Download x Probability density y 0

Probability density
y
0
μ
x
Content
Content
‹
‹ Continuous
Continuous random
random variable
variable
‹
‹ Probability
Probability density
density function
function
‹
‹ Expectation
Expectation and
and variance
variance
‹
‹ Normal
Normal distribution
distribution
‹
‹ Characteristics
Characteristics of
of the
the normal
normal distribution
distribution
‹
‹ Applications
Applications of
of the
the normal
normal distribution
distribution
Continuous
Continuous Random
Random Variable
Variable
AArandom
randomvariable
variablewhich
whichcan
cantake
takeon
onany
anyvalue
valueover
overan
an
interval
intervalof
ofreal
realnumbers
numbersdefined
definedby
byits
itssample
samplespace.
space.
Examples
Examples
‹
‹
The
Thelifetime
lifetimeof
ofaacertain
certainbrand
brandof
ofbatteries
batteries
‹
‹
The
Theheights
heightsof
ofboys
boysof
ofaaparticular
particularage
age
‹
‹
The
Theweights
weightsof
ofboxes
boxesof
ofaacertain
certainbrand
brandof
ofeggs
eggs
‹
‹
The
Thevolume
volumeof
ofmilk
milkpacked
packedin
inboxes
boxes
Activity
Activity 11
1.
1.
2.
2.
3.
3.
4.
4.
Identify
Identifywhich
whichof
ofthe
thefollowing
followingis/are
is/arecontinuous
continuousrandom
random
variables.
variables.
The
Thenumber
numberof
ofgoals
goalsscored
scoredin
inaafootball
footballmatch.
match.
The
Thewaiting
waitingtime
timefor
foraabus
bus by
bypassengers
passengersat
ataabus
busstop.
stop.
The
Theposition
positionwhere
wherethe
theroulette
roulettewheel
wheelstops.
stops.(Fig.
(Fig.1)
1)
The
Thesector
sectornumber
numberat
atwhich
whichthe
theroulette
roulettewheel
wheelstops.
stops.(Fig.
(Fig.
2)
2)
1
Fig. 1
2
Fig. 2
4
3
Probability
p.d.f.)
Probability density
density function
function ((p.d.f.)
The
Theprobability
probabilitydistribution
distributionof
ofaacontinuous
continuousrandom
randomvariable
variable XX
can
canbe
bedescribed
describedby
byits
itsp.d.f..
p.d.f..
IfIfthere
thereisisaareal-valued
real-valuedfunction
function ffsuch
suchthat
thatfor
forevery
everyinterval
interval AA
of
ofreal
realnumbers,
numbers,the
theprobability
probabilitythat
that XXtakes
takesaavalue
valuein
in AAisisthe
the
integral
integralof
offfover
overA,
A,then
thenffisiscalled
calledthe
thep.d.f.
p.d.f.of
ofX.
X.
Probability
p.d.f.)
Probability density
density function
function ((p.d.f.)
For
Foran
aninterval
intervalaa≤≤xx≤≤b,
b,the
theprobability
probabilitythat
thatXXtakes
takeson
onaavalue
valuein
inthe
the
interval
intervalisisgiven
givenby
by
b
Probability density
y
P( a ≤ x ≤ b) = ∫ a f ( x) dx
Note.
0
y = f (x)
a
b
x
1. P( X = a ) = P( X = b) = 0
2. P(a ≤ X ≤ b) = P(a < X ≤ b) = P(a ≤ X < b) = P(a < X < b)
Conditions
Conditions of
of p.d.f
p.d.f..
1.
1. ff (x)
(x) ≥≥ 00 for
for all
all real
real numbers
numbers x.
x.
2.
2.
∞
∫ − ∞ f ( x ) dx = 1
Learning
Learning Difficulty
Difficulty 11
Given X is a continuous random
Given T is a discrete random
variable and the probability
variable and the probability
density function of X is given by
function of T is given by
f ( x) = 2, 0.5 ≤ x ≤ 1
g (t ) = 0.2, for t = 1, 2, 3, 4, 5
y
0.2
f (x) = 2
2
Probability
Probability density
y
1.5
1
0.1
0.5
0
0.5
1
x
0
t
1
2
3
4
5
Probability
p.d.f.)
Probability Density
Density Function
Function ((p.d.f.)
The
Thefrequency
frequencydistribution
distributionof
ofaacontinuous
continuousvariable
variablecan
canbe
be
represented
representedgraphically
graphicallyby
byaahistogram.
histogram.
Frequency density
The frequency of each class is
represented by the area of each
rectangle.
0
f3
f2
f4
f1
a
f5
h
k
f6
b
x
the class frequency = the frequency density of the class × class width
the class frequency
frequency density of a class =
the class width
When
Wheneach
eachclass
classfrequency
frequency(f(fi)i)isisdivided
dividedby
bythe
thetotal
totalfrequency
frequency
(N),
(N),the
thehistogram
histogramshows
showsthe
therelative
relativefrequency
frequencydistribution
distributionof
of
Relative
frequency density
the
thecontinuous
continuousrandom
randomvariable.
variable.
0
f3/N
f2/N
f4/N
f1/N
a
f5/N
h
k
f6/N
b
x
As
AsNNbecomes
becomesvery
verylarge,
large, fi
N
= pi
where
theprobability
probabilityof
ofthe
theoccurrence
occurrenceof
ofclass
classi ioutcome(s)
outcome(s)of
of
whereppi iisisthe
the
thecontinuous
continuousrandom
randomvariable.
variable.
The
Thehistogram
histogramthen
thenshows
showsthe
theprobability
probabilitydistribution
distributionof
ofthe
the
Probability density
continuous
continuousrandom
randomvariable.
variable.
0
p2
p3
p4
p1
a
p5
h
k
p6
b
x
IfIfthe
thewhole
wholerange
rangeof
ofxxvalues
valuesisisdivided
dividedinto
intoan
aninfinitely
infinitelylarge
large
number
number(n)
(n)of
ofsmall
smallintervals,
intervals,the
theprobability
probabilitydistribution
distributionof
ofthe
the
continuous
continuousrandom
randomvariable
variablecan
canbe
beshown
shownby
byaagraph
graphconsisting
consisting
Probability density
of
ofaacontinuous
continuouscurve
curveabove
abovethe
the x-axis.
x-axis.
0
a
h
k
b
x
Learning
Learning Difficulty
Difficulty 22
A roulette wheel is rotated about O from A in an anticlockwise
direction and stops at B.
X is a random variable representing the length of arc AB
measured from A to B in clockwise direction.
P( X = l ) = 0
l
A
B
O
r
Learning
Learning Difficulty
Difficulty 22
Given
Given XXisisaacontinuous
continuousrandom
randomvariable,
variable,ff(x)
(x)isisthe
thep.d.f.
p.d.f.of
ofXX
and
andccisisaareal
realconstant.
constant.
0 ≤ P( X = c) ≤ P(c ≤ X ≤ c + Δx)
c + Δx
0 ≤ P( X = c) ≤ ∫ c
f ( x ) dx
Due
Dueto
toboundedness
boundednessof
ofthe
theintegratible
integratiblefunction,
function,
c + Δx
∫
Δx →0 c
0 ≤ P( X = c) ≤ lim
f ( x ) dx = 0
Conclusion.
The event impossible to occur ⇒ P(Event) = 0
P(Event) = 0 ⇒ Event impossible to occur ?
Expectation
Expectation and
and Variance
Variance
IfIf XX is
is aa discrete
discrete random
random IfIf XX is
is aa continuous
continuous
variable,
random
variable,
random variable,
variable,
∞
E( X ) = ∫ x f ( x) dx
E( X ) = ∑ x f ( x)
−∞
x
∞
Var( X ) = ∑ ( x − μ ) f ( x) Var( X ) = ∫ ( x − μ ) f ( x) dx
2
x
Var( X ) = ∑ x f ( x) − μ
2
x
Var( X ) = E ( X 2 ) − μ 2
2
−∞
∞
2
= ∫ x f ( x ) dx − μ
2
−∞
Var( X ) = E ( X 2 ) − μ 2
2
Expectation
Expectation and
and Variance
Variance
The following properties hold
E(a X + b) = a E( X ) + b
Var (a X + b) = a 2 Var ( X )
Normal
Normal Distribution
Distribution
AAcontinuous
continuousrandom
randomvariable
variable XXhas
hasaanormal
normaldistribution
distributionififthe
the
p.d.f.
p.d.f.of
ofXXisis
1
f ( x) =
σ 2π
( x−μ )2
−
2
2
σ
e
,
−∞ < x < ∞
where
andσσ are
areconstants
constantsand
andσσ>>0.
0.
where μμand
ff(x)
andσ.
σ.
(x)has
hastwo
twoparameters
parameters μμand
The
Thenormal
normaldistribution
distributionwith
withthese
theseparameters
parameterscan
canbe
bedenoted
denoted
22).
by
N(
μ
,
σ
by N(μ, σ ).
Importance
Importance of
of the
the Normal
Normal Distribution
Distribution
1.
1. The
Thenormal
normaldistribution
distributioncan
canbe
beused
usedas
asan
anapproximation
approximationto
to
some
somediscrete
discreteprobability
probabilitydistributions
distributionssuch
suchas
asbinomial
binomial
distribution
distributionand
andPoisson
Poissondistribution.
distribution.
Example
Example
An
Anunbiased
unbiasedtetrahedral
tetrahedraldie
diewith
withface
facemarked
marked1,
1,2,
2,33and
and44isis
thrown
thrown400
400times.
times. Find
Findthe
theprobability
probabilityof
ofobtaining
obtainingless
lessthan
than90
90
ones.
ones.
Let
LetXXbe
beaarandom
randomvariable
variablerepresenting
representingthe
thenumber
numberof
ofones
ones
obtained
obtainedin
in400
400throws.
throws.
1
Then
Then
X ~ Bin (400, )
4
Importance
Importance of
of the
the Normal
Normal Distribution
Distribution
Method
Method11
Using
Usingthe
thebinomial
binomialdistribution,
distribution,
P(X
P(X<<90)
90)==0.1117
0.1117
Method
Method22
Using
Usingthe
thenormal
normalapproximation
approximationto
tothe
thebinomial
binomialdistribution,
distribution,
1
X ~ Bin (400, )
4
X ~ N(100, 75)
89.5 - 100
P( X < 90) = P( X < 89.5) = P(Z <
)
75
P( Z < −1.212) = 0.1127
Importance
Importance of
of the
the Normal
Normal Distribution
Distribution
2.
2. Central
CentralLimit
LimitTheorem
Theorem
The
Thesample
samplemean
meanof
ofaalarge
largerandom
randomsample
sampleof
ofrandom
randomvariables
variables
2
with
andfinite
finitevariance
varianceσσ2has
hasapproximately
approximatelyaanormal
normal
withmean
meanμμand
2
distribution
distributionwith
withmean
meanμμand
andvariance
varianceσσ2//n.
n.
The
Theproof
proofcan
canbe
bedone
doneby
byeither
eitherusing
usingmoment
momentgenerating
generatingfunction
function
or
orprobability
probabilitydistribution
distributionfunction.
function. The
Thedetails
detailsof
ofthe
theproofs
proofscan
canbe
be
seen
seenon
on
Alexander
AlexanderM.
M.Mood,
Mood,Franklin
FranklinA.
A.Graybill
Graybilland
andDuane
DuaneC.
C.Boes
Boes(1974).
(1974).
rdrdEdition) (pp.234-235).
Introduction
to
the
Theory
of
Statistics
(3
Introduction to the Theory of Statistics (3 Edition) (pp.234-235).
Mc-Graw-Hill.
Mc-Graw-Hill.
第137頁。劉次華、萬建平
第137頁。劉次華、萬建平(1999)。概率論與數理統計。北京：施普
(1999)。概率論與數理統計。北京：施普
林格出版社。
林格出版社。
Activity
Activity 22 –– Normal
Normal Distribution
Distribution
Example
Example11––Are
Areyou
youfit?
fit?
Source
Source
http://www.hked-stat.net/common/activity_as_2.htm
http://www.hked-stat.net/common/activity_as_2.htm
Worksheet
Worksheet
Word
Worddocument
documentversion
version
pdf
pdfversion
version
BMI
BMICalculator
Calculator
http://www.ryvita.co.uk/calculatebmi.html
http://www.ryvita.co.uk/calculatebmi.html
Activity
Activity 22 –– Normal
Normal Distribution
Distribution
Example
Example22––Pieces
Piecesof
ofstring
string
Frequency
Look
Lookat
at10
10cm
cmon
onaaruler
rulerand
andthen
thentake
takeaaball
ballof
ofstring.
string. Every
Every
student
studentcuts
cutsaapiece
pieceof
ofstring
stringof
of10
10cm
cmfrom
fromthe
theball
ballof
ofstring
stringby
by
guessing.
guessing. Measuring
Measuringthe
thelengths
lengthsof
ofall
allthe
thepieces
piecesin
inmm.
mm.
0 1 2 3 4 5 6 7 8 9 10 11 cm
0
Length
Notes
Notes on
on Activity
Activity 22
The
Thefollowing
followingissues
issuescan
canbe
bediscussed
discussedafter
afterthe
theactivity.
activity.
1.
1. Describe
Describethe
thedistribution
distributionof
ofdata.
data.
2.
2. Is
Isititsymmetrical
symmetricalabout
aboutthe
themean?
mean? IfIfnot,
not,are
arethere
thereany
any
factors
factorscontributing
contributingto
tothe
theshape
shapeof
ofthe
thedistribution?
distribution?
3.
3. Can
Canwe
wemanipulate
manipulatethe
thedata
dataso
sothat
thatthese
thesefactors
factorscan
canbe
be
removed
removedafter
aftermanipulation?
manipulation?IfIfyes,
yes,how?
how?IfIfnot,
not,why?
why?
4.
4. Describe
Describethe
thedistribution
distributionof
ofthe
themanipulated
manipulateddata.
data.
History
History of
of the
the Normal
Normal Distribution
Distribution
Abraham De Moivre
(1667 – 1754)
The normal curve was developed mathematically in 1733 by De Moivre
as an approximation to the binomial distribution. His paper was not
discovered until 1924 by Karl Pearson.
Pierre Simon Laplace
(1749 – 1827)
Laplace used the normal curve in 1783 to describe the distribution of
errors. In 1810, he proved the Central Limit Theorem.
History
History of
of the
the Normal
Normal Distribution
Distribution
Johann Carl Friedrich Gauss
(1777 – 1855)
Gauss used the normal curve to analyze astronomical data in 1809. The
normal curve is often called the Gaussian distribution.
In Germany, the portrait of Gauss, the normal curve and its p.d.f. have
been put on their 10 Deutschmark Note.
Properties
Properties of
of the
the Normal
Normal Distribution
Distribution
1.
1. ff (x)
(x) >> 00 for
for all
all values
values of
of x.
x.
2.
2.
∞
∫ − ∞ f ( x ) dx = 1
3.
3. The
The parameters
parameters μμ and
and σσ of
of the
the normal
normal distribution
distribution
are
are the
the mean
mean and
and the
the standard
standard deviation
deviation of
of the
the
distribution
distribution respectively.
respectively.
Expected Value of the Normal Distribution
A random variable X follows a normal distribution with mean  and variance
2, find E(X).
f ( x) 
1
Let z 
E ( x) 


 x   2
e
 2
E( X )  
E ( x) 

2 2
,
 ∞ x  ∞
x f ( x ) dx
x

1
2
1
2
, we get
∞
∫ (z   )e
∞
∞
∫ ze
∞
z2

2
z2

2
dz
dz  
1
2
z2
∞ 
2
∫e
∞

dz 
2
2  
Variance
Variance of
of the
the Normal
Normal Distribution
Distribution
AArandom
randomvariable
variableXXfollows
followsaanormal
normaldistribution
distributionwith
withmean
meanμμand
andvariance
variance
σσ22,,find
findE(X).
E(X).
Var( X ) = ∫
Let z =
∞
2
(
x − μ)
−∞
x−μ
, we get
σ
σ2
Var ( X ) =
2π
σ2
Var ( X ) =
2π
Var ( X ) = σ 2
1
e
σ 2π
∫
z2
−
∞
2
2
z
e
−∞
2π = σ 2
dz
(
x − μ )2
−
2σ 2
dx
Properties
Properties of
of the
the Normal
Normal Distribution
Distribution
Normal curve is symmetrical
- two halves identical -
Tail
Tail
Theoretically, the
curve extends to -∞
Mean, median, and
mode are equal
Theoretically, the
curve extends to +∞
Properties
Properties of
of the
the Normal
Normal Distribution
Distribution
The
The location
location of
of the
the normal
normal distribution
distribution is
is determined
determined
by
by μμ ..
probability density
Normal
Normaldistributions
distributionswith
withdifferent
different
means
meansbut
butequal
equalstandard
standarddeviation.
deviation.
X1 ~ N(μ1, σ2)
X2 ~ N(μ2, σ2)
X3 ~ N(μ3, σ2)
μμ22>>μμ11>>μμ33
μ3
μ1
μ2
x
Properties
Properties of
of the
the Normal
Normal Distribution
Distribution
The
The dispersion
dispersion of
of the
the normal
normal distribution
distribution is
is
determined
determined by
by σσ..
Normal
Normaldistributions
distributionswith
withequal
equalmean
mean
but
butdifferent
differentstandard
standarddeviations.
deviations.
probability density
X1 ~ N(μ, σ12)
X2 ~ N(μ, σ22)
X3 ~ N(μ, σ32)
μ
x
σσ33>>σσ22>>σσ11>>00
Properties
Properties of
of the
the Normal
Normal Distribution
Distribution
X ~ N(μ, σ2)
68%
95%
99.7%
μ −3σ μ − 2σ
μ−σ
μ
μ+σ
μ + 2σ
μ + 3σ
Standardization
Standardization of
of aa Normal
Normal Variable
Variable
The
Thestandard
standardnormal
normaldistribution
distributionisisaanormal
normaldistribution
distributionwith
with
mean
mean00and
andvariance
variance1.
1.
Standardization
Standardization of
of aa Normal
Normal Variable
Variable
The
Thestandard
standardnormal
normaldistribution
distributionisisaanormal
normaldistribution
distributionwith
with
mean
mean00and
andvariance
variance1.
1.
X −μ
2
2
,,then
IfIfXX~~NN((μμ,,σσ ))and
and Z =
then
σ
E( Z ) = E(
E( Z ) =
E( Z ) =
1
σ
1
σ
E( Z ) = 0
X −μ
σ
)
[E( X ) − E(μ )]
(μ − μ )
Var ( Z ) = Var (
Var(Z ) =
Var ( Z ) =
1
σ
2
1
X −μ
[Var( X − μ )]
Var( X )
σ
σ2
Var(Z ) = 2 = 1
σ
2
σ
)
Example
Example
The life time of the batteries for a notebook
computer under normal usage is normally
distributed with mean 210 minutes and
standard deviation 15 minutes.
a) What percentage of these batteries will have a
life time between 195 minutes and 240
minutes?
b) What percentage of batteries will have a life
time less than 180 minutes?
(a)
(a) Let
LetXXbe
bethe
thelifetime
lifetimeof
ofthe
thebatteries
batteriesfor
foraanotebook
notebook
computer.
computer.
2
XX~~NN(210,
(210,15
152))
P(195
P(195≤≤XX≤≤240)
240)==??
x
150
165
180
195 210 225
240
255 270
Table
Table of
of the
the Standard
Standard Normal
Normal Distribution
Distribution
Area
Area under
under the
the Standard
Standard Normal
Normal Curve
Curve
zz
0.00
0.00
0.01
0.01
0.02
0.02
0.03
0.03
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.0
0.0
0.1
0.1
0.2
0.2
0.3
0.3
0.4
0.4
0.0000
0.0000
0.0398
0.0398
0.0793
0.0793
0.1179
0.1179
0.1554
0.1554
0.0040
0.0040
0.0438
0.0438
0.0832
0.0832
0.1217
0.1217
0.1591
0.1591
0.0080
0.0080
0.0478
0.0478
0.0871
0.0871
0.1255
0.1255
0.1628
0.1628
0.0120
0.0120
0.0517
0.0517
0.0910
0.0910
0.1293
0.1293
0.1664
0.1664
0.0160
0.0160
0.0557
0.0557
0.0948
0.0948
0.1331
0.1331
0.1700
0.1700
0.0199
0.0199
0.0596
0.0596
0.0987
0.0987
0.1368
0.1368
0.1736
0.1736
0.0239
0.0239
0.0636
0.0636
0.1026
0.1026
0.1406
0.1406
0.1772
0.1772
0.0279
0.0279
0.0675
0.0675
0.1064
0.1064
0.1443
0.1443
0.1808
0.1808
0.0319
0.0319
0.0714
0.0714
0.1103
0.1103
0.1480
0.1480
0.1844
0.1844
0.0359
0.0359
0.0753
0.0753
0.1141
0.1141
0.1517
0.1517
0.1879
0.1879
0.5
0.5
0.6
0.6
0.7
0.7
0.8
0.8
0.9
0.9
0.1915
0.1915
0.2257
0.2257
0.2580
0.2580
0.2881
0.2881
0.3159
0.3159
0.1950
0.1950
0.2291
0.2291
0.2611
0.2611
0.2910
0.2910
0.3186
0.3186
0.1985
0.1985
0.2324
0.2324
0.2642
0.2642
0.2939
0.2939
0.3212
0.3212
0.2019
0.2019
0.2357
0.2357
0.2673
0.2673
0.2967
0.2967
0.3238
0.3238
0.2054
0.2054
0.2389
0.2389
0.2704
0.2704
0.2995
0.2995
0.3264
0.3264
0.2088
0.2088
0.2422
0.2422
0.2734
0.2734
0.3023
0.3023
0.3289
0.3289
P(0 ≤ z ≤ 1.00)
0.2123
0.2123
0.2454
0.2454
0.2764
0.2764
0.3051
0.3051
0.3315
0.3315
0.2157
0.2157
0.2486
0.2486
0.2794
0.2794
0.3078
0.3078
0.3340
0.3340
0.2190
0.2190
0.2517
0.2517
0.2823
0.2823
0.3106
0.3106
0.3365
0.3365
0.2224
0.2224
0.2549
0.2549
0.2852
0.2852
0.3133
0.3133
0.3389
0.3389
1.0
1.0
1.1
1.1
1.2
1.2
1.3
1.3
1.4
1.4
0.3413
0.3413
0.3643
0.3643
0.3849
0.3849
0.4032
0.4032
0.4192
0.4192
0.3438
0.3438
0.3665
0.3665
0.3869
0.3869
0.4049
0.4049
0.4207
0.4207
0.3461
0.3461
0.3686
0.3686
0.3888
0.3888
0.4066
0.4066
0.4222
0.4222
0.3485
0.3485
0.3708
0.3708
0.3907
0.3907
0.4082
0.4082
0.4236
0.4236
0.3508
0.3508
0.3729
0.3729
0.3925
0.3925
0.4099
0.4099
0.4251
0.4251
0.3531
0.3531
0.3749
0.3749
0.3944
0.3944
0.4115
0.4115
0.4265
0.4265
0.3554
0.3554
0.3770
0.3770
0.3962
0.3962
0.4131
0.4131
0.4279
0.4279
0.3577
0.3577
0.3790
0.3790
0.3980
0.3980
0.4147
0.4147
0.4292
0.4292
0.3599
0.3599
0.3810
0.3810
0.3997
0.3997
0.4162
0.4162
0.4306
0.4306
0.3621
0.3621
0.3830
0.3830
0.4015
0.4015
0.4177
0.4177
0.4319
0.4319
1.5
1.5
1.6
1.6
1.7
1.7
1.8
1.8
0.4332
0.4332
0.4452
0.4452
0.4554
0.4554
0.4641
0.4641
0.4345
0.4345
0.4463
0.4463
0.4564
0.4564
0.4649
0.4649
0.4357
0.4357
0.4474
0.4474
0.4573
0.4573
0.4656
0.4656
0.4370
0.4370
0.4484
0.4484
0.4582
0.4582
0.4664
0.4664
0.4382
0.4382
0.4495
0.4495
0.4591
0.4591
0.4671
0.4671
0.4394
0.4394
0.4505
0.4505
0.4599
0.4599
0.4678
0.4678
0.4406
0.4406
0.4515
0.4515
0.4608
0.4608
0.4686
0.4686
0.4418
0.4418
0.4525
0.4525
0.4616
0.4616
0.4693
0.4693
0.4429
0.4429
0.4535
0.4535
0.4625
0.4625
0.4699
0.4699
0.4441
0.4441
0.4545
0.4545
0.4633
0.4633
0.4706
0.4706
X − 210
Let
Let Z = 15
P(195
P(195≤≤XX≤≤240)
240)
==
P(-1
P(-1≤≤ZZ≤≤2)
2)
==
0.3413
0.3413++0.4772
0.4772
==
0.8185
0.8185
x
150 165 180 195 210 225 240 255 270
z
-4
-3
-2
-1
0
1
2
3
4
(a)
(a) Let
LetXXbe
bethe
thelifetime
lifetimeof
ofthe
thebatteries
batteriesof
ofaanotebook
notebook
computer.
computer.
22)
XX~~NN(210,
15
(210, 15 )
X − 210
Let
Let Z =
15
P(195
P(195≤≤XX≤≤240)
240)
== P(-1
P(-1≤≤ZZ≤≤2)
2)
== 0.3413
0.3413++0.4772
0.4772
== 0.8185
0.8185
(b)
(b) P(X
P(X≤≤180)
180)==??
x
150
165
180
195 210 225
240
255 270
References
References
rdrd
‹
DeGroot,
M.H.,
&
Schervish,
M.J.
(2002).
Probability
and
statistics
(3
‹ DeGroot, M.H., & Schervish, M.J. (2002). Probability and statistics (3
ed.).
ed.). New
NewYork:
York:Addison-Wesley.
Addison-Wesley.
‹
‹ Hald,
Hald,A.
A.(1990).
(1990). AAhistory
historyof
ofprobability
probabilityand
andstatistics
statisticsand
andtheir
their
applications
applicationsbefore
before1750.
1750. New
NewYork:
York:John
JohnWiley
Wiley&&Sons.
Sons.
‹
‹ James
JamesW.
W.Tankard,
Tankard,Jr.
Jr. The
TheStatistical
StatisticalPioneers.
Pioneers. Massachusetts:
Massachusetts:
Schenkman
SchenkmanPublishing
PublishingCompany,
Company,Inc.
Inc.
‹
‹ 劉次華、萬建平
劉次華、萬建平(1999)。概率論與數理統計。北京：施普林格出版社。
(1999)。概率論與數理統計。北京：施普林格出版社。
‹
‹ Normal
NormalDistribution
Distribution
http://www.stat.wvu.edu/SRS/Modules/Normal/normal.html
http://www.stat.wvu.edu/SRS/Modules/Normal/normal.html
‹
‹ The
TheMacTutor
MacTutorHistory
Historyof
ofMathematics
Mathematicsarchive
archive
http://turnbull.mcs.st-and.ac.uk/~history/
http://turnbull.mcs.st-and.ac.uk/~history/
References
References
‹
‹ LiLiKam-yuk
Kam-yukand
andChow
ChowWai-keung
Wai-keung(2002).
(2002). New
NewWay
WayMathematics
Mathematics&&
Statistics
Statisticsfor
forHong
HongKong
KongAS-Level
AS-Level(2nd
(2ndEdition).
Edition). Hong
HongKong:
Kong:SNP
SNP
MANHATTAN
MANHATTANPRESS
PRESS(H.K.)
(H.K.)LTD.
LTD.