chapter 4-1 continuous random variables 主講人 : 虞台文
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Chapter 4-1Continuous Random Variables
主講人 :虞台文
Content Random Variables and Distribution Functions Probability Density Functions of Continuous Rando
m Variables The Exponential Distributions The Reliability and Failure Rate The Erlang Distributions The Gamma Distributions The Gaussian or Normal Distributions The Uniform Distributions
Random Variables and Distribution Functions
Chapter 4-1Continuous Random Variables
The Temperature in Taipei
今天中午台北市氣溫為25C之機率為何 ?
今天中午台北市氣溫小於或等於 25C之機率為何 ?
Renewed Definition of Random Variables
A random variable X on a probability space (, A,
P) is a function
X : Rthat assigns a real number X() to each sample point , such that for every real number x, the
set {|X() x} is an event, i.e., a member
of A.
The (Cumulative) Distribution Functions
The (cumulative) distribution function FX of a rando
m variable X is defined to be the function
FX(x) = P(X x), − < x < .
Example 1
Example 1
3 31 18 8 8 8
0 1 2 3
( )X
x
p x
18
48
78
0 0
0 1
( ) 1 2
2 3
1 3
X
x
x
F x x
x
x
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
Example 1
R
y
( ) ( )YF y P Y y 2
2
y
R
2
2
y
R 0 y R
Example 1
R
y
( ) ( )YF y P Y y
2
2
0 0
0
1 1
y
yy R
Ry
Example 1
( ) ( )YF y P Y y
2
2
0 0
0
1 1
y
yy R
Ry
0
0.5
1
y
F Y (y )
RR /20
0.5
1
y
F Y (y )
RR /2
Example 1
R
RY
R/2
( ) ( )ZF z P Z z
2
2
2
2
( )3
59 3 2
8 29 2 3
( ) 23
0 0
1 0
1
1
R z RR
R R
R R
R z RR
z
z
z
z
z R
R z
Example 1
( ) ( )ZF z P Z z
2
2
2
2
( )3
59 3 2
8 29 2 3
( ) 23
0 0
1 0
1
1
R z RR
R R
R R
R z RR
z
z
z
z
z R
R z
0
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /3
Example 1
0
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /30
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /30
0.5
1
y
F Y (y )
RR /2
0
0.5
1
y
F Y (y )
RR /2-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
( )XF x ( )YF y ( )ZF z
Properties of Distribution Functions
1. 0 F(x) 1 for all x;2. F is monotonically nondecreasing;3. F() = 0 and F() =1;4. F(x+) = F(x) for all x.
0
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /30
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /30
0.5
1
y
F Y (y )
RR /2
0
0.5
1
y
F Y (y )
RR /2-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
Definition Continuous Random Variables
0
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /30
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /30
0.5
1
y
F Y (y )
RR /2
0
0.5
1
y
F Y (y )
RR /2-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
A random variable X is called a continuous random variable if
( ) ( ) ( ) 0P X x F x F x
Example 2
0
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /30
0.5
1
z
F Z (y )
RR /2R /3 2R /3
0
0.5
1
z
F Z (y )
RR /2R /3 2R /30
0.5
1
y
F Y (y )
RR /2
0
0.5
1
y
F Y (y )
RR /2-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
-3 -2 -1 0 1 2 3 4 5 6 7
x
F X (x )1
Probability Density Functions of Continuous
Random Variables
Chapter 4-1Continuous Random Variables
Probability Density Functions of Continuous Random Variables
A probability density function (pdf) fX(x) of a continuous random variable X is a nonnegative function f such that
( ) ( )x
X XF x f u du
Probability Density Functions of Continuous Random Variables
A probability density function (pdf) fX(x) of a continuous random variable X is a nonnegative function f such that
( ) ( )x
X XF x f u du
Properties of Pdf's
1. ( ) 0;f x
2. ( ) 1;f x dx
3. ( ) ( ) ( ) ( )P a X b P a X b P a X b P a X b
( ) ( )P X b P X a ( ) ( )F b F a
( ) ( )x
X XF x f u du
( ) ( )b af x dx f x dx
( )
b
af x dx
( )4. ( ) ( ).
dF xf x F x
dx
Remark: f(x) can be larger than 1.
Example 3
Example 3
(1 ) 0 1( )
0
kx x xf x
otherwise
1 ( )f x dx
1
0(1 )kx x dx
12 3
02 3
x xk
6
k
6k
6k
Example 3
6 (1 ) 0 1( )
0
x x xf x
otherwise
6k
0
0 0
( ) 6 (1 ) 0 1
1 1
x
x
F x u u du x
x
2 3
0 0
3 2 0 1
1 1
x
x x x
x
0
0.5
1
1.5
2
-1 0 1 2
x
f (x )
0
0.2
0.4
0.6
0.8
1
1.2
-1 0 1 2
F (x )
x
0
0.5
1
1.5
2
-1 0 1 2
x
f (x )
0
0.2
0.4
0.6
0.8
1
1.2
-1 0 1 2
F (x )
x
Example 3
6 (1 ) 0 1( )
0
x x xf x
otherwise
6k
2 3
0 0
( ) 3 2 0 1
1 1
x
F x x x x
x
0
0.5
1
1.5
2
-1 0 1 2
x
f (x )
0
0.2
0.4
0.6
0.8
1
1.2
-1 0 1 2
F (x )
x
0
0.5
1
1.5
2
-1 0 1 2
x
f (x )
0
0.2
0.4
0.6
0.8
1
1.2
-1 0 1 2
F (x )
x
Example 3
6 (1 ) 0 1( )
0
x x xf x
otherwise
6k
2 3
0 0
( ) 3 2 0 1
1 1
x
F x x x x
x
2 31 1 13 3 3( ) 3( ) 2( )P X
70.25926
27
1/3
0.25926
The Exponential Distributions
Chapter 4-1Continuous Random Variables
The Exponential Distributions
The following r.v.’s are often modelled as exponential:
1. Interarrival time between two successive job arrivals.
2. Service time at a server in a queuing network.
3. Life time of a component.
The Exponential Distributions
A r.v. X is said to possess an exponential distribution and to be exponentially distributed, denoted byX ~ Exp(), if it possesses the density
( ) , 0xf x e x
The Exponential Distributions
( ) , 0xf x e x
~ ( )X Exp
: arriving rate: failure rate
cdf1 0
( )0 0
xe xF x
x
The Exponential Distributions
( ) , 0xf x e x
~ ( )X Exp
: arriving rate: failure rate
cdf1 0
( )0 0
xe xF x
x
0
0.5
1
1.5
2
2.5
3
3.5
-2 0 2 4 6
x
f (x )
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-2 0 2 4 6
F (x )
x
0
0.5
1
1.5
2
2.5
3
3.5
-2 0 2 4 6
x
f (x )
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-2 0 2 4 6
F (x )
x
~ ( )2X Exp
Memoryless or Markov Property
~ ( )X Exp
( | ) ( )P X a b X a P X b 0
0
a
b
Memoryless or Markov Property
~ ( )X Exp
( | ) ( )P X a b X a P X b 0
0
a
b
( and )( | )
( )
P X a b X aP X a b X a
P X a
( )
( )
P X a b
P X a
( ) , 0xf x e x
~ ( )X Exp
cdf1 0
( )0 0
xe xF x
x
( ) , 0xf x e x
~ ( )X Exp
cdf1 0
( )0 0
xe xF x
x
1 ( )
1 ( )
P X a b
P X a
1 ( )
1 ( )
F a b
F a
( )a b
a
e
e
be
( )P X b
Memoryless or Markov Property
~ ( )X Exp
( | ) ( )P X a b X a P X b 0
0
a
b
Exercise:
連續型隨機變數中,唯有指數分佈具備無記憶性。
The Relation Between Poisson and Exponential Distributions : arriving rate
: failure rate
Nt
Let r.v. Nt denote #jobs arriving to a computer system in the interval (0, t].
t0
?~tN ( )P t
The Relation Between Poisson and Exponential Distributions
Let X denote the time of the next arrival.
: arriving rate: failure rate
The next arrival
Nt
Let r.v. Nt denote #jobs arriving to a computer system in the interval (0, t].
t0
X
?~tN ( )P t
?~X or ( ? ?( ) )f t F t
?~tN ( )P t
The Relation Between Poisson and Exponential Distributions
Let X denote the time of the next arrival.
: arriving rate: failure rate
The next arrival
Nt
Let r.v. Nt denote #jobs arriving to a computer system in the interval (0, t].
t0
X
?~X or ( ? ?( ) )f t F t
能求出 P(X > t)嗎 ?能求出 P(X > t)嗎 ?
( ) ( 0)tP X t P N
?~tN ( )P t
The Relation Between Poisson and Exponential Distributions
Let X denote the time of the next arrival.
: arriving rate: failure rate
The next arrival
Nt
Let r.v. Nt denote #jobs arriving to a computer system in the interval (0, t].
t0
X
?~X or ( ? ?( ) )f t F t
能求出 P(X > t)嗎 ?能求出 P(X > t)嗎 ?
( ) ( 0)tP X t P N
( ) ( 0)tP X t P N 0( )
0!
tt e
te
( ) 1 tF t e
( ) tf t e
0t
0t
0t
( ) , 0xf x e x
~ ( )X Exp
cdf1 0
( )0 0
xe xF x
x
( ) , 0xf x e x
~ ( )X Exp
cdf1 0
( )0 0
xe xF x
x
?~tN ( )P t
The Relation Between Poisson and Exponential Distributions
Let X denote the time of the next arrival.
: arriving rate: failure rate
Nt
Let r.v. Nt denote #jobs arriving to a computer system in the interval (0, t].
t0
X
~ ( )X Exp
The next arrival
The Relation Between Poisson and Exponential Distributions : arriving rate
: failure ratet1 t2 t3 t4 t5
The interarrival times of a Poisson process are exponentially distributed.
Example 50
10 secs
= 0.1 job/sec = 0.1 job/sec
P(“No job”) = ?
Example 50
10 secs
= 0.1 job/sec = 0.1 job/sec
P(“No job”) = ?
Method 1:
Method 2:
Let N10 represent #jobs arriving in the 10 secs.
Let X represent the time of the next arriving job.
10( ) ( 0)P P N N o "b" o j
( ) ( 10)P P X " "No job
10 ~ (1)N P
~ (0.1)X Exp
0 11
0!
e
1e
1 ( 10)P X 101 (1 )e
1e
The Reliability and
Failure Rate
Chapter 4-1Continuous Random Variables
Definition Reliability
Let r.v. X be the lifetime or time to failure of a component. The probability that the component survives until some time t is called the reliability R(t) of the component, i.e.,
R(t) = P(X > t) = 1 F(t)
Remarks:
1. F(t) is, hence, called unreliability.
2. R’(t) = F’(t) = f(t) is called the failure density function.
The Instantaneous Failure Rate
剎那間,ㄧ切化作永恆。
( | )t X tP X tt
The Instantaneous Failure Rate
0 t
t
t+t
( | )P t X t t X t
生命將在時間 t後瞬間結束的機率
The Instantaneous Failure Rate
( | )P t X t t X t
生命將在時間 t後瞬間結束的機率
and( )
( )
P t X t t X t
P X t
( )
( )
P t X t t
P X t
( ) ( )
( )
F t t F t
R t
( ) ( )( | )
( )
F t t F tP t X t t X t
R t
The Instantaneous Failure Rate
( | )P t X t t X t
瞬間暴斃率 h(t)
( ) ( )( | )
( )
F t t F tP t X t t X t
R t
0
( | )limt
P t X t t X t
t
( )h t
0
( ) ( )lim
( )t
F t t F t
tR t
( )
( )
F t
R t
( )
( )
f t
R t
The Instantaneous Failure Rate
瞬間暴斃率 h(t)
( )( )
( )
f th t
R t
Example 6
Show that the failure rate of exponential distribution is characterized by a constant failure rate.
Show that the failure rate of exponential distribution is characterized by a constant failure rate.
( ) , 0xf x e x
~ ( )X Exp
cdf1 0
( )0 0
xe xF x
x
( ) , 0xf x e x
~ ( )X Exp
cdf1 0
( )0 0
xe xF x
x
( )( )
( )
f th t
R t
( )
1 ( )
f t
F t
t
t
e
e
0t
以指數分配來 model物件壽命之機率分配合理嗎 ?
More on Failure Rates
t
h(t)
CFR ( ) 0h t
More on Failure Rates
t
h(t)
CFR ( ) 0h t
Useful Life
CFR ( ) 0h t
DFR
( ) 0h t IFR
( ) 0h t
More on Failure Rates
t
h(t)
CFR ( ) 0h t
Useful Life
CFR ( ) 0h t
DFR
( ) 0h t IFR
( ) 0h t
Exponential
Distribution
Exponential
Distribution? ?
Relationships among F(t), f(t), R(t), h(t)
( )F t
( )f t
( )R t
( )h t
1 ( )F t
( )dF t
dt
( )
( )
f t
R t
Relationships among F(t), f(t), R(t), h(t)
( )F t
( )f t
( )R t
( )h t
( )f t 1 ( )F t
( )
( )
R t
f t
Relationships among F(t), f(t), R(t), h(t)
( )F t
( )f t
( )R t
( )h t
( )
( )
f t
R t
( )dF t
dt
1 ( )R t
Relationships among F(t), f(t), R(t), h(t)
( )F t
( )f t
( )R t
( )h t
?
?
?
Cumulative Hazard
0( ) ( )
tH t h x dx 0
( )
( )
t f xdx
R x 0
( )
( )
t R xdx
R x
0
1( )
( )
tdR x
R x 0
ln ( )t
R x
ln ( ) ln (0)R t R ln1 ln ( )R t ln ( )R t
( )( ) H tR t e 0( )
th x dx
e
Relationships among F(t), f(t), R(t), h(t)
( )F t
( )f t
( )R t
( )h t
0( )
th x dx
e 1 ( )R t
( )dF t
dt
Example 7
0( ) ( )
tH t h x dx 0 0
txdx
2
0
02
tx
20 , 02
tt
20( ) exp , 02
tR t t
The Erlang Distrib
utions
Chapter 4-1Continuous Random Variables
我的老照相機與閃光燈
它只能使用四次每使用一次後轉動九十度使用四次後壽終正寢
它只能使用四次每使用一次後轉動九十度使用四次後壽終正寢
time
The Erlang Distributions
The lifetime of my flash (X)
I(X)=? fX(t)=?[0, )
The Erlang Distributions
Consider a component subjected to an environment so that Nt, the number of peak stresses in the interval (0, t], is Poisson distributed with parameter t.
Suppose that the rth peak will cause a failure. Let X denote the lifetime of the component. Then,
Nt ~ P(t)
?( )P X t ( )tP N r1
0
( )
!
k tr
k
t e
k
1
0
( )( ) 1 , 0
!
k tr
k
t eF t t
k
cdf
The Erlang Distributions
Consider a component subjected to an environment so that Nt, the number of peak stresses in the interval (0, t], is Poisson distributed with parameter t.
Suppose that the rth peak will cause a failure. Let X denote the lifetime of the component. Then,
Nt ~ P(t)
Exercise ofChapter 2
cdf
1
( ) , 0( 1)!
r r tt ef t t
r
1
0
( )( ) 1 , 0
!
k tr
k
t eF t t
k
The r-Stage Erlang Distributions
Consider a component subjected to an environment so that Nt, the number of peak stresses in the interval (0, t], is Poisson distributed with parameter t.
Suppose that the rth peak will cause a failure. Let X denote the lifetime of the component. Then,
cdf
1
( ) , 0( 1)!
r r tt ef t t
r
1
0
( )( ) 1 , 0
!
k tr
k
t eF t t
k
The r-Stage Erlang Distributions
cdf
1
( ) , 0( 1)!
r r tt ef t t
r
1
0
( )( ) 1 , 0
!
k tr
k
t eF t t
k
( , )X Erlang r
( ) (1, )Exp Erlang
The r-Stage Erlang Distributions
1
, 0( 1)!
( )r r tt e
tr
f t
( , )X Erlang r
( ) (1, )Exp Erlang
1
, 0( 1)!
( )r r tt e
tr
f t
1
, 0( 1)!
r r tt et
r
Example 8
In a batch processing environment, the number of jobs arriving for service is 9 per hour. If the arrival process satisfies the requirement of a Poisson experiment. Find the probability that the elapse time between a given arrival and the fifth subsequent arrival is less than 10 minutes.
In a batch processing environment, the number of jobs arriving for service is 9 per hour. If the arrival process satisfies the requirement of a Poisson experiment. Find the probability that the elapse time between a given arrival and the fifth subsequent arrival is less than 10 minutes.
Let X represent the time of the 5th arrival.
= 9 jobs/hr.
16( ?hr)P X
~ (5,9)X Erlang
1
( ) , 0( 1)!
r r xx ef t x
r
( , )X Erlang r
5 4 91/ 6
0
9
4!
xx edx
4
9/ 6
0
(9 / 6)1
!
k
k
ek
0.0285
The Gamma
Distributions
Chapter 4-1Continuous Random Variables
Review
1
( ) , 0( 1)!
r r xx ef t x
r
( , )X Erlang r
r為一正整數欲將之推廣為正實數
r為一正整數欲將之推廣為正實數
Review
1
( ) , 0( 1)!
r r xx ef t x
r
( , )X Erlang r
( )
0
The Gamma Distributions
1
( ) , 0( )
xx ef x x
( , ), 0X
Review1
( ) , 0( )
xx ef t x
( , )X
1
0( ) xx e dx
1. (1) 1
2. ( ) ( 1) ( )
3. ( ) ( 1)!, N
14.
2
1 12
( 1)!5.
2 2 !n n
n n
Chi-SquareDistributions
2X
1
( ) , 0( )
xx ef t x
( , )X
1 1,
2 2
2vX 1
,2 2
v
The Gaussian or
Normal Distributions
Chapter 4-1Continuous Random Variables
The Gaussian or Normal Distributions
德國的 10馬克紙幣 , 以高斯 (Gauss, 1777-1855)為人像 , 人像左側有一常態分佈之 p.d.f.及其圖形。
The Gaussian or Normal Distributions
2( ) ( ; , )f x n x pdf
2( , )X N 2
2
( )
21
2
x
e
x
The Gaussian or Normal Distributions
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
: mean : standard deviation2: variance
Inflectionpoint
Inflectionpoint
The Gaussian or Normal Distributions
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
: mean : standard deviation2: variance
15 varying
100 varying
The Gaussian or Normal Distributions
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
: mean : standard deviation2: variance
2 / 2 2xe dx
Facts:
2 / 211
2xe dx
2 2( ) / 211
2xe dx
The Gaussian or Normal Distributions
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
: mean : standard deviation2: variance
2 2( ) / 211
2xe dx
2( 9) /8 ?xe dx
2 /18
0?xe dx
2 2
3 2
2
Standard Normal Distribution
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
2( ) ( ; , )f x n x
2( , )X N
2
2
( )
21
2
x
e
x
( ) ( ;0,1)Zf z n z
(0,1)Z N
2 / 21
2ze
z
Table of N(0, 1)
( ) ( ;0,1)Zf z n z
(0,1)Z N
2 / 21
2ze
z
z
( ) ( )ZF z z2 / 21
2
z xe dx
Table of N(0, 1)
z
( ) ( )ZF z z2 / 21
2
z xe dx
( 1.6 ?25) 0.0521
(1.62 ?5) 0.9479
Fact: ( ) 1 ( )z z
Probability Evaluation for N(, 2)
x 2
2
( )
21( )
2
tx
XF x e dt
?
2
2
( )
21( )
2
x
f x e
x
2
2
( )
21
2
tt x
te dt
2 2
2
( )
2 2
t y
t
y
t y
Probability Evaluation for N(, 2)
x 2
2
( )
21( )
2
tx
XF x e dt
?
2
2
( )
21( )
2
x
f x e
x
2
2
( )
21
2
tt x
te dt
t y
2 / 21
2
y x y
ye dy
2 / 21
2
xye dy
x
?
Probability Evaluation for N(, 2)
x
2( , )X N
( )X
xF x
Z-Score: 表 距 離 中 心 若 干 個 標 準 差
Fact: (0,1)X
N
Example 9
X ~ N(12.00, 0.202)
1. (11.92 1 27 ?2. )P X
2. ( 12.45 ?)P X
3. ( 11.70 ?)P X
Example 9X ~ N(12.00, 0.202)
1. (11.92 1 27 ?2. )P X
2. ( 12.45 ?)P X
3. ( 11.70 ?)P X
(11.92 12.27) ( 12.27) ( 11.92)P X P X P X
12.27 12 11.92 12
0.2 0.2
1.35 0.40 0.9115 0.3446 0.5669
0.5669
Example 9X ~ N(12.00, 0.202)
1. (11.92 1 27 ?2. )P X
2. ( 12.45 ?)P X
3. ( 11.70 ?)P X
( 12.45) 1 ( 12.45)P X P X
12.45 121
0.2
1 0.9878 0.0122
0.5669
1 2.25
0.0122
Example 9X ~ N(12.00, 0.202)
1. (11.92 1 27 ?2. )P X
2. ( 12.45 ?)P X
3. ( 11.70 ?)P X
11.70 12( 11.70)
0.2P X
0.0668
0.5669
1.50
0.0122
0.0668
Example 10
|X | < |X | < 2|X | < 3
2( , )X N
Example 10
|X | < |X | < 2|X | < 3
|X | < |X | < 2|X | < 3
2( , )X N
(| | )P X k ( )P k X k
XP k k
( ) ( )k k
1 ( ) ( )k k
1 2 ( )k
Example 10
|X | < |X | < 2|X | < 3
|X | < |X | < 2|X | < 3
2( , )X N
(| | )P X k 1 2 ( )k
(| | )P X 1 2 ( 1)
(| | 2 )P X 1 2 ( 2)
(| | 3 )P X 1 2 ( 3)
1 2 0.1587 0.6826
1 2 0.0228 0.9544
1 2 0.0013 0.9974
(| | )P X k
Example 10
(| | )P X k 1 2 ( )k
(| | )P X 1 2 ( 1)
(| | 2 )P X 1 2 ( 2)
(| | 3 )P X 1 2 ( 3)
1 2 0.1587 0.6826
1 2 0.0228 0.9544
1 2 0.0013 0.9974
The Uniform
Distributions
Chapter 4-1Continuous Random Variables
The Uniform Distributions
~ ( , ), X U a b a bpdf
cdf
1( ) , f x a x b
b a
0
( )
1
a x
x aF x a x b
b ab x
a bx
f(x)
1
b a
a bx
F(x)1
Summary
The Exponential Distributions The Erlang Distributions The Gamma Distributions The Gaussian or Normal Distributions The Uniform Distributions