Chapter 6 – Continuous Probability Distribution
Introduction
 A probability distribution is obtained when probability values
are assigned to all possible numerical values of a random
variable.
 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.
6.1 Normal Probability Distribution
Definition 6.1
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
6.2 Standard Normal Probability 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 6.1
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 6.2
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  0.6745)  0.25
b) P( Z  0.3585)  0.36
c) P ( Z  2.1201)  0.983
d) P( Z  1.2265)  0.89
Example 6.3
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
6.3 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.
 Definition 6.2
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 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 6.4
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
6.4 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 6.3
Given a random variable X ~ Po ( ), if   10, then X ~ N ( ,  )
with Z 
X 

Example 6.5
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