Probability Distributions
 A probability function is a function which assigns probabilities
to the values of a random variable.
 Individual probability values may be denoted by the symbol
P(X=x), in the discrete case, which indicates that the random
variable can have various specific values.
 All the probabilities must be between 0 and 1;
0≤ P(X=x)≤ 1.
 The sum of the probabilities of the outcomes must be 1.
∑ P(X=x)=1
It may also be denoted by the symbol f(x), in the continuous,
which indicates that a mathematical function is involved.
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Poisson
Continuous
Probability
Distributions
Normal
Binomial Distribution
An experiment in which satisfied the following characteristic is
called a binomial experiment:
1. The random experiment consists of n identical trials.
2. Each trial can result in one of two outcomes, which we denote
by success, S or failure, F.
3. The trials are independent.
4. The probability of success is constant from trial to trial, we
denote the probability of success by p and the probability of
failure is equal to (1 - p) = q.
Examples:
1. No. of getting a head in tossing a coin 10 times.
2. No. of getting a six in tossing 7 dice.
3. A firm bidding for contracts will either get a contract or not
A binomial experiment consist of n identical trial with probability
of success, p in each trial. The probability of x success in n trials
is given by
P( X  x)  nCx p x q n  x ;
x  0,1, 2....n
The Mean and Variance of X if X ~ B(n,p) are
Mean
Variance
  E ( X )  np
:
:
Std Deviation :
 2  V ( X )  np(1  p)  npq
  npq
where n is the total number of trials, p is the probability of
success and q is the probability of failure.
Example
Given that X
a) P ( X  2)
b) P ( X  3)
c) P ( X  4)
d) P (2  X  5)
e) E( X )
f) Var( X )
b(12, 0.4), find
Solutions:
a) P ( X  2)  12C2 (0.4) 2 (0.6)10  0.0639
b) P ( X  3)  12C3 (0.4)3 (0.6)9  0.1419
c) P ( X  4)  12C4 (0.4) 4 (0.6)8  0.2128
d) P(2  X  5)  P( X  2)  P( X  3)  P( X  4)
 0.0639  0.1419  0.2128  0.4185
e) E ( X )  np  12(0.4)=4.8
f) Var ( X )  npq  12(0.4)(0.6)= 2.88
Cumulative Binomial distribution
 When the sample is relatively large, tables of Binomial are
often used. Since the probabilities provided in the tables are in
the cumulative form P  X  k  the following guidelines can
be used:
Example:
Example:
The Poisson Distribution
 Poisson distribution is the probability distribution of the
number of successes in a given space*.
*space can be dimensions, place or time or combination of them
 Examples:
No. of cars passing a toll booth in one hour.
2. No. defects in a square meter of fabric
3. No. of network error experienced in a day.
1.
A random variable X has a Poisson distribution and it is referred to
as a Poisson random variable if and only if its probability
distribution is given by
e   x
P( X  x) 
for x  0,1, 2,3,...
x!
A random variable X having a Poisson distribution can also be
written as
X
Po ( )
with E ( X )   and Var ( X )  
Example :
Given that X
a) P( X  0);
Po (4.8) , find
b) P( X  9);
c) P( X  1)
Solution:
e 4.8 4.80
a) P( X  0)  P( X  0) 
 0.0082
0!
e 4.8 4.89
b) P( X  9) 
 0.0307
9!
or using cumulative Possion distribution table
P( X  9)  P( X  9)  P( X  8)  0.9749  0.9442  0.0307
c) P( X  1)  1  P( X  0)  1  0.0082  0.9918
Example :
Poisson Approximation of Binomial
Probabilities
Example:
The Normal Distribution
 Numerous continuous variables have distribution closely
resemble the normal distribution.
 The normal distribution can be used to approximate various
discrete prob. dist.
A continuous random variable X is said to have a normal distribution
with parameters  and  2 , where       and  2  0,
if the pdf of X is
1
e
f ( x) 
 2
1  x 
 

2  
2
  x  
X is denoted by X ~ N (  ,  2 ) with E  X    and V  X    2
The Normal Distribution
 ‘Bell Shaped’
 Symmetrical
 Mean, Median and Mode
f(X)
are Equal
Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
σ
X
μ
Mean = Median = Mode
The random variable has an
infinite theoretical range:
+  to  
Many Normal Distributions
By varying the parameters μ and σ, we obtain different
normal distributions
The Standard Normal Distribution

Any normal distribution (with any mean and standard deviation
combination) can be transformed into the standard normal
distribution (Z)
X 

Need to transform X units into Z units using Z 

The standardized normal distribution (Z) has a mean of 0,   0
2
and a standard deviation of 1,   1

Z is denoted by Z ~ N (0,1)

Patterns for Finding Areas under the Standard Normal Curve
Example
Exercises
Determine the probability or area for the portions of the Normal
distribution described.
a) P (0  Z  0.45)
b) P (2.02  Z  0)
c) P ( Z  0.87)
d) P (2.1  Z  3.11)
e) P (1.5  Z  2.55)
solutions
a) P(0  Z  0.45)  P(Z  0.45)  P(Z  0)
 0.67364  0.5
= 0.1736
b) P(2.02  Z  0)  P(0  Z  2.02)
= P(Z  2.02)  P(Z  0)
 0.97831  0.5
= 0.47831
c) P(Z  0.87)  0.80785
d)
e)
Example
Exercises
Determine Z such that
a) P( Z  Z )  0.25
b) P( Z  Z )  0.36
c) P( Z  Z )  0.983
d) P ( Z  Z )  0.89
solutions
a) P( Z  Z )  0.25;
Z  0.675
b) P( Z  Z )  0.36;
Z  0.355
c) P( Z  Z )  0.983;
Z  2.12
d) P( Z  Z )  0.89;
Z  1.225
Example
Suppose X is a normal distribution N(25,25). Find
a) P (24  X  35)
b) P ( X  20)
solutions
35  25 
 24  25
a) P(24  X  35)  P 
Z
  P(0.2  Z  2)
5 
 5
 P( Z  2)  P( Z  0.2)  P( Z  2)  P ( Z  0.2)
=0.97725  0.42074  0.55651
20  25 

b) P( X  20)  P  Z 

5 

 P( Z  1)
 P( Z  1)  0.84134
Exercises
Example
Normal Approximation of the Binomial Distribution
 When the number of observations or trials n in a binomial
experiment is relatively large, the normal probability
distribution can be used to approximate binomial
probabilities. A convenient rule is that such approximation is
acceptable when n  30, and both np  5 and nq  5.
Given a random variable X
and nq  5, then X
b(n, p), if n  30 and both np  5
N (np, npq)
with   np and  2  npq
Continuous Correction Factor
The continuous correction factor needs to be made when a
continuous curve is being used to approximate discrete
probability distributions. 0.5 is added or subtracted as a
continuous correction factor according to the form of the
probability statement as follows:
c .c
a) P ( X  x) 
 P ( x  0.5  X  x  0.5)
c .c
b) P( X  x) 
 P ( X  x  0.5)
c .c
c) P ( X  x) 
 P ( X  x  0.5)
c .c
d) P ( X  x) 
 P ( X  x  0.5)
c .c
e) P ( X  x) 
 P ( X  x  0.5)
c.c  continuous correction factor
Example
In a certain country, 45% of registered voters are male. If 300
registered voters from that country are selected at random,
find the probability that at least 155 are males.
Solutions:
X is the number of male voters.
X
b(300, 0.45)
c .c
P( X  155) 
 P( X  155  0.5)  P( X  154.5)
np  300(0.45)  135  5; nq  300(0.55)  165  5
Therefore, X
N (135, 74.25)
154.5  135 

PZ 
  P( Z  2.26)  0.01191
74.25 

Exercises
Suppose that 5% of the population over 70 years old has
disease A. Suppose a random sample of 9600 people over 70
is taken. What is the probability that fewer than 500 of them
have disease A?
Normal Approximation of the Poisson Distribution
 of a Poisson distribution is relatively
large, the normal probability distribution can be used to
approximate Poisson probabilities. A convenient rule is that
such approximation is acceptable when   10.
 When the mean
Given a random variable X
then X
N ( ,  )
Po ( ), if   10,
Example
A grocery store has an ATM machine inside. An average of 5
customers per hour comes to use the machine. What is the
probability that more than 30 customers come to use the
machine between 8.00 am and 5.00 pm?
Solutions:
X is the number of customers use the ATM machine in 9 hours.
X
Po (45);   45  10
X
N (45, 45)
c .c
P( X  30) 
 P( X  30  0.5)  P( X  30.5)
30.5  45 

PZ 
  P( Z  2.16)  0.98461
45 
