5.Discrete RV and Prob Distr
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Transcript of 5.Discrete RV and Prob Distr
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Chapter 5
Discrete Random Variables and Probability Distributions
Statistika
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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
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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
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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
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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:
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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)
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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
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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)
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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
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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
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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)]
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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)
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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
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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
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Probability Distributions
Continuous Probability
Distributions
Binomial
Hypergeometric
Poisson
Probability Distributions
Discrete Probability
Distributions
Uniform
Normal
Exponential
Ch. 5 Ch. 6
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The Binomial Distribution
Binomial
Hypergeometric
Poisson
Probability Distributions
Discrete Probability
Distributions
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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
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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
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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) ?
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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
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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
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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
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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
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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
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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
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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
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Select PHStat / Probability & Prob. Distributions / Binomial…
Using PHStat
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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
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PHStat Output
P(x = 3 | n = 10, P = .35) = .2522
P(x > 5 | n = 10, P = .35) = .0949
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The Poisson Distribution
Binomial
Hypergeometric
Poisson
Probability Distributions
Discrete Probability
Distributions
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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
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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
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Mean
Poisson Distribution Characteristics
Variance and Standard Deviation
λE(x)μ
λ]σ2 2)[( XE
λσ
where = expected number of successes per unit
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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
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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
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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
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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
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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?
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Select:PHStat / Probability & Prob. Distributions / Poisson…
Poisson Distribution in PHStat
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Complete dialog box entries and get output …
Poisson Distribution in PHStat
P(X = 2) = 0.0758