt
This would have been very tricky with manual
calculation as the factorials are very large and
the probabilities very small
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The Poisson approximation to the Binomial states
that ? will be equal to np, i.e. 100 x 0.075 so
?7.5
gt ppois(9,7.5) 1 0.7764076 gt
So it is correct to 2 decimal places. Manually,
this would have been much simpler to do than the
Binomial.