Matakuliah
Tahun
Versi
: A0064 / Statistik Ekonomi
: 2005
: 1/1
Pertemuan 10
Sebaran Normal-2
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Menghitung beberapa contoh permasalahan
yang berkaitan dengan luas daerah di bawah
kurva normal, transformasi variabel acak
normal, dan pendekatan sebaran binomial
dengan sebaran normal
2
Outline Materi
• Transformasi Variabel Acak Normal
• Pendekatan Sebaran Binomial dengan
Sebaran Normal
3
COMPLETE
4-4
BUSINESS STATISTICS
5th edi tion
4-4 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 m5and 
The transformation of X to Z:
- m x
X
Z 

Normal Distribution: m =50, =10
x
0.07
0.06
Transformation
f(x)
(1) Subtraction: (X - mx)
0.05
0.04
0.03
=10
{
0.02
Standard Normal Distribution
0.01
0.00
0.4
0
10
30
40
50
60
70
80
90 100
X
0.3
0.2
(2) Division by x)
{
f(z)
20
1.0
0.1
X  mx + Z x
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
McGraw-Hill/Irwin
The inverse transformation of Z to X:
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-5
BUSINESS STATISTICS
5th edi tion
Using the Normal Transformation
Example 4-9
X~N(160,302)
Example 4-10
X~N(127,222)
P (100  X  180)
 100 - m  X - m  180 - m 
 P

P( X < 150)
 X - m < 150 - m 
 P


 

 
 100 - 160   180 - 160

P
Z
 30
30 
(

 P -2  Z .6667
 0.4772 + 0.2475  0.7247
McGraw-Hill/Irwin
 
 
 150 - 127
 P Z <


22 
(
 P Z < 1.045
 0.5 + 0.3520  0.8520
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-6
BUSINESS STATISTICS
5th edi tion
Using the Normal Transformation Example 4-11
Normal Distribution: m = 383,  = 12
Example 4-11
X~N(383,122)
0.05
P ( 394  X  399)
 394 - m  X - m  399 - m 
 P

(

 P 0.9166  Z  1.333
 0.4088 - 0.3203  0.0885
f(X)
0.03
0.02
0.01
Standard Normal Distribution
0.00
340
0.4
390
440
X
0.3
f(z)

 

 
 394 - 383   399 - 383

P
Z
 12

12
0.04
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
Template solution
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
4-7
5th edi tion
The Transformation of Normal Random
Variables
The transformation of X to Z:
Z 
The inverse transformation of Z to X:
X - mx
X  m
+ Z
x
x
x
The transformation of X to Z, where a and b are numbers::
a - m

<

<

P ( X a ) P Z

 
b - m

>

>

P ( X b) P Z

 
b - m
a-m
<
<

<
<


P (a X b ) P
Z
 
 
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-8
BUSINESS STATISTICS
5th edi tion
Normal Probabilities (Empirical Rule)
S ta n d a rd N o rm a l D is trib utio n
• The probability that a normal
•
•
McGraw-Hill/Irwin
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.
Aczel/Sounderpandian
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
4-9
5th edi tion
4-5 The Inverse Transformation
The area within k of the mean is the same for all normal random variables. To find a
probability associated with any interval of values for any normal random variable, all that
is needed is to express the interval in terms of numbers of standard deviations from the
mean. That is the purpose of the standard normal transformation. If X~N(50,102),
 x - m 70 - m 

70 - 50 
>
  P Z >
  P( Z > 2)
P( X > 70)  P
 

 
10 
That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean
of X: 70 = m + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard normal
distribution.
Normal Distribution: m = 124,  = 12
Example 4-12
X~N(124,122)
P(X > x) = 0.10 and P(Z > 1.28) 0.10
x = m + z = 124 + (1.28)(12) = 139.36
0.04
0.03
.
.
.
. . .
. . .
. . .
.
.
.
.07
.
.
.
0.3790
0.3980
0.4147
.
.
.
McGraw-Hill/Irwin
.08
.
.
.
0.3810
0.3997
0.4162
.
.
.
.09
.
.
.
0.3830
0.4015
0.4177
.
.
.
f(x)
z
.
.
.
1.1
1.2
1.3
.
.
.
0.02
0.01
0.01
0.00
80
130
X
Aczel/Sounderpandian
180
139.36
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
4-10
5th edi tion
Template Solution for Example 4-12
Example 4-12
X~N(124,122)
P(X > x) = 0.10 and P(Z > 1.28) 0.10
x = m + z = 124 + (1.28)(12) = 139.36
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-11
BUSINESS STATISTICS
5th edi tion
The Inverse Transformation (Continued)
Example 4-14
X~N(2450,4002)
P(a<X<b)=0.95 and P(-1.96<Z<1.96)0.95
x = m  z = 2450 ± (1.96)(400) = 2450
±784=(1666,3234)
P(1666 < X < 3234) = 0.95
Example 4-13
X~N(5.7,0.52)
P(X > x)=0.01 and P(Z > 2.33) 0.01
x = m + z = 5.7 + (2.33)(0.5) = 6.865
z
.
.
.
2.2
2.3
2.4
.
.
.
.02
.
.
.
0.4868
0.4898
0.4922
.
.
.
.
.
.
. . .
. . .
. . .
.
.
.
.03
.
.
.
0.4871
0.4901
0.4925
.
.
.
.04
.
.
.
0.4875
0.4904
0.4927
.
.
.
z
.
.
.
1.8
1.9
2.0
.
.
Normal Distribution: m = 5.7 = 0.5
.06
.
.
.
0.4686
0.4750
0.4803
.
.
.07
.
.
.
0.4693
0.4756
0.4808
.
.
0.0015
Area = 0.49
0.6
.4750
.4750
0.0010
f(x)
0.5
f(x)
.05
.
.
.
0.4678
0.4744
0.4798
.
.
Normal Distribution: m = 2450  = 400
0.8
0.7
.
.
.
. . .
. . .
. . .
.
.
0.4
X.01 = m+z = 5.7 + (2.33)(0.5) = 6.865
0.3
0.0005
0.2
.0250
.0250
Area = 0.01
0.1
0.0
0.0000
3.2
4.2
5.2
6.2
7.2
8.2
1000
2000
X
-5
-4
-3
-2
-1
0
z
McGraw-Hill/Irwin
3000
4000
X
1
2
3
4
5
-5
-4
-3
-2
-1.96
Z.01 = 2.33
Aczel/Sounderpandian
-1
0
Z
1
2
3
4
5
1.96
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-12
BUSINESS STATISTICS
5th edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution: m = 2450,  = 400
0.0012
.
0.0010
.
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
0.0000
1000
2000
3000
4000
X
S tand ard Norm al D istrib utio n
0.4
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-13
BUSINESS STATISTICS
5th edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution: m = 2450,  = 400
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
S tand ard Norm al D istrib utio n
0.4
.4750
.4750
0.3
f(z)
2. Shade the area
corresponding to
the desired
probability.
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-14
BUSINESS STATISTICS
5th edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution: m = 2450,  = 400
3. From the table
of the standard
normal
distribution,
find the z value
or values.
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
2. Shade the area
corresponding
to the desired
probability.
S tand ard Norm al D istrib utio n
0.4
.4750
f(z)
z
.
.
.
1.8
1.9
2.0
.
.
.
.
.
. . .
. . .
. . .
.
.
.05
.
.
.
0.4678
0.4744
0.4798
.
.
McGraw-Hill/Irwin
.06
.
.
.
0.4686
0.4750
0.4803
.
.
.4750
0.3
.07
.
.
.
0.4693
0.4756
0.4808
.
.
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
-1.96
Aczel/Sounderpandian
1.96
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-15
BUSINESS STATISTICS
5th edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution: m = 2450,  = 400
3. From the table
of the standard
normal
distribution,
find the z value
or values.
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
2. Shade the area
corresponding
to the desired
probability.
0.4
.4750
.
.
.
. . .
. . .
. . .
.
.
.05
.
.
.
0.4678
0.4744
0.4798
.
.
McGraw-Hill/Irwin
.06
.
.
.
0.4686
0.4750
0.4803
.
.
.4750
0.3
f(z)
z
.
.
.
1.8
1.9
2.0
.
.
4. Use the
transformation
from z to x to get
value(s) of the
original random
variable.
S tand ard Norm al D istrib utio n
.07
.
.
.
0.4693
0.4756
0.4808
.
.
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
1
2
Z
-1.96
Aczel/Sounderpandian
3
4
5
x = m  z = 2450 ± (1.96)(400)
= 2450 ±784=(1666,3234)
1.96
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-16
BUSINESS STATISTICS
5th edi tion
Finding Values of a Normal Random
Variable, Given a Probability
The normal distribution with m = 3.5 and  = 1.323 is a close
approximation to the binomial with n = 7 and p = 0.50.
P(x<4.5) = 0.7749
Normal Distribution: m = 3.5,  = 1.323
Binomial Distribution: n = 7, p = 0.50
0.3
0.3
P( x 4) = 0.7734
0.2
f(x)
P(x)
0.2
0.1
0.1
0.0
0.0
0
5
10
0
1
X
2
3
4
5
6
7
X
MTB > cdf 4.5;
SUBC> normal 3.5 1.323.
Cumulative Distribution Function
MTB > cdf 4;
SUBC> binomial 7,.5.
Cumulative Distribution Function
Normal with mean = 3.50000 and standard
deviation = 1.32300
Binomial with n = 7 and p = 0.500000
x
4.5000
McGraw-Hill/Irwin
P( X <= x)
0.7751
Aczel/Sounderpandian
x
4.00
P( X <= x)
0.7734
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
4-17
BUSINESS STATISTICS
5th edi tion
4-6 The Normal Approximation of Binomial
Distribution
The normal distribution with m = 5.5 and  = 1.6583 is a closer
approximation to the binomial with n = 11 and p = 0.50.
P(x < 4.5) = 0.2732
Normal Distribution: m = 5.5,  = 1.6583
Binomial Distribution: n = 11, p = 0.50
P(x  4) = 0.2744
0.3
0.2
f(x)
P(x)
0.2
0.1
0.1
0.0
0.0
0
5
10
0
1
4
5
6
7
8
9 10 11
MTB > cdf 4;
SUBC> binomial 11,.5.
Cumulative Distribution Function
MTB > cdf 4.5;
SUBC> normal 5.5 1.6583.
Cumulative Distribution Function
Normal with mean = 5.50000 and standard deviation = 1.65830
Binomial with n = 11 and p = 0.500000
x
4.00
P( X <= x)
0.2732
McGraw-Hill/Irwin
3
X
X
x
4.5000
2
Aczel/Sounderpandian
P( X <= x)
0.2744
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
4-18
5th edi tion
Approximating a Binomial Probability
Using the Normal Distribution
 a - np
b - np 
Z
P( a  X  b) & P

 np(1 - p)
np(1 - p) 
for n large (n  50) and p not too close to 0 or 1.00
or:
 a - 0.5 - np
b + 0.5 - np 
Z
P(a  X  b) & P

 np(1 - p)
np(1 p) 
for n moderately large (20  n < 50).
If p is either small (close to 0) or large (close to 1), use the Poisson
approximation.
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
4-19
5th edi tion
Using the Template for Normal Approximation
of the Binomial Distribution
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
Penutup
• Sebaran Normal merupakan sebaran
peluang variabel acak kontinyu yang
paling banyak digunakan sebagai
landasan di dalam penarikan
kesimpulan/pengambilan keputusan
20