Chapter 5

Discrete Random Variables and Probability Distributions

Statistika

After completing this chapter, you should be able to:

Interpret the mean and standard deviation for a discrete random variable

Use the binomial probability distribution to find probabilities

Describe when to apply the binomial distribution

Use the hypergeometric and Poisson discrete probability distributions to find probabilities

Explain covariance and correlation for jointly distributed discrete random variables

Chapter Goals

Variabel Random/Random Variable Variabel random adalah suatu fungsi yang

memetakan anggota ruang sampelke bilangan real. Nilainya berubah-ubah

Introduction to Probability Distributions

Random Variables

Discrete Random Variable

ContinuousRandom Variable

Hanya dapat bernilai angka yang dapat dicacahContoh:

Melempar dadu 2x Ambil X = # mata dadu 4 muncul

(X bisa bernilai 0, 1, atau 2 )

Melempar koin 5x. X = # Muka muncul

(X = 0, 1, 2, 3, 4, or 5)

Discrete Random Variables

Eksperimen: Melempar 2 koin. X = # Head.

T

T

Discrete Probability Distribution

Nilai x Probabiliti 0 1/4 = .25 1 2/4 = .50 2 1/4 = .25

4 hasil yg mungkin

T

T

H

H

H H

Distribusi Probabilitas

0 1 2 x

.50

.25

Prob

abili

ti

Tunjukkan P(x) , P(X = x) , u semua x:

P(x) 0 for any value of x

The individual probabilities sum to 1;

(The notation indicates summation over all possible x values)

Probability DistributionRequired Properties

x

1P(x)

Cumulative Probability Function

The cumulative probability function, denoted F(x0), shows the probability that X is less than or equal to x0

In other words,

)xP(X)F(x 00

0xx

0 P(x))F(x

Expected Value (mean) dari distribusi diskrit (Weighted Average), Formula :

Contoh: Melempar 2 koin, X = # of heads, compute expected value of x:

E(x) = (0 x .25) + (1 x .50) + (2 x .25) = 1.0

Nilai Harapan/Expected Value

x P(x)

0 .25

1 .50

2 .25

x

μ E(X) xP(x)

Variance and Standard Deviation

Variance of a discrete random variable X

Standard Deviation of a discrete random variable X

x

222 P(x)μ)(xμ)E(Xσ

x

22 P(x)μ)(xσσ

Standard Deviation Example

Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(x) = 1)

x

2P(x)μ)(xσ

.707.50(.25)1)(2(.50)1)(1(.25)1)(0σ 222

Possible number of heads = 0, 1, or 2

Functions of Random Variables

If P(x) is the probability function of a discrete random variable X , and g(X) is some function of X , then the expected value of function g is

x

g(x)P(x)E[g(X)]

Linear Functions of Random Variables

Let a and b be any constants.

a)

i.e., if a random variable always takes the value a, it will have mean a and variance 0

b)

i.e., the expected value of b·X is b·E(x)

0Var(a)andaE(a)

2X

2X σbVar(bX)andbμE(bX)

Linear Functions of Random Variables

Let random variable X have mean µx and variance σ2

x Let a and b be any constants. Let Y = a + bX Then the mean and variance of Y are

so that the standard deviation of Y is

XY bμabX)E(aμ

X22

Y2 σbbX)Var(aσ

XY σbσ

X 5 6 7 8 9 10P(X) 0,1 0,15 0,25 0,25 0,15 0,1

Chap 5-14

Contoh : Data penjualan laptop perbulan

Carilah nilai harapan banyaknya laptop terjual perbulan E(X) dan variansinya V(X)

Jika untuk satu laptop diambil keuntungan 500 ribu dan biaya pemeliharaan bulanan 600 ribu, hitunglah nilai harapan keuntungan penjualan laptop perbulan

Probability Distributions

Continuous Probability

Distributions

Binomial

Hypergeometric

Poisson

Probability Distributions

Discrete Probability

Distributions

Uniform

Normal

Exponential

Ch. 5 Ch. 6

The Binomial Distribution

Binomial

Hypergeometric

Poisson

Probability Distributions

Discrete Probability

Distributions

Bernoulli Distribution

Consider only two outcomes: “success” or “failure”

Let P denote the probability of success Let 1 – P be the probability of failure Define random variable X:

x = 1 if success, x = 0 if failure Then the Bernoulli probability function is

x 1-x

P(0) (1 P) and P(1) P

P(X=x) = p (1-p) ; x 0,1

Bernoulli DistributionMean and Variance

The mean is µ = P

The variance is σ2 = P(1 – P)

P(1)PP)(0)(1P(x)xE(X)μX

P)P(1PP)(1P)(1P)(0

P(x)μ)(x]μ)E[(Xσ

22

X

222

Binomial Probability Distribution

n percobaan bernoulli e.g., 15 lemparan koin; 10 bola lampu diambil dari gudang

Antar percobaan saling asing dan independen Ada x sukses dan n-x gagal, sebanyak n

kombinasi x Kita tertarik pada VR X = # kejadian sukses P(X=x) ?

A manufacturing plant labels items as either defective or acceptable

A firm bidding for contracts will either get a contract or not

A marketing research firm receives survey responses of “yes I will buy” or “no I will not”

New job applicants either accept the offer or reject it

Possible Binomial Distribution Settings

P(x) = Peluang x sukses dari n trial, n-x gagal

Contoh: Menjawab lima soal B-S,

x = # jawaban benar:

n = 5

P = 0.5

1 - P = (1 - 0.5) = 0.5

x = 0, 1, 2, 3, 4,5

P(x)n

x ! n xP (1- P)X n X!

( ) !

Binomial Distribution Formula

Berapa probabilitas benar satu soal ?

x = 1, n = 5, and P = 0.5

4

151

XnX

.0625)(5)(0.5)(0

0.5)(1(0.5)1)!(51!

5!

P)(1Px)!(nx!

n!1)P(x

Binomial Distribution Bentuk dari distribusi binomial berdasar nilai p

dan nn = 5 P = 0.1

n = 5 P = 0.5

Mean

0.2.4.6

0 1 2 3 4 5x

P(x)

.2

.4

.6

0 1 2 3 4 5x

P(x)

0

n = 5 and P = 0.1

n = 5 and P = 0.5

Mean

Binomial DistributionMean and Variance

Variance and Standard Deviation

nPE(x)μ

P)nP(1-σ2

P)nP(1-σ

Where n = sample sizeP = probability of success(1 – P) = probability of failure

Binomial Characteristics

n = 5 P = 0.1

n = 5 P = 0.5

Mean

0.2.4.6

0 1 2 3 4 5x

P(x)

.2

.4

.6

0 1 2 3 4 5x

P(x)

0

0.5(5)(0.1)nPμ

0.67080.1)(5)(0.1)(1P)nP(1-σ

2.5(5)(0.5)nPμ

1.1180.5)(5)(0.5)(1P)nP(1-σ

Examples

Using Binomial TablesN x … p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50

10 0123456789

10

……………………………

0.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.0000

0.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.0000

0.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.0000

0.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.0000

0.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.0001

0.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.0003

0.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010

Examples: n = 10, x = 3, P = 0.35: P(x = 3|n =10, p = 0.35) = .2522

n = 10, x = 8, P = 0.45: P(x = 8|n =10, p = 0.45) = .0229

Select PHStat / Probability & Prob. Distributions / Binomial…

Using PHStat

Enter desired values in dialog box

Here: n = 10p = .35

Output for x = 0 to x = 10 will be generated by PHStat

Optional check boxesfor additional output

Using PHStat

PHStat Output

P(x = 3 | n = 10, P = .35) = .2522

P(x > 5 | n = 10, P = .35) = .0949

The Poisson Distribution

Binomial

Hypergeometric

Poisson

Probability Distributions

Discrete Probability

Distributions

Apply the Poisson Distribution when: You wish to count the number of times an event

occurs in a given continuous interval The probability that an event occurs in one

subinterval is very small and is the same for all subintervals

The number of events that occur in one subinterval is independent of the number of events that occur in the other subintervals

There can be no more than one occurrence in each subinterval

The average number of events per unit is (lambda)

The Poisson Distribution

Poisson Distribution Formula

where:x = number of successes per unit = expected number of successes per unit

e = base of the natural logarithm system (2.71828...)

x!λeP(x)

xλ

Mean

Poisson Distribution Characteristics

Variance and Standard Deviation

λE(x)μ

λ]σ2 2)[( XE

λσ

where = expected number of successes per unit

Using Poisson TablesX

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

01234567

0.90480.09050.00450.00020.00000.00000.00000.0000

0.81870.16370.01640.00110.00010.00000.00000.0000

0.74080.22220.03330.00330.00030.00000.00000.0000

0.67030.26810.05360.00720.00070.00010.00000.0000

0.60650.30330.07580.01260.00160.00020.00000.0000

0.54880.32930.09880.01980.00300.00040.00000.0000

0.49660.34760.12170.02840.00500.00070.00010.0000

0.44930.35950.14380.03830.00770.00120.00020.0000

0.40660.36590.16470.04940.01110.00200.00030.0000

Example: Find P(X = 2) if = .50

.07582!(0.50)e

!Xe)2X(P

20.50X

Graph of Poisson Probabilities

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)X

=0.50

01234567

0.60650.30330.07580.01260.00160.00020.00000.0000

P(X = 2) = .0758

Graphically: = .50

The shape of the Poisson Distribution depends on the parameter :

Poisson Distribution Shape

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x

)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)

= 0.50 = 3.00

Sebuah gerbang tol memiliki tingkat kedatangan rata-rata 360 kendaraan per jam.

Berarti permenit rata-ratanya adalah 6 kendaraan

Hitunglah peluang dalam satu menit ada 10 mobil yang datang ?

Chap 5-37

Contoh soal distribusi Poisson

Rata-rata jumlah panggilan lewat telepon yang masuk di bagian pelayanan adalah 5 buah permenit. 1.Berapa probabilitas dalam satu menit tertentu tidak terdapat panggilan yang masuk dari pelanggan? 2.Berapa probabilitas dalam satu menit lebih dari 7 panggilan masuk?

Select:PHStat / Probability & Prob. Distributions / Poisson…

Poisson Distribution in PHStat

Complete dialog box entries and get output …

Poisson Distribution in PHStat

P(X = 2) = 0.0758