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```GG 313 Lecture 6
Probability Distributions
When we sample many phenomena, we find that the
probability of occurrence of an event will be distributed in
a way that is easily described by one of several wellknown functions called “probability distributions”. We
will discuss these functions and give examples of them.
UNIFORM DISTRIBUTION
The uniform distribution is simple, it’s the probability
distribution found when all probabilities are equal. For
example: Consider the probability of throwing an “x” using
a 6-sided die: P(x), x=1,2,3,4,5,6. If the die is “fair”, then
P(x)=1/6, x=1,2,3,4,5,6.
This is a discrete probability distribution, since x can only
have integer or fixed values.
P(x) is always ≥ 0 and always ≤ 1.
If we add up the probabilities for all x, the sum = 1:
n
 P(x )  1
i
i1
In most cases, the probability distribution is not uniform,
and some events are more likely than others.
For example, we may want to know the probability of
hitting a target x times in n tries. We can’t get the solution
unless we know the probability of hitting the target in one
try: P(hit)=p. Once we know P(hit), we can calculate the
probability distribution:
P(x)=pxq(n-x),
where q=1-p. This is the number of combinations of n
things taken x at a time, nCx.
Recall that this is the probability where the ORDER of the
x hits matters, which is not what we want. We want the
number of permutations as defined earlier:
n x nx n  x
P(x)   p q   p (1 p) nx , x  1,2,
x 
x 
n
Recall that:

n 
  binomial coefficients
x 
n!

x!(n  x)!
This is known as the binomial probability distribution,
used to predict the probability of success in x events out
of n tries.
Using our example above, what is the probability of
hitting a target x times in n tries if p=0.1 and n=10?
10 x
10!
10x
P(x)   0.1  0.9

0.1x  0.910x
x!(10  x)!
 x 
We can write an easy m-file for this:
% binomial distribution
n=10
p=0.1
xx=[0:n];
for ii=0:n;
px(ii+1)=(factorial(n)/(factorial(xx(ii+1))*factorial(nxx(ii+1))))*p^xx(ii+1)*(1-p)^(n-xx(ii+1));
end
plot(xx,px)
sum(px) % check to be sure sure that the sum =1
P=0.1
The probability of 1 hit is 0.38 with p=0.1
What if we change the probability of hitting the target
in any one shot to 0.3. What is the most likely number
of hits in 10 shots?
p=0.3
Continuous Probability Distributions
Continuous populations, such as the temperature of the
atmosphere, depth of the ocean, concentration of
pollutants, etc., can take on any value with in their range.
We may only sample them at particular values, but the
underlying distribution is continuous.
Rather than the SUM of the distribution equaling 1.0, for
continuous distributions the integral of the distribution
(the area under the curve) over all possible values must
equal 1.

 P(x)dx 1

P(x) is called a probability density function, or PDF.

Because they are continuous, the probability of any
particular value being observed is zero. We discuss instead
the probability of a value being between two limits, a- and
a+:
P(a  ) 

a 
a
P(x)dx
As 0, the probability also approaches zero.
also define the cumulative probability distribution
We
giving the probability that an observation will have a value
less than or equal to a. This distribution is bounded by
0≤p(x) ≤ 1.
Pc (a) 
As a, Pc(a) 1.

a

P(x)dx
The normal distribution
While any continuous function with an area under the
curve of 1 can be a probability distribution, but in reality,
some functions are far more common than others. The
Normal Distribution, or Gaussian Distribution, is the
most common and most valued. This is the classic “bellshaped curve”. It’s distribution is defined by:
1
p(x) 
e
 2
2
1 x  
 

2   
Where µ and  are the mean and standard deviation
defined earlier.

We can make a Matlab m-file to generate the normal
distribution:
% Normal distribution
xsigma=1.;
xmean=7;
xx=[0:.1:15];
xx=(xx-xmean)/xsigma;
px=(1/(xsigma*sqrt(2*pi)))*exp(-0.5*xx.^2);
xmax=max(px)
xsum=sum(px)/10.
plot(xx,px)
Or, make an Excel spreadsheet plot:
0.4
sigma=1, mean=7
0.35
sigma=1,5, mean=8
sigma=2.25, mean=9
0.3
sigma=3.375, mean=10
sigma=5.1, mean=11
0.25
sigma=7.6, mean=12
0.2
sigma=11.4, mean=13
0.15
0.1
0.05
0
0
2
4
6
8
10
12
14
16
18
20
Note that smaller values of sigma are sharper (have
smaller kurtosis).
We can define a new variable that will normalize the
distribution to =1 and =0:
zi 
xi  

And the defining equation reduces to:
1
p(x) 
e
 2

2
1 x  
 

2   
,
1
p(z) 
e
2
z2

2
The values on the x axis are now equivalent to
standard deviations: x=±1 = ±1 standard deviation,
etc.
This distribution is very handy. We expect 68.27% of our
results to be within 1 standard deviation of the mean,
95.45% to be within 2 standard deviations, and 99.73% to
be within 3 standard deviations. This is why we can feel
reasonably confident about eliminating points that are more
than 3 standard deviations away from the mean.
We also define the cumulative normal distribution:
P(z  a) 
let :

a

p(z)dz   p(z)dz 

and dz  2du,
then
z2
u 
2
2
P(z  a)  1/2 
0

1


a
2
0
e
u2

2
a
0
1
p(z)dz 1/2 
2

a 
0
e
z2
2
dz

 a 
 a 
du  1/2 
erf   1/21 erf  
 2 
 2 
 2

1

Erf(x) is called the error function.
For any value z,

 z 
Pc (z)  1/21 erf  
 2 

1
Cumulative
Normal
distribution
0.75
Pc(z)
0.5
0.25
0
-4
-2
0
z
2
4

We easily see that the probability of a result being
between a and b is:
  b 
 a 
Pc (a  z  b)  Pc (b)  Pc (a)  1/2erf   erf  
 2 
  2 
Example: Estimates of the strength of olivine yield a
normal distribution given by µ=1.0*1011 Nm and =1.0
*1010 Nm. What is the probability that a sample
estimate will be between 9.8*1010 Nm and 1.1*1011
Nm?
xi  
First convert to normal scores
zi 

And calculate with the formula above.
DO THIS NOW, either in Excel or Matlab -

The normal distribution is a good approximation to
the binomial distribution for large n (actually, when
np and (1-p)n >5). The mean and standard deviation
of the binomial distribution become:
  np and   np(1 p)
So that:


Pb x  
1
2np1 p)
exp
1(xnp)2 


 2np(1 p ) 
I had a difficult time getting this to work in Excel
because the term -(x-np)2 is evaluated as (-(x-np))2
0.2
0.18
0.16
normal
binomial
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
10
12
14
N=20, p=0.5, µ=np=10, =np(1-p)=2.236
16
Poisson distribution
One way to think of a Poisson distribution is that it is like a
normal distribution that gets pushed close to zero but
can’t go through zero. For large means, they are virtually
the same.
The Poisson distribution is a good approximation to the
binomial distribution that works when the probability of a
single event is small but when n is large.
P(x) 
  np

xe 
x!
,
x  0,1,2,3,
n where
 Is the rate of occurrence. This is used to
evaluate the probability of rare events.
0.4
Poisson Distribution
0.35
lambda=2
lambda=4
lambda=8
0.3
lambda=12
lambda=16
0.25
lambda=20
lambda=1
0.2
0.15
0.1
0.05
0
0
2
4
6
8
10
12
14
16
18
-0.05
Note that the Poisson distribution approaches
the normal distribution for large .
20
Example: The number of floods in a 50-year period on a
particular river has been shown to follow a Poisson
distribution with =2.2. That is the most likely number of
floods in a 50 year period is a bit larger than 2.
What is the probability of having at least 1 flood in the
next 50 years?
The probability of having NO floods (x=0) is e-2.2, or 0.11.
The probability of having at least 1 flood is (1-P(0))=0.89.
The Exponential Distribution
x
P(x)  e
And:
Pc (x)  1 e


x
% exponential distribution
lambda=.5 % lambda=rate of occurrence
xx=[0:1:20];
px=lambda*exp(-1*xx*lambda);
plot(xx,px)
hold
cum=1-exp(-1*xx*lambda)
plot(xx,cum)
Pc(x)
Exponential Distribution
P(x)
Example: The height of Pacific seamounts has approximately
an exponential distribution with Pc(h)=1-e-h/340, where h is in
meters. Which predicts the probability of a height less than h
meters. What’s the probability of a seamount with a height
greater than 4 km?
Pc(4000)=1-e-4000/340 which is approximately 0.99999 so the
probability of a large seamount is 0.00001. (Which I don’t
believe….)
Log-Normal Distribution
In some situations, distributions are greatly skewed, a
situation seen in some situations, such as grain size
distributions and when errors are large and propagate as
products, rather than sums.
In such cases, taking the log of the distribution may result in
a normal distribution. The statistics of the normal distribution
can be obtained and exponentiated to obtain the actual
values of uncertainty.
```
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