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Pertemuan 10 Sebaran Normal-2

Pertemuan 10 Sebaran Normal-2. Matakuliah : A0064 / Statistik Ekonomi Tahun : 2005 Versi : 1/1. Learning Outcomes. Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu :

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Pertemuan 10 Sebaran Normal-2

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  1. Pertemuan 10Sebaran Normal-2 Matakuliah : A0064 / Statistik Ekonomi Tahun : 2005 Versi : 1/1

  2. 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

  3. Outline Materi • Transformasi Variabel Acak Normal • Pendekatan Sebaran Binomial dengan Sebaran Normal

  4. m - X x = Z s x = + m X Z s x x 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 = 50 and s = 10. The transformation of X to Z: N o r m a l D i s t r i b u t i o n :  = 5 0 ,  = 1 0 0 . 0 7 0 . 0 6 Transformation 0 . 0 5 ) 0 . 0 4 x ( f (1) Subtraction: (X - x) 0 . 0 3 { =10 0 . 0 2 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 0 . 0 1 0 . 0 0 0 . 4 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 1 0 0 X 0 . 3 ) z 0 . 2 ( f (2) Division by x) { The inverse transformation of Z to X: 1.0 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z

  5. Example 4-9 X~N(160,302) Example 4-10 X~N(127,222) £ £ < P ( 100 X 180 ) P ( X 150 ) æ ö æ ö - m - m - m - m - m 100 X 180 X 150 ç ÷ ç ÷ = £ £ = < P P è ø è ø s s s s s æ ö æ ö - - - 150 127 100 160 180 160 ç ÷ ç ÷ = < = £ £ P Z P Z è ø è ø 22 30 30 ( ) ( ) = < = - £ £ P Z 1 . 045 P 2 Z . 6667 = + = = + = 0 . 5 0 . 3520 0 . 8520 0 . 4772 0 . 2475 0 . 7247 Using the Normal Transformation

  6. £ £ P ( 394 X 399 ) æ ö - m - m - m 394 X 399 ç ÷ = £ £ P è ø 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 s s s 0 . 4 æ ö - - 394 383 399 383 ç ÷ 0 . 3 = £ £ P Z è ø ) z 0 . 2 12 12 ( f ( ) 0 . 1 = £ £ P 0 . 9166 Z 1 . 333 0 . 0 = - = - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 0 . 4088 0 . 3203 0 . 0885 Z Using the Normal Transformation - Example 4-11 Example 4-11 X~N(383,122) N o r m a l D i s t r i b u t i o n :  = 3 8 3 ,  = 1 2 0 . 0 5 0 . 0 4 0 . 0 3 ) X ( f 0 . 0 2 0 . 0 1 0 . 0 0 340 390 440 X Equivalent areas Template solution

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

  8. 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 0 . 4 0 . 3 ) z 0 . 2 ( f 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z Normal Probabilities (Empirical Rule) • The probability that a normal random variable will be within 1 standard deviationfrom its mean (on either side) is 0.6826, or approximately 0.68. • The probability that a normal random variable will be within 2 standard deviationsfrom its mean is 0.9544, or approximately 0.95. • The probability that a normal random variable will be within 3 standard deviationfrom its mean is 0.9974.

  9. - m - m - æ ö æ ö x 70 70 50 > = > = > = > ç ÷ ç ÷ P ( X 70 ) P P Z P ( Z 2 ) è ø è ø s s 10 N o r m a l D i s t r i b u t i o n :  = 1 2 4 ,  = 1 2 0 . 0 4 0 . 0 3 ) x 0 . 0 2 ( f 0 . 0 1 0 . 0 0 8 0 1 3 0 1 8 0 X 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. Example 4-12X~N(124,122) P(X > x) = 0.10 and P(Z > 1.28) 0.10 x =  + z = 124 + (1.28)(12) = 139.36 z .07 .08 .09 . . . . . . . . . . . . . . . 1.1 . . . 0.3790 0.3810 0.3830 1.2 . . . 0.3980 0.3997 0.4015 1.3 . . . 0.4147 0.4162 0.4177 . . . . . . . . . . . . . . . 0.01 139.36

  10. Template Solution for Example 4-12 Example 4-12X~N(124,122) P(X > x) = 0.10 and P(Z > 1.28) 0.10 x =  + z = 124 + (1.28)(12) = 139.36

  11. The Inverse Transformation (Continued) Example 4-14X~N(2450,4002) P(a<X<b)=0.95 and P(-1.96<Z<1.96)0.95 x =   z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234) P(1666 < X < 3234) = 0.95 Example 4-13X~N(5.7,0.52) P(X > x)=0.01 and P(Z > 2.33) 0.01 x =  + z = 5.7 + (2.33)(0.5) = 6.865 z .02 .03 .04 . . . . . . . . . . . . . . . 2.2 . . . 0.4868 0.4871 0.4875 2.3 . . . 0.4898 0.4901 0.4904 2.4 . . . 0.4922 0.4925 0.4927 . . . . . . . . . . . . . . . z .05 .06 .07 . . . . . . . . . . . . . . . 1.8 . . . 0.4678 0.4686 0.4693 1.9 . . . 0.4744 0.4750 0.4756 2.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . . N o r m a l D i s t r i b u t i o n :  = 5 . 7  = 0 . 5 N o r m a l D i s t r i b u t i o n :  = 2 4 5 0  = 4 0 0 0 0 . . 8 8 0 0 . . 0 0 0 0 1 1 5 5 Area = 0.49 0 0 . . 7 7 0 0 . . 6 6 .4750 .4750 0 0 . . 0 0 0 0 1 1 0 0 0 0 . . 5 5 ) ) x x 0 0 . . 4 4 ( f ( f 0 0 . . 3 3 X.01 = +z = 5.7 + (2.33)(0.5) = 6.865 0 0 . . 0 0 0 0 0 0 5 5 0 0 . . 2 2 .0250 .0250 Area = 0.01 0 0 . . 1 1 0 0 . . 0 0 0 0 . . 0 0 0 0 0 0 0 0 3 3 . . 2 2 4 4 . . 2 2 5 5 . . 2 2 6 6 . . 2 2 7 7 . . 2 2 8 8 . . 2 2 1 1 0 0 0 0 0 0 2 2 0 0 0 0 0 0 3 3 0 0 0 0 0 0 4 4 0 0 0 0 0 0 X X - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 z Z -1.96 1.96 Z.01 = 2.33

  12. 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 0 ,  = 4 0 0 0 . 0 0 1 2 . 0 . 0 0 1 0 . 0 . 0 0 0 8 . ) x 0 . 0 0 0 6 . ( f 0 . 0 0 0 4 . 0 . 0 0 0 2 . 0 . 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 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 0 . 4 0 . 3 ) z ( 0 . 2 f 0 . 1 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z

  13. N o r m a l D i s t r i b u t i o n :  = 2 4 5 0 ,  = 4 0 0 0 . 0 0 1 2 . .4750 .4750 0 . 0 0 1 0 . 0 . 0 0 0 8 . ) x 0 . 0 0 0 6 . ( f 0 . 0 0 0 4 . .9500 0 . 0 0 0 2 . 0 . 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 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 0 . 4 .4750 .4750 0 . 3 ) z ( 0 . 2 f 0 . 1 .9500 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z 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. 2. Shade the area corresponding to the desired probability.

  14. N o r m a l D i s t r i b u t i o n :  = 2 4 5 0 ,  = 4 0 0 0 . 0 0 1 2 . .4750 .4750 0 . 0 0 1 0 . 0 . 0 0 0 8 . ) x 0 . 0 0 0 6 . ( f 0 . 0 0 0 4 . .9500 0 . 0 0 0 2 . 0 . 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 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 0 . 4 .4750 .4750 0 . 3 ) z ( 0 . 2 f 0 . 1 .9500 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z 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. 3. From the table of the standard normal distribution, find the z value or values. 2. Shade the area corresponding to the desired probability. z .05 .06 .07 . . . . . . . . . . . . . . . 1.8 . . . 0.4678 0.4686 0.4693 1.9 . . . 0.4744 0.4750 0.4756 2.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . . -1.96 1.96

  15. N o r m a l D i s t r i b u t i o n :  = 2 4 5 0 ,  = 4 0 0 0 . 0 0 1 2 . .4750 .4750 0 . 0 0 1 0 . 0 . 0 0 0 8 . ) x 0 . 0 0 0 6 . ( f 0 . 0 0 0 4 . .9500 0 . 0 0 0 2 . 0 . 0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 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 0 . 4 .4750 .4750 0 . 3 ) z ( 0 . 2 f 0 . 1 .9500 0 . 0 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 Z 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. 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. z .05 .06 .07 . . . . . . . . . . . . . . . 1.8 . . . 0.4678 0.4686 0.4693 1.9 . . . 0.4744 0.4750 0.4756 2.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . . x =   z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234) -1.96 1.96

  16. Finding Values of a Normal Random Variable, Given a Probability The normal distribution with  = 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 N o r m a l D i s t r i b u t i o n :  = 3 . 5 ,  = 1 . 3 2 3 B i n o m i a l D i s t r i b u t i o n : n = 7 , p = 0 . 5 0 0 . 3 0 . 3 P( x 4) = 0.7734 0 . 2 0 . 2 ) ) x x ( ( P f 0 . 1 0 . 1 0 . 0 0 . 0 0 5 1 0 0 1 2 3 4 5 6 7 X X MTB > cdf 4; SUBC> binomial 7,.5. Cumulative Distribution Function Binomial with n = 7 and p = 0.500000 x P( X <= x) 4.00 0.7734 MTB > cdf 4.5; SUBC> normal 3.5 1.323. Cumulative Distribution Function Normal with mean = 3.50000 and standard deviation = 1.32300 x P( X <= x) 4.5000 0.7751

  17. 4-6 The Normal Approximation of Binomial Distribution The normal distribution with  = 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 B i n o m i a l D i s t r i b u t i o n : n = 1 1 , p = 0 . 5 0 P(x  4) = 0.2744 N o r m a l D i s t r i b u t i o n :  = 5 . 5 ,  = 1 . 6 5 8 3 0 . 3 0 . 2 0 . 2 ) x ( ) P x ( f 0 . 1 0 . 1 0 . 0 0 . 0 0 1 2 3 4 5 6 7 8 9 1 0 1 1 0 5 1 0 X X 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 MTB > cdf 4.5; SUBC> normal 5.5 1.6583. Cumulative Distribution Function Normal with mean = 5.50000 and standard deviation = 1.65830 x P( X <= x) 4.5000 0.2732

  18. - - æ ö a np b np £ £ = £ £ P ( a X b ) P Z ç ÷ & è ø - - np ( 1 p ) np ( 1 p ) ³ for n large (n 50) and p not too c lose to 0 or 1.00 - - + - æ ö a 0 . 5 np b 0 . 5 np £ £ = £ £ P ( a X b ) P Z ç ÷ & - - è ø np ( 1 p ) np ( 1 p ) £ for n moderatel y large (2 0 n < 50). Approximating a Binomial Probability Using the Normal Distribution or: If p is either small (close to 0) or large (close to 1), use the Poisson approximation.

  19. Using the Template for Normal Approximation of the Binomial Distribution

  20. Penutup • Sebaran Normal merupakan sebaran peluang variabel acak kontinyu yang paling banyak digunakan sebagai landasan di dalam penarikan kesimpulan/pengambilan keputusan

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