Download Poisson Distribution

Normal Distribution
The shaded area is the probability of z > 1
The normal distribution is actually a family
of distributions, all with the same shape
and parameterised by mean , and
standard deviation .
It is usually defined by
a reference
member of the family which is used to
define other members.
This reference member has =0 and  =1.
Definition:
A random variable Z
Gaussian) distribution
standard deviation 1,
distribution function Ф(z)
is given by
1
2
z
has a normal (or
with mean 0 and
if and only if its
(defined by p(Z  z) )
t2
2
e
dt


we write Z ~ N(0, 1) and say that Z has a
standard normal distribution
Definition:
A random variable X has a normal (or
Gaussian) distribution with mean  and
standard deviation , if and only if
Z
X 

~ N (0,1)
we write X ~ N(, 2) and say that X has a
normal distribution
The normal distribution is symmetric
about its mean .
In particular, if Z ~ N(0, 1), then
p(Z ≤ -z) = p(Z ≥ z)
i.e. Ф(-z) +Ф(z) = 1 for all z
Whatever the values of  and , the
area between  - 2 and  + 2 is
always 0.95 (95%).
Similarly, Whatever the values of  and
, the area between  -  and  +  is
always 0.68 (68%).
Example
It has been suggested IQ scores follow a
normal distribution with mean 100 and
standard deviation 15. Find the probability
that any person chosen at random will have
(a) An IQ less than 70
(b) An IQ greater than 110
(c) An IQ between 70 and 110.
In R, The function dnorm gives
the density of the normal
distribution.
Generally more useful, though, is
pnorm,
which
gives
the
cumulative distribution function.
So in the IQ example, the probability of an
IQ less than 70 is:
> pnorm(70,100,15)
[1] 0.02275013
>
Approximately 0.0228
And the probability of an IQ less than 110 is:
> pnorm(110,100,15)
[1] 0.7475075
>
Thus, the probability of an IQ more than 110
is 1 - 0.7475075
> t=pnorm(110,100,15)
> 1-t
[1] 0.2524925
>
Approximately 0.2525
Finally, for the probability of an IQ between
70 and 110, carry out a subtraction.
> pnorm(110,100,15) - pnorm(70,100,15)
[1] 0.7247573
>
Approximately 0.7248
Alternatively,
p(70  X  110)
 70  100 X  100 110  100 
 p



15
15
 15

  (0.6667)   ( 2)
> pnorm(0.6667) - pnorm(-2)
[1] 0.724768
>
These are the converted variables in the
standardised normal (z) scales. The answer
is, of course, the same.
z = -2
z =0.6667
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)
>
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!
Application
Confidence Intervals for Mean
Suppose that the random variables
Y1,Y2, …………Yn model independent
observations from a distribution with mean
µ and variance σ2 .
n
Then
1
Y   Yi
n i 1
is the sample mean.
Now by the CLT
    
Y ~ N  , 


  n 


2
This is because µ is replaced by µ/n
and σ by σ /n (for means)
Recall from Statistics 2 that, if σ2 is
estimated by the sample variance, s2, an
approximate confidence interval for µ is
given by:
s
s 

y

z
,
y

z


n
n

_
Here y is the observed sample mean, and
z is proportional to the level of confidence
required.
So for 95% confidence an approximate
interval for µ is given by:
s
s 

y

2
,
y

2


n
n

2 is approximate - an accurate value can
be obtained from tables or by using the
qnorm function on R.
> qnorm(0.975)
[1] 1.959964
> qnorm(0.995)
[1] 2.575829
> qnorm(0.025)
[1] -1.959964
>
Thus in R, an approximate 95% confidence
interval for the mean µ is given by
> mean(y)+c(-1,1)*qnorm(0.975)*sqrt(var(y)/length(y))
where y is the vector of observations.
A more accurate confidence interval,
allowing for the fact that s2 is only an
estimate of σ2,is given by use of the
function t.test.