Transcript Chapter 3 - Portal UniMAP

ROHANA
ROHANABINTI
BINTIABDUL
ABDULHAMID
HAMID
INSTITUT
INSTITUTE FOR
E FORENGINEERING
ENGINEERINGMATHEMATICS
MATHEMATICS(IMK)
(IMK)
UNIVERSITIMALAYSIA
MALAYSIAPERLIS
PERLIS
UNIVERSITI
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CHAPTER 3
PROBABILITY DISTRIBUTION
PROBABILITY DISTRIBUTION
• 3.1
Introduction
Binomial
distribution
• 3.2
• 3.3
Poisson
distribution
Normal
distribution
• 3.4
3.1 INTRODUCTION

A probability distribution is obtained when
probability values are assigned to all possible
numerical 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.

It may also be denoted by the symbol
f(x), in the continuous, which indicates
that a mathematical function is involved.

The sum of the probabilities for all the
possible numerical events must equal 1.0.
3.2 THE BINOMIAL DISTRIBUTION
Definition 3.1 :
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.
Definition 3.2 :
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)  Cx p q
n

x = 0, 1, 2, ......, n
x
n x
Definition 3.3 :The Mean and Variance of X
If X ~ B(n,p), then
Mean
Variance
where
 n is the total number of trials,
 p is the probability of success and
 q is the probability of failure.
Standard
deviation
EXAMPLE 3.1
Given that X ~ b(12, 0.4), find
a) P ( X  2)
b) P ( X  3)
c) P ( X  4)
d) P (2  X  5)
e) E( X )
f) Var( X )
SOLUTIONS
a) P ( X  2) 
12
C2 (0.4) 2 (0.6)10
 0.0639
b) P ( X  3) 
12
C3 (0.4)3 (0.6)9
 0.1419
c) P ( X  4) 
12
C4 (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
3.3 The Poisson Distribution
Definition 3.4
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!


λ (Greek lambda) is the long run mean
number of events for the specific time or space
dimension of interest.
A random variable X having
distribution can also be written as
X ~ Po ( )
with E ( X )   and Var ( X )  
a
Poisson
EXAMPLE 3.2
Given that X ~ Po (4.8), find
a) P( X  0)
b) P( X  9)
c) P( X  1)
SOLUTIONS
a) P ( X  0) 
b) P( X  9) 
e
4.8
0
4.8
9
e
4.8
 0.0082
0!
4.8
 0.0307
9!
c) 1  P ( X  0)  1  0.0082
= 0.9918
EXAMPLE 3.3
Suppose that the number of errors in a piece of
software has a Poisson distribution with
parameter   3 . Find
a) the probability that a piece of software has no
errors.
b) the probability that there are three or more
errors in piece of
software .
c) the mean and variance in the number of errors.
SOLUTIONS
e 3  30
a) P( X  0) 
0!
 e3  0.050
b)P( X  3)  1  P( X  0)  P( X  1)  P( X  2)
e 3  30 e 3  31 e 3  32
 1


0!
1!
2!
3 9
3  1
 1 e    
1 1 1 
 1  0.423  0.577
3.4 The Normal Distribution
Definition 3.5
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
f ( x) 
e
 2
1  x 
 

2  
2
  x  
If X ~ N (  ,  2 ) then E  X    and V  X    2
The Standard Normal Distribution

The normal distribution with parameters
  0 and  2  1 is called a standard normal
distribution.

A random variable that has a standard normal
distribution is called a standard normal random
variable and will be denoted by
.
Z ~ N (0,1)
Standardizing A Normal Distribution
If X is a normal random variable with E ( X )   and V ( X )   2 ,
the random variable
X 
Z

is a normal random variable with E ( Z )  0 and V ( Z )  1.
That is Z is a standard normal random variable.
EXAMPLE 3.4
Determine the probability or area for the portions
of the Normal distribution described. (using the
table)
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 )  0.5  P( Z  0.45 )
 0.5  0.3264  0.1736
b) P(2.02  Z  0)  0.5  P( Z  2.02 )
 0.5  0.0217  0.4783
c) P( Z  0.87 )  1  P( Z  0.87 )
 1  0.1922  0.8078
d ) P(2.1  Z  3.11)  1  P( Z  2.1)  P( Z  3.11)
 1  0.0179  0.009  0.9731
e) P(1.5  Z  2.55)  P( Z  1.5)  P( Z  2.55)
 0.0668  0.0054  0.0614
EXAMPLE 3.5
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  0.675)  0.25
b) P( Z  0.355)  0.36
c) P( Z  2.12)  0.983
d) P( Z  1.225)  0.89
EXAMPLE 3.6
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

5
5


 P(0.2  Z  2)
= 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
3.4.1 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.
Definiton 3.6
Given a random variable X ~ b(n, p), if n  30 and both np  5
and nq  5, then X ~ N (np, npq)
X  np
with Z 
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
c .c
follows:
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 3.7
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

154.5  300(0.45) 
154.5  135 

PZ 
  P  Z 


300(0.45)(0.55) 
74.25 


 P ( Z  2.26)
 0.01191
3.4.1 Normal Approximation of the
Poisson Distribution



When the mean  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.
Definition 3.7
Given a random variable X ~ Po ( ), if   10, then X ~ N ( ,  )
with Z 
X 

Example 3.8
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 come to 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)
45 

 0.98461