Matakuliah
Tahun
: I0134 – Metode Statistika
: 2007
Pertemuan 09
Peubah Acak Kontinu
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Learning Outcomes
Pada 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.
y  f ( x)
Area = 1
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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,
P  a  X  b    f ( x)dx
b
a
The graph of f is the density curve.
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Probability Density Function
P(a  X  b) is given by the area of the shaded
region.
y  f ( x)
a
b
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Important difference of pmf and pdf
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.
b
P( A)   f ( x)dx   f ( x)dx
A
a
k
P( X  k )   f ( x)dx  0
k
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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.
0.5 x 0  x  2
f ( x)  
otherwise
0
1 21
f ( x)dx   0.5 xdx  x 0.25
0
4 0
1
1
a. P ( x  1)  

1.5
b. P (0.5  x  1.5)   0.5 xdx 0.5
0. 5
c. P x  1.5   0.5 xdx 0.4375
2
1.5
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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
 1

f ( x; a, b)   b  a
0
a xb
otherwise
X ~ U (a,b)
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Exponential distribution
X ~ Exp( )
X is said to have the exponential distribution
  0,
if for some
 1  x
 e
f ( x)   

 0
x0
x0
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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

• The expected or mean value of a discrete rv X
with pmf f (x) is
E( X )   X 
 x  p ( x)
xD
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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
E  h( x )    h ( X ) 

 h( x)  f ( x)dx

• If X is a discrete rv with pmf f(x) and h(x) is any
function of X, then
E[h( X )]   h( x)  p( x)
D
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Variance and Standard Deviation
The variance of continuous rv X with
pdf f(x) and mean  is
2
X

 V ( x) 
 (x  )

2
 f ( x)dx
 E[ X    ]
2
The standard deviation is  X  V ( x).
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Short-cut Formula for Variance
    E ( X )
V (X )  E X
2
2
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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
F ( x)  P  X  x    f ( y)dy
x

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
Let X be a continuous rv with pdf f(x)
and cdf F(x). Then for any number a,
P  X  a   1  F (a )
and for any numbers a and b with a < b,
P  a  X  b   F (b)  F (a)
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Ex 6 (Continue). X = length of time in
remission, and
1 2
f ( x)  x , 0  x  3
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What is the probability that a malaria
patient’s remission lasts long than one year?
P( X  1)  
3
1
1 2
1 x3 3 1
x dx 
 (27  1)  96.29%
9
9 3 1 27
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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) exists,
F ( x)  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  ( p ), is
defined by
p  F  ( p)   
 ( p)

f ( y)dy
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Median
The median of a continuous distribution,
denoted by  , is the 50th percentile. So 
satisfies 0.5  F (  ). That is, half the area
under the density curve is to the left of  .
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• Selamat Belajar Semoga Sukses.
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