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SPECIAL PROBABILITY DISTRIBUTION Budiyono 2011

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BINOMIAL DISTRIBUTION (Bernoulli Distribution)

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Solution:

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NORMAL DISTRIBUTION (Gaussian Distribution)

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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

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STANDARD NORMAL DISTRIBUTION N(0,1)

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z=0 1 area =

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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

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Example using a standard normal distribution table Area = ? P(Z|0

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Example using a standard normal distribution table Area =? Area = =0.7290

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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:

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Area = – = P(73.6

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STANDARD NORMAL DISTRIBUTION N(0,1) sumbu simetri z=0 1 luas =

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STANDARD NORMAL DISTRIBUTION N(0,1) z= z z z z z

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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 }

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Getting z α for α = 25% zαzα α = 25% z 0.25 = ?

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Getting z α for α = 10% zαzα α = 10% z 0.10 = ?

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Getting z α for α = 5% zαzα α = 5% z 0.05 = ?

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The Important Values z α zαzα Z 0.01 = 2.33 Z 0.05 = Z = 1.96 Z = 2.575

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

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STUDENT’S t DISTRIBUTION

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Critical Values for t distribution t α ; Ʋ t 0.10 ; 12 = t 0.05 ; 12 = t 0.01 ; 24 = t ; 28 = α Seen from the table

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

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THE CHI-SQUARE DISTRIBUTION

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

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THE F DISTRIBUTION

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

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

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