1 pertemuan 8 variabel acak-2 matakuliah: a0064 / statistik ekonomi tahun: 2005 versi: 1/1

1

Pertemuan 8Variabel Acak-2

Matakuliah : A0064 / Statistik Ekonomi

Tahun : 2005

Versi : 1/1

2

Learning Outcomes

Pada akhir pertemuan ini, diharapkan mahasiswa

akan mampu :

• Menghitung beberapa persoalan yang berkaiatan dengan sebaran geometris, poisson, dan sebaran seragam

3

Outline Materi

• Sebaran Geometris

• Sebaran Poisson

• Variabel Acak Kontinyu

• Sebaran Seragam

COMPLETE 5 t h e d i t i o nBUSINESS STATISTICS

Aczel/SounderpandianMcGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2002

3-4

Geometric distribution:

where x = 1,2,3, . . . and and are the binomial parameters.

The mean and variance of the geometric distribution are:

P x pqx

p

q

p

( )

1

1 22

p q

Within the context of a binomial experiment, in which the outcome of each of n independent trials can be classified as a success (S) or a failure (F), the geometric random variable counts the number of trials until the first success..

3-6 The Geometric Distribution



3-5

Example:A recent study indicates that Pepsi-Cola has a market share of 33.2% (versus 40.9% for Coca-Cola). A marketing research firm wants to conduct a new taste test for which it needs Pepsi drinkers. Potential participants for the test are selected by random screening of soft drink users to find Pepsi drinkers. What is the probability that the first randomly selected drinker qualifies? What’s the probability that two soft drink users will have to be interviewed to find the first Pepsi drinker? Three? Four?

PPPP

( ) (. )(. ) .( ) (. )(. ) .( ) (. )(. ) .( ) (. )(. ) .

( )

( )

( )

( )

1 332 668 0 3322 332 668 0 2223 332 668 01484 332 668 0 099

1 1

2 1

3 1

4 1

The Geometric Distribution - Example



3-6

Calculating Geometric Distribution Probabilities using the Template



3-7

The hypergeometric probability distribution is useful for determining the probability of a number of occurrences when sampling without replacement. It counts the number of successes (x) in n selections, without replacement, from a population of N elements, S of which are successes and (N-S) of which are failures.

The hypergeometric probability distribution is useful for determining the probability of a number of occurrences when sampling without replacement. It counts the number of successes (x) in n selections, without replacement, from a population of N elements, S of which are successes and (N-S) of which are failures.

Hypergeometric Distribution:

P x

S

X

N S

n x

N

n

( )

The mean of the hypergeometric distribution is: where

The variance is:

np pS

NN n

Nnpq

,

2

1

3-7 The Hypergeometric Distribution



3-8

Example:Suppose that automobiles arrive at a dealership in lots of 10 and that for time and resource considerations, only 5 out of each 10 are inspected for safety. The 5 cars are randomly chosen from the 10 on the lot. If 2 out of the 10 cars on the lot are below standards for safety, what is the probability that at least 1 out of the 5 cars to be inspected will be found not meeting safety standards?

P

P

( )

!

! !

!

! !

!

! !

.

( )

!

! !

!

! !

!

! !

.

1

2

1

10 2

5 1

10

5

2

1

8

4

10

5

2

1 1

8

4 4

10

5 5

5

90 556

2

2

1

10 2

5 2

10

5

2

1

8

3

10

5

2

1 1

8

3 5

10

5 5

2

90 222

Thus, P(1) + P(2) =

0.556 + 0.222 = 0.778.

The Hypergeometric Distribution - Example



3-9

Calculating Hypergeometric Distribution Probabilities using the

Template



3-10

The Poisson probability distribution is useful for determining the probability of a number of occurrences over a given period of time or within a given area or volume. That is, the Poisson random variable counts occurrences over a continuous interval of time or space. It can also be used to calculate approximate binomial probabilities when the probability of success is small (p0.05) and the number of trials is large (n20).

Poisson Distribution:

P xex

x

( )!

for x = 1,2,3,...

where is the mean of the distribution (which also happens to be the variance) and e is the base of natural logarithms (e=2.71828...).

3-8 The Poisson Distribution



3-11

Example 3-5:Telephone manufacturers now offer 1000 different choices for a telephone (as combinations of color, type, options, portability, etc.). A company is opening a large regional office, and each of its 200 managers is allowed to order his or her own choice of a telephone. Assuming independence of choices and that each of the 1000 choices is equally likely, what is the probability that a particular choice will be made by none, one, two, or three of the managers? n = 200 = np = (200)(0.001) = 0.2 p = 1/1000 = 0.001

Pe

Pe

Pe

Pe

( ).

!

( ).

!

( ).

!

( ).

!

.

.

.

.

02

0

12

1

22

2

32

3

0 2

1 2

2 2

3 2

= 0.8187

= 0.1637

= 0.0164

= 0.0011

The Poisson Distribution - Example



3-12

Calculating Poisson Distribution Probabilities using the Template



3-13

• Poisson assumptions:The probability that an event will occur in a short interval of time or space

is proportional to the size of the interval.In a very small interval, the probability that two events will occur is close to

zero.The probability that any number of events will occur in a given interval is

independent of where the interval begins.The probability of any number of events occurring over a given interval is

independent of the number of events that occurred prior to the interval.

The Poisson Distribution (continued)



3-14

20191817161514131211109876543210

0.15

0.10

0.05

0.00

X

P(x

)

= 10

109876543210

0.2

0.1

0.0

X

P( x

)

= 4

76543210

0.4

0.3

0.2

0.1

0.0

X

P( x

)

= 1.5

43210

0.4

0.3

0.2

0.1

0.0

X

P( x

)

= 1.0

The Poisson Distribution (continued)



3-15

• A discrete random variable:– counts occurrences – has a countable number of possible

values– has discrete jumps between

successive values– has measurable probability

associated with individual values– probability is height

• A continuous random variable:– measures (e.g.: height, weight,

speed, value, duration, length)– has an uncountably infinite number

of possible values– moves continuously from value to

value– has no measurable probability

associated with individual values– probability is area

For example: Binomial n=3 p=.5

x P(x)0 0.1251 0.3752 0.3753 0.125

1.0003210

0.4

0.3

0.2

0.1

0.0

C1

P(x)

Binomial: n=3 p=.5 For example:In this case, the shaded area epresents the probability that the task takes between 2 and 3 minutes.

654321

0.3

0.2

0.1

0.0

Minutes

P(x)

Minutes to Complete Task

Discrete and Continuous Random Variables - Revisited



3-16

6.56.05.55.04.54.03.53.02.52.01.51.0

0.15

0.10

0.05

0.00

Minutes

P(x)

Minutes to Complete Task: By Half-Minutes

0.0. 0 1 2 3 4 5 6 7

Minutes

P(x)

Minutes to Complete Task: Fourths of a Minute

Minutes

P(x)

Minutes to Complete Task: Eighths of a Minute

0 1 2 3 4 5 6 7

The time it takes to complete a task can be subdivided into:

Half-Minute Intervals Quarter-Minute Intervals Eighth-Minute Intervals

Or even infinitesimally small intervals:When a continuous random variable has been subdivided into infinitesimally small intervals, a measurable probability can only be associated with an interval of values, and the probability is given by the area beneath the probability density function corresponding to that interval. In this example, the shaded area represents P(2 X ).

When a continuous random variable has been subdivided into infinitesimally small intervals, a measurable probability can only be associated with an interval of values, and the probability is given by the area beneath the probability density function corresponding to that interval. In this example, the shaded area represents P(2 X ).

Minutes to Complete Task: Probability Density Function

76543210

Minutes

f(z)

From a Discrete to a Continuous Distribution



3-17

A continuous random variable is a random variable that can take on any value in an interval of numbers.

The probabilities associated with a continuous random variable X are determined by the probability density function of the random variable. The function, denoted f(x), has the following properties.

1. f(x) 0 for all x. 2. The probability that X will be between two numbers a and b is equal to the area

under f(x) between a and b. 3. The total area under the curve of f(x) is equal to 1.00.

The cumulative distribution function of a continuous random variable:

F(x) = P(X x) =Area under f(x) between the smallest possible value of X (often -) and the point x.

A continuous random variable is a random variable that can take on any value in an interval of numbers.

The probabilities associated with a continuous random variable X are determined by the probability density function of the random variable. The function, denoted f(x), has the following properties.

1. f(x) 0 for all x. 2. The probability that X will be between two numbers a and b is equal to the area

under f(x) between a and b. 3. The total area under the curve of f(x) is equal to 1.00.

The cumulative distribution function of a continuous random variable:

F(x) = P(X x) =Area under f(x) between the smallest possible value of X (often -) and the point x.

3-9 Continuous Random Variables



3-18

F(x)

f(x)x

x0

0

ba

F(b)

F(a)

1

ba

}

P(a X b) = Area under f(x) between a and b = F(b) - F(a)

P(a X b)=F(b) - F(a)

Probability Density Function and Cumulative Distribution Function



3-19

The uniform [a,b] density:

1/(a – b) for a X b f(x)= 0 otherwise

E(X) = (a + b)/2; V(X) = (b – a)2/12

{

bb1ax

f(x)

The entire area under f(x) = 1/(b – a) * (b – a) = 1.00

The area under f(x) from a1 to b1 = P(a1Xb) = (b1 – a1)/(b – a)

3-10 Uniform Distribution

a1

Uniform [a, b] Distribution



3-20

The uniform [0,5] density:

1/5 for 0 X 5 f(x)= 0 otherwise

E(X) = 2.5

{

6543210-1

0.5

0.4

0.3

0.2

0.1

0.0.

x

f(x)

Uniform [0,5] Distribution

The entire area under f(x) = 1/5 * 5 = 1.00

The area under f(x) from 1 to 3 = P(1X3) = (1/5)2 = 2/5

Uniform Distribution (continued)



3-21

Calculating Uniform Distribution Probabilities using the Template



3-22

The exponential random variable measures the time between two occurrences that have a Poisson distribution.Exponential distribution:

The density function is:

for

The mean and standard deviation are both equal to 1

The cumulative distribution function is:

for

f x e x

F x e x

x

x

( )

.

( ) .

0, 0

1 03210

2

1

0

f (x)

Exponential Dis tribution: = 2

Time

3-11 Exponential Distribution



3-23

Example

The time a particular machine operates before breaking down (time between breakdowns) is known to have an exponential distribution with parameter = 2. Time is measured in hours. What is the probability that the machine will work continuously for at least one hour? What is the average time between breakdowns?

F x e P X x eP X e

x x( ) ( )( )

.

( )( )

1

11353

2 1

E X( ) . 1 12

5

Exponential Distribution - Example



3-24

Calculating Exponential Distribution Probabilities using the Template

25

Penutup

• Materi Variabel Acak ini pada hakekatnya adalah dasar-dasar untuk pemahaman pola sebaran data, mengingat penarikan kesimpulan/pengambilan keputusan mempunyai sifat ketidakpastian, dan pada umumnya didasarkan pada suatu sampel yang dipilih

1 pertemuan 8 variabel acak-2 matakuliah: a0064 / statistik ekonomi tahun: 2005 versi: 1/1

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