Transcript Pertemuan 15 Aplikasi Sebaran Normal – Metoda Statistika Matakuliah

Matakuliah
Tahun
Versi
: I0134 – Metoda Statistika
: 2005
: Revisi
Pertemuan 15
Aplikasi Sebaran Normal
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat menggunakan tabel
normal dan menghitung peluang Binom
dengan pendekatan normal.
2
Outline Materi
• Kaidah Peluang normal
• Ketaksamaan Chebyshev
• Pendekatan normal pada Binomial
3
The Normal Distribution as an Approximation
to Other Probability Distributions
The normal distribution with  = 5.5 and  = 1.6583 is a closer
approximation to the binomial with n = 11 and p = 0.50.
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Hampiran Normal
P(x<4.5) = 0.2732
Normal Distribution:  = 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
2
3
4
Normal with mean = 5.50000 and standard deviation = 1.65830
x P( X <= x)
4.5000
0.2732
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7
8
9 10 11
X
X
MTB > cdf 4.5;
SUBC> normal 5.5 1.6583.
Cumulative Distribution Function
5
MTB > cdf 4;
SUBC> binomial 11,.5.
Cumulative Distribution Function
Binomial with n = 11 and p = 0.500000
x P( X <= x)
4.00
0.2744
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Approximating a Binomial
Probability Using the Normal
Distribution
b  np 
 a  np
P( a  X  b)  P
Z

 np(1  p)
np(1  p) 
for n large (n  50) and p not too close to 0 or 1.00
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Approximating a Binomial
Probability Using the Normal
Distribution
b  0.5  np 
 a  0.5  np
P(a  X  b)  P
Z

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.
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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
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Using the Normal Transformation
- Minitab Solutions
MTB > cdf 100;
SUBC> normal 160,30.
Cumulative Distribution Function
Normal with mean = 160.000 and standard
deviation = 30.0000
x P( X <= x)
100.0000
0.0228
MTB > cdf 180;
SUBC> normal 160,30.
Cumulative Distribution Function
Normal with mean = 160.000 and standard
deviation = 30.0000
MTB > cdf 150;
SUBC> normal 127,22.
Cumulative Distribution Function
Normal with  = 127.000 and  = 22.0000
x P( X <= x)
150.0000
0.8521
x P( X <= x)
180.0000
0.7475
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Using the Normal
Transformation
Normal Distribution:  = 383,  = 12
Example 4-3
X~N(383,122)
0.05
P ( 394  X  399)
 394   X   399   
 P



f(X)
0.03
0.02
 
399  383


12


 P 0.9166  Z  1.333
 0.4088  0.3203  0.0885
0.01
Standard Normal Distribution
0.00
340
0.4
390
440
X
0.3
f(z)

 

 394  383
P
Z 
 12
0.04
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
MTB > cdf 394;
SUBC> normal 383,12.
MTB > cdf 399;
SUBC> normal 383,12.
Cumulative Distribution Function
Cumulative Distribution Function
Normal with mean = 383.000 and standard deviation = 12.0000
Normal with mean = 383.000 and standard deviation = 12.0000
x P( X <= x)
394.0000
0.8203
x P( X <= x)
399.0000
0.9088
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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
11
• Selamat Belajar Semoga Sukses.
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