SPECIAL PROBABILITY DISTRIBUTION Budiyono 2011

BINOMIAL DISTRIBUTION (Bernoulli Distribution)

Solution:

NORMAL DISTRIBUTION (Gaussian Distribution)

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

STANDARD NORMAL DISTRIBUTION N(0,1)

STANDARD NORMAL DISTRIBUTION N(0,1)

STANDARD NORMAL DISTRIBUTION N(0,1) area = 1 • • • • • • • -3 -2 -1 z=0 1 2 3

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)

Example using a standard normal distribution table Area = ? 0.4115 P(Z|0<Z<1.35) = 0.4115 1.35 P(Z|Z>1.35) = 0.5000-0.4115 = 0.0885 0.05 .4115 1.3

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

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:

Area = 0.4913 – 0.2642 = 0.2271 0.72 2.38 P(73.6<X<81.9) = P(Z|0.72<Z<2.38) = 0.2271 So, the number of students having score between 73.6 and 81.9 is 0.2271 x 1000 = 227 student

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

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

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}

Getting zα for α = 25% • zα z0.25 = ? 0.67 .07 0.6 α = 25% 0.25 0.25 0.2486 0.2500

Getting zα for α = 10% • zα z0.10 = ? 1.28 .08 1.2 α = 10% 0.40 0.10 0.3997 0.4000

Getting zα for α = 5% • zα z0.05 = ? 1.645 .04 .05 1.6 α = 5% 0.45 0.4495 0.4500 0.4505

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

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

STUDENT’S t DISTRIBUTION

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

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

THE CHI-SQUARE DISTRIBUTION

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

THE F DISTRIBUTION

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

Critical Values for F distribution 0.05 • F0.95; 2, 15 F0.95; 2, 15 = = 0.051