Matakuliah
Tahun
Versi
: I0134 – Metoda Statistika
: 2005
: Revisi
Pertemuan 14
Sebaran Normal
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat menghitung peluang
peubah acak berdistribusi normal.
2
Outline Materi
• Fungsi peluang normal
• Peubah acak normal baku
• Penggunaan tabel normal
3
Distribusi Normal









Using Statistics
The Normal Probability Distribution
The Standard Normal Distribution
The Transformation of Normal Random Variables
The Inverse Transformation
More Complex Problems
The Normal Distribution as an Approximation to
Other Probability Distributions
Using the Computer
Summary and Review of Terms
4
Introduction
As n increases, the binomial distribution approaches a ...
n=6
n = 10
Binomial Distribution: n=10, p=.5
Binomial Distribution: n=14, p=.5
0.3
0.3
0.2
0.2
0.2
0.1
P(x)
0.3
P(x)
P(x)
Binomial Distribution: n=6, p=.5
n = 14
0.1
0.0
0.1
0.0
0
1
2
3
4
5
6
0.0
0
1
x
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
x
x
Normal Probability Density Function:
f ( x) 
1




 x 
2
2
where e  2.7182818... and   314159265
.
...
0.4
0.3
f(x)
x 2

e 2 2 for




Normal Distribution:  = 0, = 1
0.2
0.1
0.0
-5
0
x
5
5
The Normal Probability
Distribution
The normal probability density function:
1
e
0.4
x 2




2 2
0.3
for
 x 
2 2
where e  2.7182818... and   314159265
.
...
f(x)
f ( x) 





Normal Distribution:  = 0, = 1
0.2
0.1
0.0
-5
0
5
x
6
The Normal Probability
Distribution
• The normal is a family of
– Bell-shaped and symmetric distributions.
because the distribution is symmetric, one-half
(.50 or 50%) lies on either side of the mean.
– Each is characterized by a different pair of
mean, , and variance, . That is: [X~N()].
– Each is asymptotic to the horizontal axis.
– The area under any normal probability density
function within k of  is the same for any
normal distribution, regardless of the mean
and variance.
7
Normal Probability
Distributions
All of these are normal probability density functions, though each has a different mean and variance.
Normal Distribution:  =40, =1
Normal Distribution:  =30, =5
0.4
Normal Distribution:  =50, =3
0.2
0.2
0.2
f(y)
f(x)
f(w)
0.3
0.1
0.1
0.1
0.0
0.0
35
40
45
0.0
0
10
20
30
w
40
x
W~N(40,1)
X~N(30,25)
50
60
35
45
50
55
65
y
Y~N(50,9)
Normal Distribution:  =0, =1
Consider:
0.4
f(z)
0.3
0.2
0.1
0.0
-5
0
z
Z~N(0,1)
5
P(39  W  41)
P(25  X  35)
P(47  Y  53)
P(-1  Z  1)
The probability in each
case is an area under a
normal probability density
function.
8
The Standard Normal
Distribution
The standard normal random variable, Z, is the normal random
variable with mean  = 0 and standard deviation  = 1: Z~N(0,12).
Standard Normal Distribution
0 .4
=1
{
f( z)
0 .3
0 .2
0 .1
0 .0
-5
-4
-3
-2
-1
0
1
2
3
4
5
=0
Z
9
Finding Probabilities of the Standard
Normal Distribution: P(Z < -2.47)
To find P(Z<-2.47):
z ...
.
.
P(0 < Z < 2.47) = .4932
.
P(Z < -2.47) = .5 - P(0 < Z < 2.47)2.3 ...
2.4 ...
= .5 - .4932 = 0.0068
2.5 ...
.
.
.
Find table area for 2.47
0.4909
0.4931
0.4948
.06
.
.
.
0.4911
0.4932
0.4949
.07
.
.
.
0.4913
0.4934
0.4951
.08
.
.
.
Standard Normal Distribution
Area to the left of -2.47
P(Z < -2.47) = .5 - 0.4932
= 0.0068
0.4
Table area for 2.47
P(0 < Z < 2.47) = 0.4932
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
10
Finding Probabilities of the Standard
Normal Distribution: P(1< Z < 2)
To find P(1  Z  2):
1. Find table area for 2.00
F(2) = P(Z  2.00) = .5 + .4772 =.9772
2. Find table area for 1.00
F(1) = P(Z  1.00) = .5 + .3413 = .8413
3. P(1  Z  2.00) = P(Z  2.00) - P(Z  1.00)
z
.
.
.
0.9
1.0
1.1
.
.
.
1.9
2.0
2.1
.
.
.
.00
.
.
.
0.3159
0.3413
0.3643
.
.
.
0.4713
0.4772
0.4821
.
.
.
...
...
...
...
...
...
...
= .9772 - .8413 = .1359
Standard Normal Distribution
0.4
Area between 1 and 2
P(1  Z  2) = .4772 - .8413 = 0.1359
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
11
The Transformation of Normal Random
Variables
The area within k of the mean is the same for all normal random variables. So an area
under any normal distribution is equivalent to an area under the standard normal. In this
example: P(40  X  P(-1  Z     since 5and 
The transformation of X to Z:
X x
Z
x
Normal Distribution:  =50, =10
0.07
0.06
Transformation
f(x)
(1) Subtraction: (X - x)
0.05
0.04
0.03
=10
{
0.02
Standard Normal Distribution
0.01
0.00
0.4
0
20
30
40
50
60
70
80
90 100
X
0.3
0.2
(2) Division by x)
{
f(z)
10
1.0
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
The inverse transformation of Z to X:
X  x  Z x
12
Using the Normal
Transformation
Example 4-1
X~N(160,302)
Example 4-2
X~N(127,222)
P (100  X  180)
 100    X    180   
 P

P ( X  150)
 X    150   
 P



 

 
100  160
180  160

P
Z

 30
30 


 P 2  Z  .6667
 0.4772  0.2475  0.7247
 
 
 150  127 
P Z 


22 


 P Z  1.045
 0.5  0.3520  0.8520
13
The Transformation of Normal
Random Variables
The transformation of X to Z:
Z 
X  x
x
The inverse transformation of Z to X:
X  
 Z
x
x
The transformation of X to Z, where a and b are numbers::
a  

P( X  a)  P Z 


 
b  

P( X  b)  P Z 


 
b  
a 
P(a  X  b)  P
Z

 
 
14
Normal Probabilities
S ta n d a rd N o rm a l D is trib utio n
• The probability that a normal
•
•
0.4
0.3
f(z)
random variable will be within 1
standard deviation from its mean
(on either side) is 0.6826, or
approximately 0.68.
The probability that a normal
random variable will be within 2
standard deviations from its mean
is 0.9544, or approximately 0.95.
The probability that a normal
random variable will be within 3
standard deviation from its mean is
0.9974.
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
15
• Selamat Belajar Semoga Sukses.
16