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SPECIAL PROBABILITY DISTRIBUTION Budiyono 2011. BINOMIAL DISTRIBUTION (Bernoulli Distribution)

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Presentation on theme: "SPECIAL PROBABILITY DISTRIBUTION Budiyono 2011. BINOMIAL DISTRIBUTION (Bernoulli Distribution)"— Presentation transcript:

1 SPECIAL PROBABILITY DISTRIBUTION Budiyono 2011

2 BINOMIAL DISTRIBUTION (Bernoulli Distribution)

3 Solution:

4 NORMAL DISTRIBUTION (Gaussian Distribution)

5 Normal Distribution Curve symetri axes x=µ area = 1 x=µ+σx=µ+2σ x=µ+3σ x=µ-σ x=µ-2σ x=µ-3σ

6 STANDARD NORMAL DISTRIBUTION N(0,1)

7

8 z=0 1 area = 1 2 3 -2 -3

9 Standard Normal Distribution Table 0z This area can be found by using a standard normal distribution tablel This area can be thought as a probability appearing Z between 0 and z, written as P(Z|0 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/11/3227725/slides/slide_9.jpg", "name": "Standard Normal Distribution Table 0z This area can be found by using a standard normal distribution tablel This area can be thought as a probability appearing Z between 0 and z, written as P(Z|0

10 Example using a standard normal distribution table 0 1.35 Area = ? P(Z|01.35) = 0.5000- 0.4115 = 0.0885 1.3 0.05.4115 0.4115

11 Example using a standard normal distribution table 0 -1.24 0.98 Area =? Area =0.3925 +0.3365 =0.7290

12 On a group of 1000 students, the mean of their score is 70.0 and the standard deviation is 5.0. Assuming that the score are normally distributed. How many students have score between 73.6 dan 81.9? Problem Solution µ = 70.0;σ = 5.0;X 1 = 73.6;X 2 = 81.9; We transform X into z by using the formulae:

13 0 0.72 Area = 0.4913 – 0.2642 = 0.2271 P(73.6 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/11/3227725/slides/slide_13.jpg", "name": "0 0.72 Area = 0.4913 – 0.2642 = 0.2271 P(73.6

14 STANDARD NORMAL DISTRIBUTION N(0,1) sumbu simetri z=0 1 luas = 1 2 3 -2 -3

15 STANDARD NORMAL DISTRIBUTION N(0,1) z=0 1 2 3 -2 -3 0.4772 0.4987 0.0013 z 0.0013 z 0.0228 z 0.1587 z 0.5000 z 0.8413 0.3413

16 Critical Value and Crtitical Region on N(0,1) Significance level, usually denoted by α It is called critical value (nilai kritis) (CV), denoted by z α It is called critical region (daerah kritis), denoted by CR CR = {z | z > z α }

17 Getting z α for α = 25% zαzα α = 25% z 0.25 = ? 0.67 0.25 0.25000.2486 0.6.07

18 Getting z α for α = 10% zαzα α = 10% z 0.10 = ? 1.28 0.10 0.40 0.40000.3997 1.2.08

19 Getting z α for α = 5% zαzα α = 5% z 0.05 = ? 1.645 0.05 0.45 0.45000.4495 1.6.04 0.4505.05

20 The Important Values z α zαzα Z 0.01 = 2.33 Z 0.05 = 1.645 Z 0.025 = 1.96 Z 0.005 = 2.575

21 Properties of z α zαzα α α z 1- α z 1- α = -z α

22 STUDENT’S t DISTRIBUTION

23 Critical Values for t distribution t α ; Ʋ t 0.10 ; 12 = 1.356 t 0.05 ; 12 = 1.782 t 0.01 ; 24 = 2.492 t 0.005 ; 28 = 2.763 α Seen from the table

24 Properties of t α;n t α ; n α α t 1- α; n t 1- α; n = -t α; n

25 THE CHI-SQUARE DISTRIBUTION

26 Critical Value for Chi-Square Distribution α Properties: Example Seen from the table 11.070 48.278 α

27 THE F DISTRIBUTION

28 Critical Values for F distribution α Properties: Examples: Seen from the table 3.29 26.87 α

29 Critical Values for F distribution 0.05 F 0.95; 2, 15 = F 0.95; 2, 15 = 0.051


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