Transcript Poisson Distribution

Report Writing
A report should be self-explanatory.
It should be capable of being read
and understood without reference
to the original project description.
Thus, for each question, it should
contain all of the following:
(a)a statement of the problem;
(b) a full and careful description of how it is
investigated;
(c) All relevant results, including graphical and
numerical analyses; variables should be
carefully defined, and figures and tables should
be properly labelled, described and referenced;
(d) relevant analysis, discussion, and conclusions.
It should be written in the third person.
NOT: I think the Central Limit Theorem is
true for this example because I see that the
graph is normal.
INSTEAD: It can be clearly seen that the
graph displays a normal distribution
confirming that the Central Limit Theorem
holds.
The Central Limit
Theorem
Let X1, X2………. Xn be independent
identically distributed random variables with
mean µ and variance σ 2.
Let S = X1,+ X2+ ………. +Xn
Then elementary probability theory tells us
that E(S) = nµ and var(S) = nσ 2 .
The Central Limit Theorem (CLT) further
states that, provided n is not too small, S has
an approximately normal distribution with the
above mean nµ, and variance nσ 2.
In other words,
S
approx
~ N(nµ, nσ 2)
The approximation improves as n increases.
We will use R to demonstrate the CLT.
Let X1,X2……X6 come from the
Uniform distribution, U(0,1)
1
0
1
For any uniform distribution on [A,B],
µ is equal to A  B
2
2
(
B

A
)
and variance, σ2, is equal to
12
So for our distribution, µ= 1/2 and
σ2 = 1/12
The Central Limit Theorem therefore
states that S should have an
approximately normal distribution with
mean nµ (i.e. 6 x 0.5 = 3)
and var nσ2 (i.e. 6 x 1/12 = 0.5)
This gives standard deviation 0.7071
In other words,
S approx ~ N(3, 0.70712)
Generate 10 000 results in each of six
vectors for the uniform distribution on
[0,1] in R.
> x1=runif(10000)
> x2=runif(10000)
> x3=runif(10000)
> x4=runif(10000)
> x5=runif(10000)
> x6=runif(10000)
>
Let S = X1,+ X2+ ………. +X6
> s=x1+x2+x3+x4+x5+x6
> hist(s,nclass=20)
>
Consider the mean and standard deviation
of S
> mean(s)
[1] 3.002503
> sd(s)
[1] 0.7070773
>
This agrees with our earlier calculations
A method of examining whether the distribution
is approximately normal is by producing a
normal Q-Q plot.
This is a plot of the sorted values of the vector
S (the “data”) against what is in effect a
idealised sample of the same size from the
N(0,1) distribution.
If the CLT holds good, i.e. if S is approximately
normal, then the plot should show an
approximate straight line with intercept equal
to the mean of S (here 3) and slope equal to
the standard deviation of S (here 0.707).
> qqnorm(s)
>
> qqnorm(s)
>
> qqnorm(s)
>
4.4 – 1.8
4
= 0.7 to 1 DP
From these plots it seems that
agreement with the normal distribution
is very good, despite the fact that we
have only taken n = 6, i.e. the
convergence is very rapid!