Chapter 3 - Normal Distribution

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Transcript Chapter 3 - Normal Distribution

Normal Distribution
*Numerous
continuous variables have distribution closely
resemble the normal distribution.
*The
normal distribution can be used to approximate various
discrete probability distribution.
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
f ( x) 
1
e
 2
1  x 
 

2  
2
  x  
X is denoted by X ~ N (  ,  2 ) with E  X    and V  X    2
CHARACTERISTICS OF NORMAL
DISTRIBUTION
*‘Bell Shaped’
* Symmetrical
* Mean, Median and Mode
are Equal
f(X)
Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
The random variable has an infinite
theoretical range:
+  to  
σ
X
μ
Mean = Median = Mode
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)
 Need to transform X units into Z units using
Z
X 


The standardized normal distribution (Z) has a mean of ,
and a standard deviation of 1,  2  1

Z is denoted by Z ~ N (0,1)

Thus, its density function becomes
 0
Calculating Probabilities for a General Normal Random Variable
*Mostly, the probabilities involved x, a normal random variable
with mean,  and standard deviation, 
*Then, you have to standardized the interval of interest, writing it
in terms of z, the standard normal random variable.
*Once this is done, the probability of interest is the area that you
find using the standard normal probability distribution.
*Normal probability distribution, X ~ N ( , 2 )
*Need to transform x to z using Z  X  

*
Patterns for Finding Areas under the Standard Normal Curve
Example 3.1:
a)
Find the area under the standard normal curve of P(0  Z  1)
a)
Find the area under the standard normal curve of P(2.34  Z  0)
Exercise 3.1
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)
Answer : a) 0.1736, b) 0.4783, c) 0.8078, d) 0.9812, e) 0.0614
Example 3.2
(a ) P  Z  Z   0.16
(b) P  Z  Z   0.48
(c) P ( Z  Z )  0.80
(d ) P  Z  Z   0.95
Exercise 3.2
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
Answers:
a) P( Z  Z )  0.25;
Z  0.675
b) P( Z  Z )  0.36;
Z  0.3585
c) P( Z  Z )  0.983;
Z  2.12
d) P( Z  Z )  0.89;
Z  1.2265
Example 3.3
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
  P(0.2  Z  2)
5 
 5
20  25 

b) P( X  20)  P  Z 

5 

Exercise 3.3:
1.
A normal random variable x has mean, 10, and standard
deviation, 2. Find the probabilities below:
(a) P  X  13.5  (b) P  X  8.2  (c) P  9.4  X  10.6 
2.
Hupper Corporation produces many types of soft
drinks including Orange Cola. It has been observed
that the net amount of soda in such a can
has
a
normal distribution with a mean of 12 ounces and a
standard deviation of 0.015 ounce. What
Is
the
probability that a randomly
selected can of Orange
Cola contains between
11.97 and 11.99 ounces of
soda?
3. The random variable X is normally distributed. Given   54 and
P  X  80   0.02 . Find the value of 
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 b(n, p), if n  30 and both np  5
and nq  5, then X
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
How do calculate Binomial Probabilities Using the Normal
Approximation?
* Find the necessary values of n and p. Calculate   np and   npq
* Write the probability you need in terms of x.
* Correct the value of x with appropriate continuous correction factor
(ccf).
* Convert the necessary x-values to z-values using
x  0.5  np
z
npq
* Use Standard Normal Table to calculate the approximate probability.
Example 3.5
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)  0.5  0.4881  0.0119
74.25 

Exercise 3.5
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 less than 500 of them have disease
A?
Answer: 0.8186/0.8194
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 3.6
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?
Solution:
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)  0.5  0.4846  0.9846
45 

Exercise 3.6
The average number of accidental drowning in United States per
year is 3.0 per 100000 population. Find the probability that in a city
of population 400000 there will be less than 10 accidental drowning
per year.
Answer : 0.2358