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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 area = 1 • • • • • • • x=µ x=µ-σ x=µ+σ x=µ+2σ x=µ-2σ x=µ+3σ x=µ-3σ

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**STANDARD NORMAL DISTRIBUTION N(0,1)**

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**STANDARD NORMAL DISTRIBUTION N(0,1)**

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**STANDARD NORMAL DISTRIBUTION N(0,1)**

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

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**Standard Normal Distribution Table**

This area can be found by using a standard normal distribution tablel z This area can be thought as a probability appearing Z between 0 and z, written as P(Z|0<Z<z)

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**Example using a standard normal distribution table**

Area = ? 0.4115 P(Z|0<Z<1.35) = 1.35 P(Z|Z>1.35) = = 0.05 .4115 1.3

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**Example using a standard normal distribution table**

Area =? -1.24 0.98 Area = = 0.7290

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**On a group of 1000 students, the mean of their score is 70**

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; X1 = 73.6; X2 = 81.9; We transform X into z by using the formulae:

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Area = – = 0.72 2.38 P(73.6<X<81.9) = P(Z|0.72<Z<2.38) = So, the number of students having score between 73.6 and 81.9 is x 1000 = 227 student

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**STANDARD NORMAL DISTRIBUTION N(0,1)**

sumbu simetri luas = 1 • • • • • • • -3 -2 -1 z=0 1 2 3

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**STANDARD NORMAL DISTRIBUTION N(0,1)**

0.3413 0.4772 0.0013 0.4987 • • • • • • • -3 -2 -1 z=0 1 2 3 z0.0013 z0.0228 z0.8413 z0.5000 z0.1587

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**Critical Value and Crtitical Region on N(0,1)**

Significance level, usually denoted by α • It is called critical region (daerah kritis), denoted by CR It is called critical value (nilai kritis) (CV), denoted by zα CR = {z | z > zα}

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**Getting zα for α = 25% • zα z0.25 = ? 0.67 .07 0.6 α = 25% 0.25 0.25**

0.2486 0.2500

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**Getting zα for α = 10% • zα z0.10 = ? 1.28 .08 1.2 α = 10% 0.40 0.10**

0.3997 0.4000

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**Getting zα for α = 5% • zα z0.05 = ? 1.645 .04 .05 1.6 α = 5% 0.45**

0.4495 0.4500 0.4505

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**The Important Values zα**

• zα Z0.025 = 1.96 Z0.01 = 2.33 Z0.005 = 2.575 Z0.05 = 1.645

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Properties of zα α α • • z1-α zα z1-α = -zα

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**STUDENT’S t DISTRIBUTION**

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**Critical Values for t distribution**

α Seen from the table • tα ; Ʋ t0.10 ; 12 = 1.356 t0.05 ; 12 = 1.782 t0.005 ; 28 = 2.763 t0.01 ; 24 = 2.492

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Properties of tα;n α α • • t1-α; n tα ; n t1-α; n = -tα; n

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**THE CHI-SQUARE DISTRIBUTION**

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**Critical Value for Chi-Square Distribution**

α α Seen from the table • • Properties: Example 48.278 11.070

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THE F DISTRIBUTION

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**Critical Values for F distribution**

α α Seen from the table • • Properties: Examples: 3.29 26.87

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**Critical Values for F distribution**

0.05 • F0.95; 2, 15 F0.95; 2, 15 = = 0.051

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Chapter Normal Probability Distributions 1 of 25 5 2012 Pearson Education, Inc. All rights reserved.

Chapter Normal Probability Distributions 1 of 25 5 2012 Pearson Education, Inc. All rights reserved.

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