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

SPECIAL PROBABILITY DISTRIBUTION

Budiyono2011

BINOMIAL DISTRIBUTION(Bernoulli Distribution)

Solution:

NORMAL DISTRIBUTION(Gaussian Distribution)

Normal Distribution Curve

symetri axes

x=µ•

area = 1• • ••••

x=µ+σ x=µ+2σ x=µ+3

σ

x=µ-σx=µ-2σ

x=µ-3σ

STANDARD NORMAL DISTRIBUTION N(0,1)

STANDARD NORMAL DISTRIBUTION N(0,1)

STANDARD NORMAL DISTRIBUTION N(0,1)

z=0 1•

area = 1• •

2•3

•-1

•-2

•-3

Standard Normal Distribution Table

0 z

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<Z<z)

Example using a standard normal distribution table

0 1.35

Area = ?

P(Z|0<Z<1.35) = 0.4115

P(Z|Z>1.35) = 0.5000-0.4115 = 0.0885

1.3

0.05

.4115

0.4115

Example using a standard normal distribution table

0-1.24

0.98

Area =?

Area =

0.3925 +

0.3365 =

0.7290

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:

0 0.72

Area = 0.4913 – 0.2642 = 0.2271

P(73.6<X<81.9) = P(Z|0.72<Z<2.38) = 0.2271

2.38

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

z=0 1•

luas = 1• •

2•3

•-1

•-2

•-3

STANDARD NORMAL DISTRIBUTION N(0,1)

z=0 1• • •

2•3

•-1

•-2

•-3

0.4772

0.4987

0.0013

z0.0013z0.0228z0.1587z0.5000z0.8413

0.3413

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 zα

It is called critical region (daerah kritis), denoted by CR

CR = {z | z > zα}

Getting zα for α = 25%

•zα

α = 25%

z0.25 = ? 0.67

0.25

0.25

0.25000.24860.6

.07

Getting zα for α = 10%

•zα

α = 10%

z0.10 = ?1.28

0.10

0.40

0.40000.39971.2

.08

Getting zα for α = 5%

•zα

α = 5% z0.05 = ?1.645

0.05

0.45

0.45000.44951.6

.04

0.4505

.05

The Important Values zα

•zα

Z0.01 = 2.33Z0.05 = 1.645

Z0.025 = 1.96Z0.005 = 2.575

Properties of zα

•zα

αα

•z1-α

z1-α = -zα

STUDENT’S t DISTRIBUTION

Critical Values for t distribution

•tα ; Ʋ

t0.10 ; 12 =

1.356 t0.05 ; 12 =

1.782

t0.01 ; 24 =

2.492 t0.005 ; 28 =

2.763

α

Seen from the table

Properties of tα;n

•tα ; n

αα

•t1-α; n

t1-α; n = -tα; n

THE CHI-SQUARE DISTRIBUTION

Critical Value for Chi-Square Distribution

•

α

Properties:Example

Seen from the table

11.070 48.278

α

•

THE F DISTRIBUTION

Critical Values for F distribution

•

α

Properties:

Examples:

Seen from the table

3.29 26.87

α

•

Critical Values for F distribution

2,15;05.0F1

0.05

•

F0.95; 2, 15 =

F0.95; 2, 15

43.191

= 0.051