2. Sebaran Normal dan Normal Baku Pertemuan 4
Matakuliah : D Statistika dan Aplikasinya
Tahun : 2010
Sebaran Normal dan Normal Baku Pertemuan 4

3. Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu :
menerapkan sifat-sifat sebaran normal, sebaran normal baku dan tabel kurva normal baku
dapat menghitung Inverse transformasi dan pendekatan sebaran binomial dengan sebaran normal

4. The Normal Distribution
Using Statistics
The Normal Probability Distribution
The Standard Normal Distribution
The Transformation of Normal Random Variables
The Inverse Transformation
The Normal Distribution as an Approximation to Other Probability Distributions
Summary and Review of Terms

5. Introduction Normal Probability Density Function:
As n increases, the binomial distribution approaches a ...
n = 6
n = 10
n = 14 6 5 4 3 2 1 . x P ( ) B i n o m a l D s t r b u : = , p 1 9 8 7 6 5 4 3 2 . x P ( ) B i n o m a l D s t r b u : = , p 1 4 3 2 9 8 7 6 5 . x P ( ) B i n o m a l D s t r b u : = , p
Normal Probability Density Function: 5 - . 4 3 2 1 x f ( ) N o r m a l D i s t b u n :  = , 
... . 3
and 2
where
for 1 ) ( =
< - ÷ ø ö ç è æ p s m ps e x f

6. The Normal Probability Distribution
The normal probability density function: 5 - . 4 3 2 1 x f ( ) N o r m a l D i s t b u n :  = ,  f x e ( ) . = -
< æ è ç ö ø ÷ 1 2 3 ps m s p
for
where
and

7. Properties of 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.

8. Properties of the Normal Probability Distribution (continued)
If several independent random variables are normally distributed then their sum will also be normally distributed.
The mean of the sum will be the sum of all the individual means.
The variance of the sum will be the sum of all the individual variances (by virtue of the independence).

9. Normal Probability Distributions
All of these are normal probability density functions, though each has a different mean and variance. 4 5 3 . 2 1 w f ( ) N o r m a l D i s t b u n :  = ,  6 5 4 3 2 1 . x f ( ) N o r m a l D i s t b u n :  = ,  6 5 4 3 . 2 1 y f ( ) N o r m a l D i s t b u n :  = , 
W~N(40,1)
X~N(30,25)
Y~N(50,9) 5 - . 4 3 2 1 z f ( ) N o r m a l D i s t b u n :  = , 
Consider:
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.
Z~N(0,1)

10. 4-3 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 . 4 . 3 { z )
=1 ( f . 2 . 1 . - 5 - 4 - 3 - 2 - 1 1 2 3 4 5
=0 Z

11. Finding Probabilities of the Standard Normal Distribution: P(0 < Z < 1.56)
Standard Normal Probabilities 5 4 3 2 1 - . Z f ( z ) S t a n d r N o m l D i s b u
1.56 { z
Look in row labeled 1.5 and column labeled .06 to find P(0  z  1.56) = .4406

12. 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 = 50 and s = 10.
The transformation of X to Z: Z X x = - m s N o r m a l D i s t r i b u t i o n :  = 5 ,  = 1 . 7 . 6
Transformation . 5 ) ( x . 4
(1) Subtraction: (X - x) f . 3 . 2
=10 { S t a n d a r d N o r m a l D i s t r i b u t i o n . 1 . . 4 1 2 3 4 5 6 7 8 9 1 X . 3 z ) f ( . 2 {
(2) Division by x)
The inverse transformation of Z to X:
1.0 . 1 X x Z = + m s . - 5 - 4 - 3 - 2 - 1 1 2 3 4 5 Z

13. Using the Normal Transformation
Example 4-9
X~N(160,302)
Example 4-10
X~N(127,222) ( ) P X Z .
100
180
160 30 2
6667
4772
2475
7247 = - + æ è ç ö ø ÷ m s ( ) P X Z .
< = - + æ è ç ö ø ÷
150
127 22 1
045 5
3520
8520 m s

14. The Transformation of Normal Random Variables
The transformation of X to Z:
The inverse transformation of Z to X:
The transformation of X to Z, where a and b are numbers:: Z X x = - m s X x Z = + m s P X a Z b ( )
< = - æ è ç ö ø ÷
> m s

15. 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),
That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean of X: 70 =  + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard normal distribution. P X x Z ( )
> = - æ è ç ö ø ÷ 70 50 10 2 m s
Example X~N(124,122)
P(X > x) = and P(Z > 1.28) 0.10
x =  + z = (1.28)(12) = 1 8 3 . 4 2 X f ( x ) N o r m a l D i s t b u n :  = ,  z
0.01
139.36

16. Finding Values of a Normal Random Variable, Given a Probability
1. Draw pictures of the normal distribution in question and of the standard normal distribution. N o r m a l D i s t r i b u t i o n :  = 2 4 5 ,  = 4 . . 1 2 . . 1 . . 8 x ) ( f . . 6 . . 4 . . 2 . 1 2 3 4 X S t a n d a r d N o r m a l D i s t r i b u t i o n . 4 . 3 z ) f ( . 2 . 1 . - 5 - 4 - 3 - 2 - 1 1 2 3 4 5 Z

17. Finding Probabilities of the Standard Normal Distribution: P(0 < Z < 1.56)
Standard Normal Probabilities 5 4 3 2 1 - . Z f ( z ) S t a n d r N o m l D i s b u
1.56 { z
Look in row labeled 1.5 and column labeled .06 to find P(0  z  1.56) = .4406

18. Finding Values of a Normal Random Variable, Given a Probability
1. Draw pictures of the normal distribution in question and of the standard normal distribution. 4 3 2 1 . 8 6 X f ( x ) N o r m a l D i s t b u n :  = 5 , 
.4750
.9500
2. Shade the area corresponding to the desired probability. 5 4 3 2 1 - . Z f ( z ) S t a n d r N o m l D i s b u
.4750
.9500

19. Finding Values of a Normal Random Variable, Given a Probability4 3 2 1 . 8 6 X f ( x ) N o r m a l D i s t b u n :  = 5 , 
.4750
.9500
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
3. From the table of the standard normal distribution, find the z value or values.
2. Shade the area corresponding to the desired probability. 5 4 3 2 1 - . Z f ( z ) S t a n d r N o m l D i s b u
.4750
.9500 z
-1.96
1.96

20. Finding Values of a Normal Random Variable, Given a Probability4 3 2 1 . 8 6 X f ( x ) N o r m a l D i s t b u n :  = 5 , 
.4750
.9500
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
3. From the table of the standard normal distribution, find the z value or values.
2. Shade the area corresponding to the desired probability.
4. Use the transformation from z to x to get value(s) of the original random variable. 5 4 3 2 1 - . Z f ( z ) S t a n d r N o m l D i s b u
.4750
.9500 z
x =   z = 2450 ± (1.96)(400)
= 2450 ±784=(1666,3234)
-1.96
1.96

21. Approximating a Binomial Probability Using the Normal Distributionnp p Z ( )
& = - æ è ç ö ø ÷ 1
for
large (n
50) and
not too c
lose to 0
or 1.00 n
or: P a X b np p Z ( )
& . = - + æ è ç ö ø ÷ 5 1
for
moderatel
y large (2 n
<
50).
If p is either small (close to 0) or large (close to 1), use the Poisson approximation.

22. RINGKASAN
Sebaran Normal
Sifat-sifat sebaran normal
Sebaran normal baku
Inverse transformasi
Pendekatan sebaran binom