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Normal Distribution
*Numerous continuous variables have distribution closely resemble the normal distribution.
*The normal distribution can be used to approximate various discrete probability distribution.
2
2 2
1
2
2
A continuous random variable is said to have a normal distribution
with parameters and , where and 0,
if the pdf of is
1( )
2
is denoted by ~ ( , ) wit
x
X
X
f x e x
X X N
2h and E X V X
CHARACTERISTICS OF NORMAL DISTRIBUTION*‘Bell Shaped’* Symmetrical
* Mean, Median and Mode are Equal
Location is determined by the mean, μ
Spread is determined by the standard deviation, σ
The random variable has an infinite theoretical range: + to
Mean = Median = Mode
X
f(X)
μ
σ
By varying the parameters μ and σ, we obtain different normal distributions
Many 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
The standardized normal distribution (Z) has a mean of , and a standard deviation of 1,
Z is denoted by
Thus, its density function becomes
0 2 1
~ (0,1)Z N
XZ
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,
*Need to transform x to z using
2~ ( , )X N X
Z
Example 1.30
a) Find the area under the standard normal curve of
b)Find the area under the standard normal curve of
(0 1)P Z
( 2.34 0)P Z
Exercise 1.11
Determine the probability or area for the portions of the Normal distribution described.
Answer : a) 0.1736, b) 0.4783, c) 0.8078, d) 0.9812, e) 0.0614
a) (0 0.45)
b) ( 2.02 0)
c) ( 0.87)
d) ( 2.1 3.11)
e) (1.5 2.55)
P Z
P Z
P Z
P Z
P Z
Determine such that
a) ( ) 0.25
b) ( ) 0.36
c) ( ) 0.983
d) ( ) 0.89
Z
P Z Z
P Z Z
P Z Z
P Z Z
Exercise 1.12
a) ( ) 0.25; 0.675
b) ( ) 0.36; 0.3585
c) ( ) 0.983; 2.12
d) ( ) 0.89; 1.2265
P Z Z Z
P Z Z Z
P Z Z Z
P Z Z Z
Answers:
Suppose X is a normal distribution N(25,25). Find
a) (24 35)
b) ( 20)
P X
P X
Solutions
24 25 35 25a) (24 35) ( 0.2 2)
5 5
20 25b) ( 20)
5
P X P Z P Z
P X P Z
Example 1.33
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?
Exercise 1.13
1. Suppose X is a normal distribution, N(70,4). Find
a)
b)
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
(67 75)P X ( 74)P X
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
30, and both 5 and 5.n np nq
2
Given a random variable ( , ), if 30 and both 5
and 5, then ( , )
with and
X b n p n np
nq X N np npq
np 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:
.
.
.
.
.
a) ( ) ( 0.5 0.5)
b) ( ) ( 0.5)
c) ( ) ( 0.5)
d) ( ) ( 0.5)
e) ( ) ( 0.5)
. continuous correction factor
c c
c c
c c
c c
c c
P X x P x X x
P X x P X x
P X x P X x
P X x P X x
P X x P X x
c c
How do calculate Binomial Probabilities Using the Normal Approximation?
*Find the necessary values of n and p. Calculate
*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
*Use Standard Normal Table to calculate the approximate probability.
and np npq
0.5x npz
npq
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:
.
is the number of male voters.
(300,0.45)
( 155) ( 155 0.5) ( 154.5)
300(0.45) 135 5; 300(0.55) 165 5
Therefore, (135,74.25)
154.5 135( 2.26) 0.5 0.4881 0.0119
74.25
c c
X
X b
P X P X P X
np nq
X N
P Z P Z
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
Exercise 1.14
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.
Given a random variable ( ), if 10,
then ( , )
oX P
X N
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:
.
is the number of customers use the ATM machine in 9 hours.
(45); 45 10
(45,45)
( 30) ( 30 0.5) ( 30.5)
30.5 45( 2.16) 0.5 0.4846 0.9846
45
o
c c
X
X P
X N
P X P X P X
P Z P Z
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) Between $1000 and $1100.
b) Between $790 and $1000.
c) Between $840 and $1200.
Answer : a) 0.3413, b) 0.4821, c) 0.9224