ページ3:

f(x)
76200
0,5
04-
03-
02-
A negand 01 +
A=0.7
Section 2.7 The Poisson Distribution 95
f(x)
0.5+
入 = 1.3
golindo.0.4-
0.3
1x
02
oldano al 01-
Two 20
0
1
2
3
4
6 P
0
1
boweit
2
3 4
5
6
ni hod
f(x)
$10f(x)
0.16 -
入=6.5
0.16+
入=10.5
da 0.14-
730.0- 0.10
0.14
A
0.12
adtoonsbl0.12
big dibe and
0.10
Site 0.08
1001 or
0.08
0.06
idnaib no
0.06 -
0.04.
0.04
0.02.
0.02
x
0 2 4 6 8 10 12 14 16 18 20
024 6 8 10 12 14 16 18 20
Figure 2.7-1 Poisson probability histograms
Remarks
1. Note that the mgf of X is
00
M (t) = E [e²x] = 2 + 1 * ^
x=0
etx
Thus, M'(t)=e^(e²-1)
x!
x
(λe²)
=
= e
x=0 x!
-λ ^e² 1 (e²-1)
e
= e
and M"(t) = (^e²)² e^(e²-1)+(e²)
Hence, We have μ = M'(0) = λ
and 0²= M"(0) - (M²(01)² = ( 2²+ 1) − 2² = 2.
-
2. Poisson distribution can he used to approximate probabilities for a
binomial distribution (as n
22-2
large)
P(X =x) = *** ~ (*) (*)* (1-4)**
x!
Let p = x/n, then (mp)²="p ~ (") px (1-p) "-x
.
x!
e
as
n
lange..

ページ4:

Example 2.7-4
In a
average of
two
large city, telephone calls to 911 come on the
every 3 mmutes. If one assumes an approximate Poisson process.
What is the probability of five or more calls arriving in a 9 mins.
period?
Let X denote the number of calls in a 9 minute period.
Then 22.3 = 6.
Hence, PCX ) = 1- P(X ≤4) ·
= 1 -
$ 676
x=0
= 1 - 0.285 = 0.715.
Example 2.7-5
box
A manufacturer of Christmas tree light bulbs knows that 2% of
its bulbs are defective. Assume the number of defective bulbs in
100 bulbs has a binomial distribution with parameters n = 100 and p = 0.02.
PIX < 3) =
Note that
3
100) (0.02) (0.98)
x
100-x
= 100-(0,02) = 2.
x=0
λ = np
3
ㄩㄩˋˊ (100) (0.02)* (0.98)
100-x
x=0
=
?
3
2 e
2e2
=
:0.857.
~ㄩ
x=0 x!
a
of

ページ5:

鄰魯文化、教師機
Table The Poison Destrib
0.20
1.01
Append 509
08
42
4.4
46
48
0.15
50
Poisson, A-3.8
52
06
010-
Poisson-3.8
0.015
0.012
0.010
0.00
54
0
F
0.078
0.066
0.056
600
56
58
0048
04
0.210
0.185
0.05
0305 0.359 0.33%
0.163
0143
0.040
6006
60
0.590
0.551
0294
0.125
0014
6005
0.004
6309
6009
0004
0005
0002
0.513
02
4
0.476
0265
0.238
6095
0021
0017
0440
0.753
0.720 0.686
0651
0406
0213
6072
0062
0373
0170
0151
5
0.867
0.844
0.818
0.791
0516
0342
6
10
12
10
12
0936 0.921
0905
0.38
0762
0.581
0313
0.285
6730
0546
0512
0.478
0.972
0.964
0967
0446
0.955
0944
0.989
0.985
0932
0.845
0670
04
0822
0.606
0797
0.980
F(x) = P(X)=
9
0.975
0968
0908 0903
771 0744
10
0.996
0.994
0.992
0.960
06
QM
0347
0951
0.990
0.999
11
0.998 0.997
0.996
0941
0929
0916
0.996 0.995
0.982
0.977
A-E(X)
12
1,000
0.999
0.999
0.999
0.993
0972
0.965
0.957
1000
1.000
1,000
0.998
0.990
0.1
02
03
104.
0.5
0.6
0.7
13
1000
0997
0988 0984
0980
0.999
0.996
0.995
08
09
1000
1,000
14
1.000
0.999
0.993
0991
1000
0.999
18
1000
0.998
0997
0.996
1000
0.999
0
0905
0819
0.741
0.670
0.607
0.549
0.497
0.449
4
15
1.000
1,000
0.999
0999
1000
1000
0.999
1000
1
0.995
0982
0.963
0908
0.910
0.878
0.844
0.809
8.3
772
16
1.000
1000
1000
1000
1000
L000
1000
1000 1,000
0.999
2
1000
0.999
0.996
0.992
0986
0.977
0.966
0.953
0.937
4.T
1.000
1000
L000
1000
1000
L000
1.000
1.000
0.999
0.998
0997
0.994
0.991
0987
0920
65
20
25
1000
1000
1000
1,000
1000.
1000
0.999
0.999 0.998
80
x
85
9.0
95
100
105
1.P
11.0
S
1.000
1.000
1.000
1,000
1,000
1000
1.000
1.000
6
1.000
1,000
1,000
1.000
1,000
1,000
1000
1000
1000
0.002
1000
6.999
0
10.001 0.001 0.000 0.000
0.000
0.000
1.000
1
0.011
10.007
0.005
0.003
0.002
0.043
0.001
0.001
0.030
0.020
0.014
0.009
x
1.1
1.2
13
14
15
16
1.7
18
19
0.112
0.006
0.004 0.003
0.082
0.059
0.042
0.000 0.000 0.000
0.000
0.000
0.000
0.002 0.001
0.030
20
0.021
0.015
0010 0.007
6
CANZA Sera
4
0.224
0.005
0.173
0.132
0.100 0.074
0.065
0.333
0301
10.273
0.247
10.223
-0.202
10.183
0.165
0.040
0.029 0.021
0.015
0.150
1
0699
0.663
0.627
0.592
0.558
0.525
0.493
0.463
044
0135
5
10.369
0.301 10.241
0.191
0150
0.116
0.089 0.067
0.050
0038
2
6406
0,900
0.879
0857
0.833
0.809
0.783
0.757
0.731
01704
6
0.527
0,450
0.378
0313
0.25%
0.207
0.365 0.130
0.102
0.079
0.974
0.966
0.957
0.946
0.934
0.921
0.907
0.891
075
047
0.673
0.599
0.525
0.453 0.386
0.324
0.369
0.220
0.179
0.143
0.995
0.992
0.989
0986
0.981
0.976
0.970
0964
0956
0.792
0.729
0.662
0.593
047
0.877
0.999
0.998
0.998
0.997
0.996
0.994
0.992
0990
0.830 0.776
0.987 M3
1.000
1000
1,000
0.999
0.999
0.999
0.998
0.997
0997
0995
10
0933
0.901
0.862
1,000
1.000
1.000
1.000
1.000
1000
1.000
0.999
11
0.999
0.966
0.947
0.921
0.816
0.888
0.523
0.717 0.683 0.587
0.763 0.706
10.849 0.803
0.456 0.392 0.333
0.279
0.232
0.522 0.458
10.397
0.341
0.645
0.583
0521
0460
0752 0.697
0.639
0.579
0.999
8
1000
1.000
1,000
1.000
1.000
1.000
1000
1.000
1000
1000
12
0.984
13
0.993
0.987 0.978 0.966 0949 -0.926
0.973 0.957 0936 0.909 0.876 0.836
0.898
0.792
0.742
0.864
0825
0.781
x
22
24
2.6
12.8
30
3.2
34
36
13.8
14
0.997
0.994 0.990
0.983 0.973 0.959 0.940 0.917 0.888 0854
40
0
0.111
0,091
0.074
0,061
0.050
0.041
0.033
0027
15
0.999
0.998
0.995
0.992 0.986 0978 0.967 0.951 0.902
0907
0.022
0018
16
1,000
0.999
0.998
0.996
0.993 0989
0982
0.973
0.960
0.944
0.355
0.308
0.267
0.231
0.199
0.171
0.347
0.126
0.107
0092
1000
1000
0.999
0.998 0.997
0.995
0.991
0.996
0.978
0.968
2
0.623
0.570
0.518
0.469
10.423
0.380
10.340
0.300
0.269
02M
18
1.000
1.000
1.000
0.999 0.999
0.998
0.0%
0993
0.9%
0982
3
0819
0.779
0.736
0.692
0.647
0.603
0.558
0.515
0.473
0.433
19
1.000
1000
1000
1,000
0.999
0.999 0.998
0997
0994
0991
0.928
0.904
0.877
0.848
0.815
0.781
0.744
0.706
0.668
0629
20
1.000
1.000
1000
1,000
L000
1000
0.999 0.998
0997
0.995
5
0.975
0964
0.951
0.935
0.916
0.895
0.871
0.8441
0816
0.785
21
1,000
1000 1000 1000
1000
L000
L000
0999
0.999
0.998
6
0.993
0988
0983
0976
0.966
0.955
10.942
0.927
0.909
09
22
1000
1,000
1.000
1000
1000
1000
1000
0.999
1000
0.999
7
0.998
0997
0.995
0992
0.988
0.983
0.977
0.969
0.960
0.949
23
1.000
1.000
1000 L000
1000
1000
1000
1000
1000 1000
8
1.000
0.999
0.999
0.998
0.996
0.994
0.992
0.988
0.984
0.979
9
1.000
1.000
1.000
0.999
0.999
0.998
0.997
0.996
0.994
0.992
10
1.000
1000
1.000
1,000
1000
1000
0.999
0.999
0.998
0.997
1.000
1.000
1.000
1.000
1000
1.000
1,000
1.000
0.999
0.999
12
1.000
1.000
1000
1.000
1000
1.000
1.000
1000
1,000
1.000

ページ6:

§3 Continuous Distributions.
§3.1 Random Variables of the Continuous Type
Ootype if it takes
A random variable X is of continuas
value in some intervals of real number.
A function f is the probability density function (pdf) of X
any
a.be R. a<b.
if for
pla≤ X ≤b) = (³f(+) dt.
y↑
a
b
x
a
b
The cumulative distribution function (cdf.) of X is
F(x) = P(X ≤x) = (x fit) dt, x ER
-Do
Property
ff is
a
pdf of a random variable X, then
(i) fix) >o for all x = R ;
00
(li) so f(x) dx = 1;
00-
(iii) pla≤ X ≤b) = p(a < X ≤b) = p ( a < x <b) = p ( a< x <b)
b
-fro
S°³ fit) dt
=

ページ7:

If F is the cdf of X, then
(i) 0 ≤ Fix) ≤ 1 for all x ER;
(ii) lim F(x) = 0 & lim F(x) = 1;
X-10
X47
(iii) Fis monotone increasing & F is right continuous; that is,
Example
=
lim Fit Flx) for each & ER
t→x+
x
Consider a <b. Let X he a random variable with its pdf
1
f(x) = a
b-a
'
for all x [a,b]
0
if x ≤a
F(x) = { fl+) dt
Sx =
-DO
X-a if aɛxeb.
b-a
1 if xzb
Such random variable has a
it by X ~ Unif (a.b).
b-a
f
uniform distribution. We denote
F
a
b
a
x
Definition
μ = E[X] := √x fix dx
the
mean or
the
expected Value of X
-Do
00
6² = Van (X) = E[(X-4)] = [" (x-4) = f(x) dx the variance of X
-DO

ページ8:

6 = Van (X)
tx
the standard derivation of X
00
+x
M (t) = E [ e²x] = foo etx f(x) dx the moment generating function of X
Remarks
00-
1. E[g(X)] = √ g(x) fix dx
2
Van (X) = E[X³] - (E[X])² = √ x² fix dx - ( S xfxdx)²
2. X~ Unif (a,b).
E[X]=x
Sbx. 11
2
|
dx = b²-a² 1₁ = a+b
2
0² = f² x² + a dx = b² - a³
a
Example 3.1-2
b-a
3
Let X have the pdf f(x) = 100
b-a
2
1 -a - (a+b)² = (b-a)
+
0 < x < 100 ( X ~ Unif (0, 1001)
then
μ = 10+100)
5²= (100-0)²
= 50
10000
=
12
12.
Example 3.1-3
Let Y he a random variable with pdf gly) = 2y, o< y <l
Then the cdf Gly) =
0
if yo
f12tdt = y² if o≤ y ≤ 1
1
0
if yal

ページ9:

5
16
Moreover, P( ± < Y≤ ¾³) = G(¾) - G(±) = ( ¾)²- (#)² =
P(Y < 2) = G (2) - 6 ( 2 ) = 1 - ( 4 )² = 15
2
|μ = [[Y] = [ y. (zy) dy = — — y³ |'
· S' ' = — — 4
。
o² = Van (Y) = { ' y² (zy) dy - (±)* =
Example 3.1-4
Let
+
18
3
have the pdf f(x) = 1x|, −| <x < |
Thom P( ± < x < 4) = {* 1×2 dx = f * *
5%
Then =
½
E[X] = { ' x · 1×1 dx = 5° = x² dx + [ ' x²
-
16
S.
dx + (x de f
dx =
13
32
dx = 0
S
Van (X) = {"² x²+ 1x1 dx = 0² = { " - x² dx + [ ' x² dx = = =
-1
Example 3.1-5
Let X have the pdf fix) = xex, 0 ≤x<00
Then Mit) = So
100
etxx. ex dx = lim b
0
ba
So
0
-(1-+)x
xe
T
l-t
(1-1)2
ē-(1-1)x
xe
u
dv
324] 16 = (1-4) provided that +<1
0
t<l
xx (11)
dx
=
·lim
6100
[-
Note that M(0) = 1 and
6
M'(t) = 2 & M"(t) = (1-0)4
Thus, μ = M'lo) = 2 & 0₁ = M" (0) - (M²(01) = 6-2'=2
M
0²

ページ10:

Notation
dt
Typ: the percentile, is a mumber top such that p = 5th fit at
In other words, F(πp) = P
00-
In particular, π0.5 is called the median (Second quartile)
T0.25 is called the first quartile
Example 3.1-6.
TU0.75 is called the third quartile.
f(x) = 1x1, -1<x<1. the pdf of X
Since f is symmetric, We have so fit dt = [' fit) dt
Hence, = = 5° fit dt, which implies πos = 0.
π10.25
In addition, + = 5th 0.25 fit dt = [ta* (-t) dt =
Th0.25
=
言
1-
π0.9
S.
-1
π0.9
2
So" t dt = 0.9-05
10.9 = 50.9 findt = ± ± + [10.9 f() dt => So"
-1
0
π0.9
=
4/5.

ページ11:

§ 3.2 The exponential, Gamma, and Chi- Aquare
Recall
Distributions
The Poisson distribution describes the probability of observing
Certain number
P(X₁ = k) =
The
k
of
(at)e-at
events within a fixed time interval t.
k = 0 1,2,3, etc
k!
exponential distribution describes the waiting time between
consecutive events in a Poisson process.
Let W he the time until first event occurs.
-λt
p ( w > t ) = P( X ₁ = 0) = (λt)° e²²±
Therefore, the cdf of W is
F(t) = P( W <t) =
b!
e-λt
= e
1- P(W>t) = 1 - ent
-λt
,
tzo
Hence the pdf of W is f (t) = F'(t) = ^ e^t, tzo
Definition
an
exponential distribution
x = 0.
-22
The random variable X has
(X ~ exp (x)) if it's pdf flx) = ^e^z
Remarks
ae
1. If we set λ = 10, then X ~ explo
.
with
a
flx) = ± √ e³, 120
= fe

ページ12:

2. The mgf of X is given by
.00
tx
-2x
M (t) = E[ex] = √ ex ^e^x dx = lim √ °^e'
D
provided that t<λ.
Thus, M'(+) = Ya
(1-%2
⇒ μ = M'(0) = ½
and σ² = M" (0) - (M²(01)² =
819
b
So
ne
2
72
(t-a)x
& M" (t) =
=
(1-1)³
2
—
22
-
I
22
0 & 6² = 0²
In other words, μ = 0
3. Note that F(x) =
0
{° if x <o
-7x
=
1- e if x 20
|
22.
Then the median Thas can he computed
as
dx
=
1-Яt
-2π0.5
-271015
=
- e
e
=
± <=> π0,5 = — In 2
Type
Meaning
Distribution
Poisson
P
discrete (Counting process) Number of events within time t
Xt
Exponential continuous (Timing Process)
W₁. W₂-W1, ... etc
[How many
events occur in a
given time?
Waiting time between events
wait)
How long do we have to wait
for the next event?

ページ13:

Example 3.2-1
Let X have an
exponential distribution with
Then the pdf of X is
PCX <18) = 200
.18
0
-x
x
20
a
f(x) = 10 e²², xx0.
20
-18
20
20 dx = 1-e
e
=0.593.
Example 3.2-2
mean of 0=20.
Customers arrive in a certain shop according to an
Poisson process at a
What
ng
mean rate of 20
per-
hour.
approximate
the probability that the shopkeeper will have to wait
more than 5 mins for the arrival of the first customer?
Let X denote the waiting time in mins until the first customer
arrives.
expected
Note that λ = 1/3 ( the
Then the pdf of X is given by
=
3
-x
e ×20
number of amivals per-mins)
f(x)
Hence,
.00
-x
P(X > 5) = √²² = e ³ ³ dx = e
=
0.1889
5
Moreover, the median time until the first arrival
is
π015
=
-3-142 = 2.0794.

ページ14:

Recall that in the Poisson
the Poisson process
with
mean 2, the waiting
time until the first occurrence
has
an
exponential distribution
Let W denote the waiting time until the 2-th
Then cdf of W is given by
F(w) = P(Ww) = 1-P(W >w)
=
Occurrence.
: 1- Pl fewer than & Occurrences in [0, w]) for w 30
"α-1" (74) ké
(765)*
Then
F'(w) = ^ e
(1.174)°r
0!
土 +
= Ae
k=1
k!
-2w
-Aw
= e te
-7w
k=0 k!
K=1
k!
d-1
kt
-aw
-λw
k (aw) i
- e
(74) 2 ]
K!
(aw) λ
To!
21
31
α-2
(d-1) (21)
-20
-
(α-))
e
-20
+
217
24
|
-
(74)2
(2-1))
[
α-+
a (aw)
alaw)°
2+
-
=
(d-+)!
(d-1)!
The random variable W is said to he
a
gamma
distribution
if the pdf is
fiw) =
{
712418-1
-2w
e
(d-17!
,
0
0731

ページ15:

Remark
The
gamma
function is defined by [(t) =
So y they dy, t>o
Note that " (t) = ( -yth e-0)/00 + S (+-1) y ²² = "dy
DO
=
(t-1) for yt-² e-³ dy
= (t-1) (t-1)
In addition,
F(1) = √
e³ dy = 1.
Thus, P (n) = (n-1) P(n-1) = (n-1) (n-2) P (n-2)
= (n-1)!
We observe that law) -1 -aw
=
=
(2-1)!
P(d)α
e
d-1
w e
=
(n-1) (n-2) 3.2.1. P(1)
2d 2-4-20
W e
=
α--20
W e
=
(d-1)!
P(d)
M-

ページ16:

Definition
X has
f(x) =
=
a
gamma distribution if it pdf is defined by
1
P(a)
x e
d
x≥0, α>0, 0>0.
We denote it by X ~ Gamma (d,o).
Remark
Let us check that
00
00
00
So fix dx = So
-40
dx
Jo Pla) od
2-1-y
00
(oy) e
T12)
11
DO
=
y= 0 Ma) ga dy = 111 So you a dy
x
In addition, the mgf of X is
M (t) = E[ ex] =
y=
00
So
e
(d)
= 1.
1-to
So
Pla)pa 0
...dy
00
x
tx
1
x e
dx
=
(2)8%
∞
d-1
-y e
e
1-8t
11-04) Praja So
where 1-ot >0
=
60
[(d) (1-0+)" So ya+ - dy
Thus, M'(+) =
M'(+) = 10
(1-01)α1
= M'(0) = α0
=> Sμ=.
ye
||
|
=
(1-0+)9
P(d)
& M" (t) =
2(4+1)02
(1-0)4+2
{"6² = M" (0) - (M²(0)² = 2 (2+1) 0² - α²0² = α0?
1-p
e
(10-1)x
dx

ページ17:

f(x)
0.25 a 1/4
f(x)
0.25
9=5/6
0
a=4
0.20
0.20
201
0=1
0.15a=10=4
10.15.
8 be
Lop
0.10:
a=2
a=3
0.05
a=4
5 10 15 20 25
30
0.10.
0=2
0=3
0.05
0=4
x
5 10 15 20 25 30
Figure 3.2-2 Gamma pdfs: 0 = 4, α = 1/4, 1, 2, 3, 4; α = 4,0 = 5/6, 1, 2, 3, 4
Example 3.2-4
Example
Suppose the number of customers.
hour at a
per
shop follows
a Poisson process with mean 20. What is the Probability that
the second customer amives more than 5 mins after the shop
opens for the day?
Let X he the waiting time in mins until the second
customer arrives, then X ~ Gamma (d. A) where
d = 2.0 = 1/4 = 3 ( 20 / hour <=> 1/3 / min
Thus, P(X > 5) =
=
00
√5
1
P(2) 32
= ½ | | (-3) x e
<=> λ = ½)
00
2-1743
x e
x
e
dx =
dx
5
9
-5
8
3
- 9 e* 1100 = 1 e²³ ³~≈ 0.504
5

ページ18:

Example 3.2-5
Telephone calls arrive at an
minute according
per
office
at
a mean rate of 1=2
to a Poisson process.
Let X denote the waiting time in minute until the fifth
call arrives.
Then X ~ Gamma (d.0) where α = 5 & 0 = = = ½⁄2
入
μ = 20 = 5/2 & σ² = α0² = 5. (±)² = 5/4.
Note that if X ~ Gamma (0,α) where 0 = 2 & α = 1½ for
1
12-1-742
then the pdf of X is
fix =
x e
(½).2%
some Y>O
01X18.
,
(*)
We say X has a chi-square distribution with r degrees.
of freedom ( demote it by X ~ X (r)) if its pdf is (*)
Note that if X ~ x²(r), then μl = 20 = = · 2 = r
P²² = α 8² = 2² = 2r
|
=
(1-2+) 1/2 <1/2
and the mgf is Mit) =
(1-04)α

ページ19:

f(x)
0.5.
0.4-
r=2
0.3-
0.2+
r=3
r=5
0.1
r=8
2
4
6
8
10
12
14
Figure 3.2-3 Chi-square pdfs with r = 2,3,5,8
Example 3.2-6.
Let X ~ X²(5), then.
P ( 1.145 ≤ X ≤ 12.83) = F(12.83) - F( 1.145) = 0.975 -0.05 = 0.925
P(X> 15.09) = 1- F(15.09) = 1-0.99 = 0.01
Example 3.2-7
If X ~ x²-17), then pla< x <b) = 0.95
Example 3.2-9
=> F(b) - F(a) = 0.95 => a = 1.690
ww
0.975 0.025
b = 16.01
NOT a unique pair
average of 30 per
hour in
customer arrive at a shop on the
accordance with a Poisson process. What is the probability that
the shoe keeper will have to wait longer than 9.39 minutes for
the first nine customers to arrive ?

ページ20:

Let X he the waiting time until the ninth arrival.
Note that λ = 1/2 ( mean rate
per
min),
α = 9 => r = 2α = 18
then 0=2" =2
Then X ~ X² (18). Hence, P(X > 9.390) = 1-P(X = 9.390)
Remark
= 1-0.05 = 0.95.
Let o≤ d = 1 & r>o. We consider the following number K² (r)
f(x)
both areas
are d
which is defined by
P(X = x²₁₂m) = d
100 (1-2)th percentile
<=> P(X ≤ X² (n) = d
x
X² (r)

ページ21:

Table IV The Chi-Square Distribution
0.10
0.05
x²(8)
0.10
0.05
x²(8)
02
4
a
8
10 12 x14 16
18 20
0
2 4
10
x²(8)
16 18 20
P(X ≤ x) =
-
= √
1
T(r/2)2r/2
w/2-1/2 dw
P(X ≤ x)
0.010
0.025
0.050
0.100
0.900
0.950
0.975
0.990
r
X₁₁yy(r)
X0,975 (r)
X95 (r)
Xu90(r)
xàio(r)
xius(r)
x²)
x()
12345
678910 112345 689 20 21 22 23 24 25 26 27 28 29 30 40 50 60 70 80
0.000
0.001
0.004
0.016
2.706
0.020
0.051
DER 3.841
5.024
0.103
6.635
0.211
4.605
0.115
0.216
THE 5.991
7378
0.352
9.210
0.584
電話:6.251
7.815
0.297
9.348
0.484
11.34
0.711
1.064
7.779
9.488
0.554
11.14
13.28
0.831
1.145
1.610
9.236
11.07
12.83
15.09
0.872
1.237
1.635
2.204
10.64
12.59
14.45
16.81
1.239
1.690
2.167
2.833
12.02
14.07
16.01
18.48
1.646
2.180
2.733
3.490
13.36
15.51
1754
20.09
2.088
2.700
3.325
4.168
14.68
16.92
19.02
21.67
2.558
3.247
3.940
4.865
15.99
18.31
20.4880
23.21
3.053
3.816
4.575
5.578
17.28
19.68
21.92
The 24.72
3.571
4,404
5.226
6.304
18.55
電話:21.03
23.34
26.22
4.107
5.009 0
5.892
7042
19.81
22.36
24.74
2769
4.660
5.6290
6.571
7.790
21.06
23.68
26.12
29.14
5.229
6.262
7.261
8.547
22.31
25.00
2749
30.58
16
5.812
6.908
7962
9.312
23.54
26.30
28.84
32.00
17
6.408
7.564
8.672
10.08
24.77
27.59
30.19
33.41
18
7.015
8.231
9.390
10.86
25.99
28.87
31.53
34.80
19 7.633
8.907
10.12
11.65
27.20
30.14
32.85
36.19
8.260
9.591
10.85
12.44
28.41
31.41
34.17
37.57
8.897
10.28
11.59
13.24
29.62
32.67
35.48
38.93
9.542
10.98
12.34
14.04
30.81
33.92
36.78
40.29
10.20
11.69
13.09
14.85
32.01
35.17
38.08
41.64
10.86
12.40
13.85
15.66
33.20
36.42
39.36
42.98
11.52
13.12
14.61
16.47
34.38
37.65
40.65
44.31
12.20
13.84
15.38
17.29
35.56
38.88
41.92
45.64
12.88
14.57
16.15
18.11
36.74
140.11
43.19
46.96
13.56
15.31
16.93
18.94
37.92
41.34
44.46
48.28
14.26
16.05
17.71
19.77
39.09
42.56
45.72
49.59
14.95
16.79
18.49
20.60
40.26
43.77
46.98
50.89
22.16
24.43
26.51
29.05
51.80
55.76
59.34
63.69
29.71
32.36
34.76
37.69
63.17
67.50
71.42
76.15
37.48
40.48
43.19
46.46
74.40
79.08
83.30
88.38
55.33
85.53
90.53
95.02
100.4
45.44
48.76
51.74
53.34
57.15
60.39
This table is abridged and adapted from Table III in Biometrika Tables for Statisticians, edited by E.S.Pearson and H.O.Hartley.
64.28
96.58
101.9
106.6
112.3

ページ22:

$3.3 The Normal Distribution
Definition
The random variable X has
is defined by
1
f(x) =
√27 e
a
normal distribution if its pdf
-(x-1)²
202
-XX<∞0.
,
where -∞s<μ<∞s & 0 <σ<∞. We denote it by X ~ Niμ, s³)
Remarks
1° f(x) 20 for all x ЄR.
00
Let 1 = S_~ fix)dx, then we shall show I = 1.
-00
Firstly. We set 3
Observe that
I² = 1
270
S
00
-00
e
-3
2
0
(x-4)
=
then I = 1.00
∞
1
e
dz.
00
60
-(3+33)
"d3₂ =
=
2
d3d32
27
-00
-Do
27
00
0
So
-²
276
e
2 r drdo
1
=
do = 1
I=1.
2TU
0
0
- dz, · S.
00
e
-32
3₁=80050
3₂ = Vsine
1
2πC
2. The mgf of X is
ex exp (-(x-1)²).
M(+) = S
·
=
So
1-00 σ√2π
1
exp
dx
[² - (+2)+µ?] } dz

ページ23:

Note that x² - 2 (1μ4 + 6 + ) x + y² = [ X− (µ+o*ts]²= 2µo°*t −6*t²
Hence, M(t) = exp ( 240 + +±0 +" ). 500
262
exp
{ = = = = = (x + 1)] } =
dx
=
exp (246²+ +0² ²). 1
pdf of N ( μ+ o²±² 6²)
=
exp (μt + 0²²)
Then, M'(t) = (u+o't). exp (μut + o`t²)
and _ _M"(t) = [(4+o't)²+ σ'] exp (μut + + +¹² )
Hence, E[X] = M'(0) = μ and Var (X) = M" (0) - (M'(0)²==²
Condusion: X~NCA. 6) Variance.
Example 3.3-1.
mean
if the pdf of X is f(x) = √17/1/14 exp [ -(x+7)² ]
√32πt
32
184X48.
Observe that fix) = 1 exp [ -(x-(-1) ² ] ; ;.e, X ~ N(-7.16).
4√270
2.42
and the mgf of X is M(t) = exp (−7++ 8+²).
Example 3.3-2.
If the mgf of X is
Mlt) = exp (5t + 12t³), then X ~N(5,24)
and its pdf is
1
f(x) =
-exp [-(x-5)² ]
181X18.
48TG
48

ページ24:

Definition
We
Jay Z has
a standard normal distribution if X ~ NCO, D
(ie, μ = 0 & 6² = 1).
Moreover, the cdf of Z is
3
$ (3) = P(Z ≤ 3) = (³00 √//
e
dt
2TC
symmetric
₤(30)
Example 3.3-3.
If Z ~ NCO,1), then
子。
-30
(-30)
30
(-3)-1-(3) P(Z<-3)=PIZ >30)
P(Z = 1.24) = (1.24) = 0.8925
P(1.24 Z 2.37) = 1 (2.37) - (1.24) = 0.9911-0.8925 = 0.0986.
P(-2,37 ≤ Z ≤ -1.24) = P( 1.24 ≤ Z ≤ 2,37) = 0.0986.
Also, P(Z>1.24) = 0.1075
PIZ-214) = P(Z≥2.14) = 0.0162.
S
Moreover, pl-2.14 ≤ Z ≤ 0.77) = p(2 ≤ 0.77) P(Z = -2.14)
=
0.7794 -0.0162 = 0.7632.

ページ25:

Table Va The Standard Normal Distribution Function
-3
f(z)
0.44
0.3
0(20)
0.2-
0.1
0
1 zo
2
-w²/2 dw
P(Z≤z)=$(z) =
(-2) = 1 - (2)
ㄥˋ
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.5000
0.5040
0.5080
0.5120
0.5160
0.5199
0.5239
0.5279
0.5319
~0.5359
0.1
0.5398 0.5438
0.5478
0.5517
0.5557
0.5596
0.5636
0.5675
0.5714
0.5753
0.2
0.5793
0.5832
0.5871
0.5910
0.5948
0.5987
0.6026
0.6064
0.6103
0.6141
0.3
0.6179 0.6217
0.6255
0.6293
0.6331
0.6368
0.6406
0.6443
0.6480
0.6517
0.4
0.6554
0.6591
0.6628
0.6664
0.6700
0.6736
0.6772
0.6808
0.6844
0.6879
0.5
0.6915 0.6950
0.6985
0.7019
0.7054
0.7088
0.7123
0.7157
0.7190
0.7224
0.6
0.7257
0.7291
0.7324
0.7357
0.7389
0.7422
0.7454
0.7486
0.7517
0.7549
0.7
0.7580 0.7611
0.7642
0.7673
0.7703
0.7734
0.7764
0.7794
0.7823
0.7852
0.8
0.7881 0.7910
0.7939
0.7967
0.7995
0.8023
0.8051
0.8078
0.8106
0.8133
0.9
0.8159
1.0 0.8413
0.8186 0.8212
0.8238
0.8264
0.8289
0.8315
0.8340
0.8365
0.8389
0.8438
0.8461
0.8485
0.8508
0.8531
0.8554
0.8577
0.8599
0.8621
1.1
0.8643
0.8665 0.8686
0.8708
0.8729
0.8749
0.8770
0.8790
0.8810
0.8830
1.2
0.8849
0.8869
0.8888
0.8907
0.8925
0.8944
0.8962
0.8980
0.8997 0.9015
1.3
0.9032
0.9049
0.9066
0.9082
0.9099
0.9115
0.9131
0.9147
1.4
0.9192 0.9207
0.9222
0.9236
0.9251
0.9265
0.9162 10.9177
0.9279 0.9292 0.9306 0.9319
1.5
0.9332
0.9345
0.9357
0.9370
0.9382
0.9394
0.9406
1.6
0.9452 0.9463
0.9474
0.9484
0.9495
0.9505
0.9515
0.9418 0.9429 0.9441
0.9525 0.9535 0.9545
1.7
0.9554 0.9564
0.9573
0.9582
0.9591
0.9599
0.9608
0.9616 0.9625 0.9633
1.8
0.9641 0.9649
0.9656
0.9664
0.9671
0.9678
1.9
0.9713
0.9719
0.9726
0.9732 0.9738
0.9744
2.0
0.9772 0.9778
0.9783
0.9788
0.9793
0.9798
2.1
0.9821 0.9826 0.9830 0.9834
0.9838
0.9842
0.9686 0.9693
0.9750 0.9756
0.9803 0.9808
0.9846 0.9850
0.9699 0.9706
0.9761 0.9767
0.9812 0.9817
0.9854 0.9857
2.2
0.9861
0.9864 0.9868
0.9871
0.9875
0.9878
0.9881
0.9884 0.9887
0.9890
2.3
0.9893
0.9896 0.9898
0.9901
0.9904
0.9906
0.9909
0.9911
0.9913
0.9916
2.4 0.9918
0.99200.9922 0.9925 0.9927
0.9929
0.9931
0.9932 0.9934
0.9936
2.5
0.9938
0.9940 0.9941 0.9943
0.9945
0.9946
0.9948
0.9949 0.9951
0.9952
2.6
0.9953
0.9955
0.9956
0.9957
0.9959
0.9960
0.9961
0.9962
0.9963
0.9964
2.7
0.9965 0.9966
0.9967
0.9968
0.9969
2.8 0.9974 0.9975
0.9976
0.9977
0.9977
0.9978
2.9 0.9981 0.9982
0.9982
0.9983
0.9984
0.9984
3.0 0.9987 0.9987
0.9987 0.9988
0.9988
0.9989
0.9970 0.9971 0.9972
0.9979 0.9979 0.9980
0.9985 0.9985
0.9989
0.9973 0.9974
0.9981
0.9986 0.9986
0.9989
0.9990 0.9990
a
0.400
0.300
0.200
0.100
0.050
0.025
0.020
0.010
0.005
0.001
Za
0.253
0.524
0.842
1.282
1.645
1.960
2.054
2.326
2.576
3.090
Za 2
0.842
1.036
1.282
1.645
1.960
2.240
2.326
2.576
2.807
3.291

ページ26:

Table Vb The Standard Normal Right-Tail Probabilities
(2)
0.44
0.3.
0.2.
0.1
04
a
-2
-1 0
1 za 2
3
P(Z > za) = x
P(Z > z) = 1 - 0(z)=b(-2)
Za
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.5000
0.4960
0.4920
0.4880
0.4840
0.4801
0.1
0.4602
0.4761
0.4562
0.4721
0.4681
0.4641
0.4522
0.4483
0.4443
0.4404
0.2
0.4364
0.4207
0.4325
0.4168
0.4286
0.4247
0.4129
0.4090
0.4052
0.4013
0.3
0.3821
0.3783
0.3974 0.3936
0.3897
0.3859
0.3745 0.3707
0.3669
0.3632
0.3594
0.4
0.3557
0.3520
0.3483
0.3446
0.3409
0.33720.3336
0.3300
0.3264
0.3228
0.3192
0.3156
0.3121
0.5
0.3085
0.3050
0.3015 0.2981
0.2946
0.2912
0.2877
0.2843
0.2810
0.2776
0.6
0.2743
0.2709
0.2676
0.2643
0.2611
0.2578
0.2546
0.2514
0.2483
0.2451
0.7
0.2420 0.2389
•
0.2358
0.2327
0.2296
0.2266
0.2236
0.2206
0.2177
0.2148
0.8
0.2119
0.2090
0.2061
0.2033
0.2005
0.1977
0.1949
0.1922 0.1894
0.1867
0.9
0.1841
0.1814
0.1788
0.1762
0.1736
0.1711
0.1685
0.1660
0.1635
0.1611
10
0.1587
0.1562 0.1539
0.1515 0.1492
0.1469
0.1446
0.1423
0.1401
0.1379
1.1
0.1357
0.1335
0.1314
0.1292
0.1271
0.1251
0.1230
0.1210
0.1190
0.1170
1.2
0.1151
0.1131
0.1112
0.1093
0.1075
0.1056
0.1038
0.1020
0.1003
0.0985
1.3
0.0968
0.0951
0.0934
0.0918
0.0901
0.0885
0.0869
0.0853
0.0838
0.0823
1.4
0.0808
0.0793
0.0778
0.0764
0.0749
0.0735
0.0721
0.0708
0.0694
0.0681
1.5
0.0668
0.0655
0.0643
0.0630
0.0618
0.0606
0.0594
0.0582
0.0571
0.0559
1.6
0.0548
0.0537
0.0526
0.0516
0.0505 0.0495 0.0485
0.0475
0.0465
0.0455
1.7
0.0446
0.0436
0.0427 0.0418
0.0409
0.0401
0.0392 0.0384
0.0375
0.0367
1.8
0.0359
0.0351
0.0344
0.0336
0.0329
0.0322
0.0314
0.0307
0.0301
0.0294
1.9
0.0287
0.0281
0.0274
0.0268
0.0262
0.0256
0.0250
0.0244
0.0239 0.0233
2.0
0.0228
0.0222
0.0217 0.0212
0.0207
0.0202
0.0197
0.0192
0.0188
0.0183
2.1
0.0179
0.0174
0.0170
0.0166
0.0162
0.0158
0.0154
0.0150
0.0146
0.0143
2.2
0.0139
0.0136
0.0132
0.0129
0.0125
0.0122
-0.0119
0.0116
0.0113
0.0110
2.3
0.0107 0.0104
0.0102
0.0099
0.0096
0.0094
0.0091
0.0089
0.0087
0.0084
2.4
0.0082
0.0080 0.0078
0.0075 0.0073
0.0071
0.0069
0.0068
0.0066
0.0064
2.5 0.0062 0.0060 0.0059
2.6 0.0047 0.0045 0.0044
0.0057
0.0055
0.0054
0.0052
0.0051
0.0049 0.0048
0.0043
0.0041
0.0040
0.0039
0.0038
0.0037
0.0036
2.7
0.0035 0.0034
0.0033 0.0032
0.0031
0,0030 0.0029
0.0028
0.0027
0.0026
2.8
0.0026 0.0025 0.0024
0.0023
0.0023
0.0022 0.0021
0.0021
0.0020
0.0019
2.9
0.0019 0.0018 0.0018
0.0017
0.0016
0.0016
0.0015
0.0015
0.0014
0.0014
3.3
3.4
3.0 0.0013 0.0013 0.0013
3.1 0.0010
3.2 0.0007 0.0007 0.0006 0.0006
0.0005 0.0005 0.0005 0.0004
0.0003
0.0012
0.0012
0.0011
0.0011
0.0011
0.0010
0.0010
0.0009 0.0009 0.0009
0.0008
0.0008
0.0008
0.0008
0.0007
0.0007
0.0006
0.0006
0.0006
0.0005
0.0005 0.0005
0.0004
0.0004
0.0004
0.0003 0.0003 0.0003
0.0003
0.0003
0.0004
0.0003 0.0003 0.0003
0.0004
0.0003
0.0002

ページ27:

Remark
In statistic, people
such that
are interested in
P(Z32) = 2
finding "32"
where Z ~ N(0.1).
3d is the 100 (1-d)th percentile.
or
upper
100d
percent point.
2
-3-2
-1
T
2 32 3
Area = &
Since f is symmetric. We have 31-d.
=-
-3d.
0.05 = 5%
0.025 = 2.5%
30.95 30.5 30.05 30.025
11
-30.05
]]
1.645 1.96
Thearem (standardization)
If X ~ Neu, o²), then Z = X-M ~ Nco.1)
6
proof Note that P(Z = 3) = P( X-μ ≤ 3) = P(X ≤ 35+ μ)
= √30+μ
1
√2TU exp
|
202
-(x-1)²/dx
00-
•
6
1=12-136
I
x = 50+μ
z
Son 1/1/= exp(-w³) dw
87-

ページ28:

Remark
If X ~ NCμ, o²), then
a-μ
P( a≤ X ≤b) = p( a-1 = X-1 b-^^) = (b-μ) - (a-^^)
Example 3.3-6
If X ~ N (3.16), then
4-3
♡
<
S
N(0,1)
N
1
P( 4 ≤ X ≤ 8) = P( 4+ 3 = Z ≤ 87 ) = ( 1.25) - § (0.25)
4
= 0.8944 -0.5987 = 0.2957
_0-3
P(0 ≤ X ≤ 5) = P( 0 = 3 ≤ Z ≤ 5-³ ) = 1 (05) - (-0.75)
4
= 0.6915-0.2266 = 0.4649.
-2-3
P ( -2 ≤ X ≤ 1) = p( -2 - 3 ≤ Z ≤ 1-3 ) = 1 (-05) - 1 (-1.25)
= 0.3085 - 0.1056
Example 3.3-7
4
= 0.2029
If X ~ N(25, 36), then find c such that P(IX-25|≤ 0) = 0.9544
Note that pl = ≤ Z ≤ ÷ )
≤ Z ≤ — ) = 0.9544
||
五(卡)-五(卡)
Since ( ) = 1 - ( ). We have 1 (7)-(1-1 (+)) = 0.9544
( 7 ) = 0.9772 = % = 2 = c = 12.

ページ29:

The arem
If X ~ NCμ, o²), then V = (*-^^;² = 2² ~ X²(1)
proof
Let G he the cdf of V, Then for v=o, G(2)=0
For v≥0,
G(v) = P(V≤ v) = P(2² <v) = p(-√ √ ≤ Z ≤ √ )
=
3=√y
√ 1
-√ √270
v
-3²
-3²
e
好
dz = 2
e
ds
办
So Jazy
dz = dy
Thus, the pdf of V is G'(v) =
Let x = %, then the pdf of V is
--x T
(2x) e =
e
啦
-y
dy
V e
ó0.
+
-x
X
e
~ x²(1).