1 pertemuan 09 peubah acak kontinu matakuliah: i0134 – metode statistika tahun: 2007
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Pertemuan 09Peubah Acak Kontinu
Matakuliah : I0134 – Metode Statistika
Tahun : 2007
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Learning OutcomesPada akhir pertemuan ini, diharapkan mahasiswa akan mampu :
• Mahasuswa akan dapat menghitung sifat-sifat peluang peubah acak kontinu.
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Outline Materi
• Fungsi kepekatan peubah acak kontinu• Fungsi distribusi peubah acak kontinu• Nilai harapan peubah acak kontinu• Varians dan simpangan baku peubah acak
kontinu
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Continuous Random Variables
A random variable X is continuous if its set of possible values is an entire interval of numbers (If A < B, then any number x between A and B is possible).
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Probability Density Function
For f (x) to be a pdf
1. f (x) > 0 for all values of x.
2.The area of the region between the graph of f and the x – axis is equal to 1.
Area = 1
( )y f x
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Probability Distribution
Let X be a continuous rv. Then a probability distribution or probability density function (pdf) of X is a function f (x) such that for any two numbers a and b,
( )b
aP a X b f x dx
The graph of f is the density curve.
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Probability Density Function
is given by the area of the shaded region.
( )y f x
ba
( )P a X b
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Important difference of pmf and pdf
k
k
b
aA
dxxfkXP
dxxfdxxfAP
0)()(
)()()(
Y, a discrete r.v. with pmf f(y)X, a continuous r.v. with pdf f(x);
• f(y)=P(Y = k) = probability that the outcome is k.
• f(x) is a particular function with the property that for any event A (a,b), P(A) is the integral of f over A.
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Ex 1. (4.1) X = amount of time for which a book on 2-hour reserve at a college library is checked out by a randomly selected student and suppose that X has density function.
otherwise
xxxf
0
205.0)(
4375.05.05.1.
5.05.0)5.15.0(.
25.00
1
4
15.0)()1(.
2
5.1
5.1
5.0
1 1
0
2
xdxxPc
xdxxPb
xxdxdxxfxPa
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Uniform Distribution
A continuous rv X is said to have a uniform distribution on the interval [a, b] if the pdf of X is
otherwise
bxaabbaxf
0
1),;(
X ~ U (a,b)
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Exponential distribution
X is said to have the exponential distribution if for some
00
01
)(
,0
x
xexf
x
)(~ ExpX
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Probability for a Continuous rv
If X is a continuous rv, then for any number c, P(x = c) = 0. For any two numbers a and b with a < b,
( ) ( )P a X b P a X b
( )P a X b
( )P a X b
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Expected Value
• The expected or mean value of a continuous rv X with pdf f (x) is
( )X E X x f x dx
( ) ( )Xx D
E X x p x
• The expected or mean value of a discrete rv X with pmf f (x) is
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Expected Value of h(X)
• If X is a continuous rv with pdf f(x) and h(x) is any function of X, then
( )( ) ( ) ( )h XE h x h x f x dx
[ ( )] ( ) ( )D
E h X h x p x
• If X is a discrete rv with pmf f(x) and h(x) is any function of X, then
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Variance and Standard Deviation
The variance of continuous rv X with pdf f(x) and mean is
2 2( ) ( ) ( )X V x x f x dx
2
[ ]E X
The standard deviation is
( ).X V x
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Short-cut Formula for Variance
22( ) ( )V X E X E X
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The Cumulative Distribution Function
The cumulative distribution function, F(x) for a continuous rv X is defined for every number x by
( ) ( )x
F x P X x f y dy
For each x, F(x) is the area under the density curve to the left of x.
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Using F(x) to Compute Probabilities
( ) ( )P a X b F b F a
Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any number a,
and for any numbers a and b with a < b,
1 ( )P X a F a
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Ex 6 (Continue). X = length of time in remission, and
30,9
1)( 2 xxxf
What is the probability that a malaria patient’s remission lasts long than one year?
%29.96)127(27
1
1
3
39
1
9
1)1(
3
1
32
xdxxXP
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Obtaining f(x) from F(x)
If X is a continuous rv with pdf f(x) and cdf F(x), then at every number x for which the derivative
( ) ( ).F x f x
( ) exists, F x
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Percentiles
Let p be a number between 0 and 1. The (100p)th percentile of the distribution of a continuous rv X denoted by , is defined by
( )p
( )
( ) ( )p
p F p f y dy
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Median
The median of a continuous distribution, denoted by , is the 50th percentile. So satisfies That is, half the area under the density curve is to the left of
.
0.5 ( ).F
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• Selamat Belajar Semoga Sukses.
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