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
PZ
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
PZ
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