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?
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
1
Binomial distribution with n=100 and a= 6
This can be done on R using the command
rbinom.
For example, let’s simulate 1000 students.
> xsim=rbinom(1000,100,1/6)
> 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
[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
[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
It makes sense now to look at properties of
these 1000 simulations which have been
placed in the vector “xsim”.
> mean(xsim)
[1] 16.624
> median(xsim)
[1] 17
> sd(xsim)
[1] 3.778479
> var(xsim)
[1] 14.2769
>
Now compare the actual values from the
simulations, with the theoretical values from
the probability distribution.
SIMULATION THEORETICAL
MEAN
16.624
16.66667
VARIANCE
14.2769
13.88889
A full summary of the results of the simulation
is given with:
> 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
>
A Histogram can also be plotted of this:
> hist(xsim)
Notice that a BARPLOT of xsim does
NOT produce a useful graph!
> barplot(xsim)
A barplot of the TABLE of xsim does
work,though.
> barplot(table(xsim))
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
 x
e 
p( x ) 
x!
 is the shape parameter which indicates the
average number of events in the given time
interval.
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.
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.
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.
The probabilities of 0 up to 2 arrivals can
be calculated directly from the formula
e 
p( x ) 
x!

p(0) 
e
4.5
x
with  =4.5
0
4.5
0!
So p(0) = 0.01111
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
R Code
As with the Binomial distribution, the
codes
dpois
and
ppois
will do the calculations for you.
> x=dpois(0:20,4.5)
>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
>
> barplot(x,names=0:20)
Now check that ppois gives the same answer
(ppois is a cumulative distribution).
> ppois(2,4.5)
[1] 0.1735781
>
Consider a collection of graphs for
different values of 
=3
=4
=5
=6
=10
In the last case, the probability of 20
arrivals is no longer negligible, so
values up to, say, 30 would have to be
considered.
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.
Approximation:
If n is large and p is small, then the
Binomial distribution with parameters n and
p, ( B(n;p) ), is well approximated by the
Poisson distribution with parameter np, i.e.
by the Poisson distribution with the same
mean
Example
Binomial situation, n= 100, p=0.075
Calculate the probability of fewer than
10 successes.
> pbinom(9,100,0.075)
[1] 0.7832687
>
This would have been very tricky with
manual calculation as the factorials
are very large and the probabilities
very small
The Poisson approximation to the
Binomial states that  will be equal
to np, i.e. 100 x 0.075
so =7.5
> ppois(9,7.5)
[1] 0.7764076
>
So it is correct to 2 decimal places.
Manually, this would have been much
simpler to do than the Binomial.