11th Material Subject: Special Distribution of Continuous Random Variable

11 ^th Material Subject: Special Distribution of Continuous Random Variable

Undergraduate of Telecommunication Engineering

MUH1F3 - PROBABILITY AND STATISTICS

Telkom University

Center of eLearning & Open Education Telkom University Jl. Telekomunikasi No.1, Bandung - Indonesia

http://www.telkomuniversity.ac.id

Lecturer: Nor Kumalasari Caecar Pratiwi, S.T., M.T. (caecarnkcp@telkomuniversity.ac.id)

TABLE OF CONTENTS:

1. Uniform 2. Exponential 3. Normal

LEARNING OBJECTIVES:

After careful study of this chapter, student should be able to do the following:

1. Understand the assumptions for some common continuous probability distributions

2. Calculate continuous probability distribution to calculate probabilities in specific applications 3. Calculate probabilities and determine means and variances for some common continuous

probability distributions

UNIFORM

Uniformrandom variables appear in situations where all values in a certain interval

(

a

,

b

)

have same proba- bility.

• IfXis a random variable that is uniformly distributed at interval

(

a

,

b

)

then:

X

⇒

UNI

(

a

,

b

)

(1)

• TheProbability Density Functionof Uniform Distribution:

fX

(

x

) = (

1

b−1

,

^a

≤

x

≤

b 0

,

otherwise

(2)

UNIFORM

• TheMeanof Uniform Distribution:

E

(

X

) = Z

∞

−∞

x

·

fX

(

x

)

dx

= Z

b

a

x

·

¹

b

−

adx

=

^x

2

2

(

b

−

a

)

b

a

=

^b

2

−

a²

2

(

b

−

a

) = (

b

−

a

)(

b

+

a

)

2

(

b

−

a

)

E

(

X

) =

^b

+

a

2 (3)

• TheVarianceof Uniform Distribution:

E

(

X²

) = Z

∞

−∞

x²

·

fX

(

x

)

dx

= Z

b

a

x²

·

¹

b

−

adx

=

^x

3

3

(

b

−

a

)

b

a

=

^b

3

−

a³ 2

(

b

−

a

)

E

(

X²

) = (

b

−

a

)(

b²

+

a²

+

ab

)

3

(

b

−

a

) =

^b

2

+

a²

+

ab 3

UNIFORM

Var

(

X

) =

E

(

X²

) − (

E

(

X

))

²

=

b²

+

a²

+

ab 3

−

b

+

a 2

2

= (

b

−

a

)

²

12 (4)

• TheMoment Generation Functionof Uniform Distribution:

fX

(

x

) =

(

e^bt−e^at

(b−a)t

,

fort

6=

0 1

,

fort

=

0

(5)

UNIFORM

Example:LetXbe a random variable that states the amount of fading which occurs due to the number of obstacles in the propagation environment. Fading that occurs uniformly distributed in the range of80 dBup to 95 dB. Determine:

a. Probability Density Function (PDF) ofX b. Mean / Expected Value ofX

c. Variance random variable ofX d. Moment Generation Function ofX

e. P

(

X

≤

85

)

^,P

(

X

<

85

)

^,P

(

X

>

85

)

^andP

(

X

≥

85

)

Answer:

a. The PDF can be written:

fX

(

x

) = (

1

95−80

,

80

≤

x

≤

95 0

,

^otherwise

UNIFORM

b. Mean / Expected Value of random variableX

E

(

X

) =

^b

+

a 2

=

¹⁷⁵

2 c. Variance of random variableX

Var

(

X

) = (

b

−

a

)

²

12

= (

95

−

80

)

²

12

=

²²⁵

12 d. Moment Generation Function of random variableX

fX

(

x

) =

(

e^95t−e^80t

15t

,

fort

6=

0 1

,

fort

=

0

UNIFORM

d. P

(

X

≤

85

)

,P

(

X

<

85

)

,P

(

X

>

85

)

andP

(

X

≥

85

)

P

(

X

≤

85

) =

P

(

X

<

85

) = Z

85

80

1

15dx

=

⁵

15

P

(

X

≥

85

) =

P

(

X

>

85

) = Z

95

85

10

15dx

=

⁵

15

EXPONENTIAL

• The random variableXthat equals the distance between successive events from a Poisson process with mean number of events

λ >

0per unit interval is anExponentialrandom variable with parameter

λ

^.

X

⇒

EXP

(λ)

(6)

• The probability density function ofXis:

fX

(

x

) =

( λ

e^−λ

,

0

≤

x

< λ

0

,

otherwise

(7)

• TheMeanof Exponential Distribution:

E

(

X

) =

¹

λ

⁽⁸⁾

EXPONENTIAL

• TheVarianceof Exponential Distribution:

Var

(

X

) =

¹

λ

^{2 (9)}

• TheMoment Generation Functionof Binomial Distribution:

E

(

e^tx

) = λ

λ −

t , for t

< λ

(10)

EXPONENTIAL

Example:The average of telephone calls coming to the information center has a waiting time of 5 minutes / call. IfXis random variable which states the average length of waiting time between call, specify:

a. Probability Density Function (PDF) of random variableX b. Mean / Expected Value random variableX

c. Variance random variableX

d. Moment Generation Function random variableX e. The probability of waiting no more than 1 minute

f. The probability of waiting between 1 minute - 3 minutes g. The probability of waiting more than 2 minute

EXPONENTIAL

Answer:

a. The PDF can be written:

fX

(

x

) = (

5e⁻⁵

,

0

≤

x

< ∞

0

,

otherwise

(11)

b. Mean / Expected Value of random variableX

E

(

X

) =

¹

λ =

¹

5 c. Variance of random variableX

Var

(

X

) =

¹

λ

²

=

¹

25 d. Moment Generation Function of random variableX

E

(

e^tx

) = λ

λ

t

=

⁵

5

−

t

,

, for t

< λ

EXPONENTIAL

e. The probability of waiting no more than 1 minute

P

(

t

≤

1

) = Z

1

0

5

·

e^−5xdx

= −

e^−5x

1

0

=

0

.

9933 f. The probability of waiting between 1 minute - 3 minutes

P

(

1

≤

t

≤

3

) = Z

2

1

5

·

e^−5xdx

= −

e^−5x

3

1

=

0

.

0067 g. The probability of waiting more than 2 minute

P

(

t

>

2

) = Z

∞

2

5

·

e^−5xdx

= −

e^−5x

∞

2

=

4

.

54

×

10⁻⁵

NORMAL

Undoubtedly, the most widely used model for a continuous measurement is aNormalrandom variable. The Normaldistribution often referred to as the Gauss distribution, taken from the German mathematician Carl Friedrich Gauss.

• A Normal random variableXis with an average valueE

(

x

) = µ

and varianceVar

(

x

) = σ

^2:

X

⇒

NOR

(µ, σ

²

)

⁽¹²⁾

• TheProbability Density Functionof Normal Distribution:

fX

(

x

) = (

1

σ√

2πe⁻

1

2

(

^x−µ_σ

)

²

, −∞ <

x

< ∞

0

,

otherwise

(13)

(14)

NORMAL

• TheMeanof Normal Distribution:

E

(

X

) = µ

(15)

• TheVarianceof Normal Distribution:

Var

(

X

) = σ

² (16)

• TheMoment Generation Functionof Normal Distribution:

Mx

(

t

) =

E

(

e^tx

) = Z

∞

−∞

e^tx 1

σ

X

√

2

π

^e

−¹

2

(

^x−µσ

)

²

=

e^µt+

σ²t2

2 (17)

NORMAL

A normal random variable with

µ =

0and

σ

²

=

1is called aStandard Normal random variable and is denoted asZ. The cumulative distribution function of a standard normal random variable is denoted as:

Φ(

z

) =

P

(

Z

≤

z

)

(18)

Probabilities that are not of the form

Φ(

z

) =

P

(

Z

≤

z

)

are found by using the basic rules of probability and the symmetry of the normal distribution along with Appendix Table. IfXis a normal random variable with E

(

X

) = µ

^andVar

(

X

) = σ

², the random variable:

Z

=

^X

− µ

σ

⁽¹⁹⁾

is a normal random variable withE

(

Z

) =

0andVar

(

Z

) =

1. That is,Zis a standard normal random variable.

NORMAL

Example: lamp life produced by PT. PIJAR JAYA normally distributed with an average of 1000 hours and standard deviations 100 hours. If X is a random variable state the lamp’s lifespan as in the conditions above, determine:

a. The Probability Density Function ofX b. Mean, Standard Deviation and Variance ofX

c. fX

(

900

)

,P

(

X

≤

850

)

,P

(

X

≥

1254

)

, andP

(

975

<

x

<

1378

)

d. If the company advertises that 95% of the lights can ignite for at least 900 hours. Are these advertisements honest?

NORMAL

Answer:

a. The Probability Density Function ofX

fX

(

x

) = (

1

100

√

2πe⁻

1

2

(

^x−100¹⁰⁰⁰

)

²

, −∞ <

^x

< ∞

0

,

otherwise

b. Mean, Standard Deviation and Variance ofX

µ =

1000 ,

σ =

100 and

σ

²

=

100²

=

10000

NORMAL

Answer:

c. fX

(

900

)

,P

(

X

≤

850

)

,P

(

X

≥

1254

)

, andP

(

975

<

x

<

1378

)

fX

(

900

) =

¹

100

√

2

π

^e

−¹₂

(

⁹⁰⁰⁻100¹⁰⁰⁰

)

²

=

0

.

0024

P

(

X

≤

850

) =

P

Z

≤

⁸⁵⁰

−

1000 100

= Φ(−

1

.

5

) =

0

.

0681

P

(

X

>

1254

) =

1

−

P

(

X

<

1254

) =

1

−

P

Z

≤

¹²⁵⁴

−

1000 100

=

1

− Φ(

2

.

54

) =

0

.

00554 P

(

975

<

x

<

1254

) =

P

(

X

<

1254

) −

P

(

X

<

975

)

=

P

Z

≤

¹²⁵⁴

−

1000 100

−

P

Z

≤

⁹⁷⁵

−

1000 100

= Φ(

2

.

54

) − Φ(−

0

.

25

) =

0

.

59863

NORMAL

d. If the company advertises that 95% of the lights can ignite for at least 900 hours. Are these advertisements honest?

Advertising is said to be honest if:

P

(

X

≥

900

) =

0

.

95 Lets count:

P

(

X

≥

900

) =

1

−

P

(

X

<

900

) =

1

−

P

Z

<

⁹⁰⁰

−

1000 100

=

1

− Φ(

1

) =

1

−

0

.

15866

=

0

.

84134 It turns out that there are only about 84.134% of light that can light up for at least 900 hours, so the

advertisement is dishonest.

Thank You