Download Chapter 1 - Probability Distribution (Normal)

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 1.30
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 1.11
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 1.32
Exercise 1.12
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 1.33
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 

Example 1.34
An assessment test is used to measure a person’s readiness for
college. The mathematics scores in the test are scaled to have a
normal distribution with mean 500 and standard deviation 100.
i.
What is the probability that the people taking the test will
score below 350?
ii.
Remedial assistance will be given to students in the bottom
10%. What is the maximum score of this group of students?
Solution:
i. P( X  350) 
ii. P( Z 
max  500
)
100
Exercise 1.13
1. Suppose X is a normal distribution, N(70,4). Find
a) P (67  X  75)
b) P ( X  74)
2. Suppose the test scores of 600 students are normally distributed with a
mean of 76 and standard deviation of 8. The number of scoring is from
70 to 82 is:
Answer : 1. a) 0.927 b) 0.0228 2. 328 students
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 1.35
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 1.14
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
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 1.36
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 1.15
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
Exercise 1.16
Reported that the mean weekly income of a shift foreman in the
glass industry is normally distributed with a mean of $1000 and
standard deviation of $100. What is the probability of selecting
a shift foreman in the glass industry whose income is
a)
b)
c)
Between $1000 and $1100.
Between $790 and $1000.
Between $840 and $1200.
Answer : a) 0.3413, b) 0.4821, c) 0.9224