課程片段︰
https://youtube.com/playlist?list=PLzDe9mOi1K8ptJB8WVTLLcbAfUw6kkMi7
筆記內容︰
Binomial Theorem
二項式定理︰
p.2 ~ 9
Differentiation
微分︰
p.10 ~ 39
Tangent
切線︰
p.40 ~ 42
Rate of change
改變率︰
p.43 ~ 46
Maximum, minimum, greatest, least
極大、極小、最大、最小︰
p.47 ~ 49
Integration
積分︰
p.50 ~ 64
Area
面積︰
p.65 ~ 67
Trapezoidal Rule
梯形法則︰
p.68 ~ 69
Curve  Straight line
曲線  直線︰
p.70 ~ 72
Bayes’ Theorem
貝葉斯定理︰
p.73 ~ 76
Normal Distribution
正態分佈︰
p.77 ~ 84
Binomial Distribution
二項分佈︰
p.85 ~ 87
Poisson Distribution
泊松分佈︰
p.88 ~ 89
E(X), Var (X)
E(X), Var (X)︰
p.90 ~ 92
Central limit theorem
中心極限定理
p.93 ~ 94
Confidence interval
置信區間
p.95 ~ 96
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Maths (M1)
Content
Binomial Theorem
二項式定理
A.
Factorial
階乘
B.
Exponential Series
指數數列
C.
nCr
nCr
D.
Binomial Theorem
二項式定理
E.
Question Demo
題目示範
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Maths (M1)
A. Factorial 階乘
n !  n  n  1  n  2  ...  1
where n is non-negative integer
其中 n 為非負整數
B. Exponential Series 指數數列
ex  1 
x x2 x3 x 4



 ...
1! 2! 3! 4 !
e f x   1 
e ax  1 
f x  f x 
f x   f x   ...


1!
2!
3!
4!
2
3
4
ax ax 2 ax 3 ax 4



 ...
1!
2!
3!
4!
Calculator Program 計數機程式
?X:1
X
X2  2
X3  6
X^(4)  24
Review 重溫
1
ap

a p 

a p  a q  a p q

ap
 a p q
q
a
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Maths (M1)
C. nCr nCr
Crn 
n!
r ! n  r  !
Properties of Crn / Crn 的特徵:
 n, r  non-negative integer 非負整數
 nr
 C 0n  Cnn  1
 C1n  n
 Crn  Cnnr
秘技:
上下㇐齊數、件數睇 r 盧
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Maths (M1)
D. Binomial Theorem 二項式定理
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Maths (M1)
D. Binomial Theorem 二項式定理

Pascal’s Triangle 楊輝三角
(a  b) 0 = 1 ..….…………………………………………………...
(a  b)1 =
(1)a  (1)b ……………………………………………
(a  b)2 =
(1)a 2  (2)ab  (1)b 2 ……………………………….
(a  b)3 =
(1)a 3  (3)a 2 b  (3)ab 2  (1)b 3 ………………….
( a  b) 4 =
(1)a 4  (4)a 3 b  (6)a 2 b 2  (4)ab 3  (1)b 4 ……..
1
1
1
2
1
1
1
3
4
1
3
1
6 4
……

Expand by Binomial Theorem 用二項式定理展開
C 00 
……...….…………………………………………….
C 01 
C11 
C 02 
C12 
C 22 
……………………………..
C 03 
C13 
C 23 
C 33 
………………..
C 04 
C14 
C 24 
C 34 
C 44 
C 00
………………………………………….
…
1
C 01
C 02
C 03
C 04
C11
C12
C 22
C13 C 23
C14
C 24 C 34
……
C 33
C 44
Binomial Theorem 二項式定理:
(a  b)n  C0na n  C1na n 1b  C2na n  2b 2    Crna n  rbr    Cnn1ab n1  Cnnbn
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Maths (M1)
D. Binomial Theorem 二項式定理

Calculator Programme 計算機程式
ClrMemory : ?  A : ?  B : ?  C : Lbl 1 :
CCX x A^(C – X) x B^(X X + 1  X : C  X : Goto 1 : 0
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Maths (M1)
E. Various Examination Format 不同考試類型
Expand 1  2x   e
6
term containing x
3
1
x
3
in ascending powers of x up to and including the
term.
按 x 的升冪次序，展開 1  2x   e
6
1
x
3
至 x 3 項。
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Maths (M1)
E. Various Examination Format 不同考試類型
1  3x 5

Expand
in ascending powers of x up to and including the term
e 2x
containing x 3 term.
1  3x 5

按 x 的升冪次序，展開
至 x 3 項。
e 2x
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Maths (M1)
Content
Differentiation
微分
A.
Concept of Limit
極限的概念
B.
Concept of Slope
斜率的概念
C.
Concept of Differentiation
微分的概念
D.
Rule of Differentiation
微分法則
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Maths (M1)
A. Concept of Limit 極限的概念
Draw the graph of 描繪圖像 y 
x2  x  6
.
x 2
x
0
1
2
3
y
3
4
Undef
6
x
1.9999
y
4.9999
2
2.0001
5.0001
y
x
x2  x  6
 lim

x 2
x 2
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Maths (M1)
A. Concept of Limit 極限的概念
Rules of Limit 極限的法則:
f x  , g x  are functions 函數.
c is a constant 常數.
Assume 假設 lim f x   F , lim g x   G
x a
x a
(1) lim c  c
x a
(2) lim cf x   c lim f x   cF
x a
x a
(3) lim f x   g x   lim f x   lim g x   F  G
x a
x a
x a



(4) lim f x g x   lim f x  lim g x   F  G
x a
x a
x a
f x  F
f x  lim
 x a

if G  0
x  a g x 
lim g x  G
(5) lim
x a


(6) lim f g x   f lim g x   f G 
x a
x a
Remarks 備註︰
After a very long time 經過㇐段⾧時間 : t  
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Maths (M1)
A. Concept of Limit 極限的概念
Find the following limits. 求以下的極限。

x3
6
x  1  x

x3
6
x  1  x

2


lim  2
 7
x  3x  24x  49




1.5 
lim 4  t

t 
2 8  0.5 


lim

t


1


 
2
t  2t  1  3 
1 
 1
lim
ln 2

5
t  2 2
ln 3
ln 3 
t  2t  1




lim
lim
32
t  2
8
5  4t
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Maths (M1)
A. Concept of Limit 極限的概念
Find the following limits. 求以下的極限。
[Hint : You may use the fact that
t
lim mt  0 for any positive constant m.]
t  e
t
[提示︰可利用以下事實 – 對任何正常數 m， lim mt  0 。]
t  e

t
t


20
20

lim 9630  480te  9600e  
t 



 25

lim
ln 1  4te 0.04t  10  
t   2



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Maths (M1)
B. Concept of Slope 斜率的概念
Slope 斜率  m 
y
y2  y1
x 2  x1
B (x2, y2)
A (x1, y1)

x
y
(0, 3)
(1, 2)
(2, 1)
(3, 0)
(2, –1)
x
(1, –2)
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Maths (M1)
C. Concept of Differentiation 微分概念
(a) Find the slopes of the chords joining the points on the curve y  x 2
for which
求在曲線 y  x 2 上由以下兩點所組成的弦的斜率
(i) x  3 and x  3.1 ,
(ii) x  3 and x  3.01 .
(b) (i) Expand 展開 3  h  .
2
2
3  h   32

(ii) Simplify 簡化
.
h
(iii) By (b)(ii), find the slope of y  x 2 at x  3 .
根據 (b)(ii)，求於 x  3 時， y  x 2 的斜率。
(c) Find the slope of the curve y  x 2 .
求曲線 y  x 2 的斜率。
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Maths (M1)
C. Concept of Differentiation 微分概念
Let A(1, 12) and B(1.1, 1.12)
y 1.12  12

 2.1
x
1.1  1
Let A(1, 12) and B(1.01, 1.012)
y 1.012  12

 2.01
x
1.01  1
B
∆y
y  x2
A
Let A(1, 12) and B(1.001, 1.0012)
y 1.0012  12

 2.001
x
1.001  1
∆x
Slope 斜率 of AB 
y
x
y
x 0 x
Slope of tangent at A 在 A 點的切線的斜率  lim
e.g. Find slope 求斜率: y  3  4x .
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Maths (M1)
C. Concept of Differentiation 微分概念
Demo 1 (Out-of-M1-Syllabus)
Find the slope of y  3 .
d
3 .
Find
dx
dy
If y  3 , find
.
dx
求 y  3 的斜率。
d
3 。
求
dx
dy
若 y  3 ，求
。
dx
Solution:
dy
d
33
0

 lim  0
3  lim
h

0
h

0
dx dx
h
h
Demo 2 (Out-of-M1-Syllabus)
Find the slope of y  4x .
d
Find
4x  .
dx
dy
If y  4x , find
.
dx
求 y  4x 的斜率。
d
求
4x  。
dx
dy
若 y  4x ，求
。
dx
Solution:
dy
d
4x  h   4x
4h

 lim
 lim 4  4
4x   lim
h

0
h

0
h 0
dx dx
h
h
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Maths (M1)
C. Concept of Differentiation 微分概念
Demo 3 (Out-of-M1-Syllabus)
Find the slope of y  x 3 .
d
x 3  .
Find
dx
dy
If y  x 3 , find
.
dx
求 y  x 3 的斜率。
d
x 3  。
求
dx
dy
若 y  x 3 ，求
。
dx
Solution:
3
dy
d
x  h  x 3

3

x   lim
h 0
dx dx
h
3
2
2
x  3x h  3xh  h 3  x 3
 lim
h0
h
2
2
3x h  3xh  h 3
 lim
h0
h
2
 lim3x  3xh  h 2 
h0
 3x 2
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Maths (M1)
C. Concept of Differentiation 微分概念
Demo 4 (Out-of-M1-Syllabus)
Find the slope of y  e x .
d
e x .
Find
dx
dy
If y  e x , find
.
dx
求 y  e x 的斜率。
d
e x 。
求
dx
dy
若 y  e x ，求
。
dx
Solution:
dy
d x
e x h   e x
e x eh  1



e  lim
 lim
h 0
h 0
dx dx
h
h
2
3


h h
h
e x  1  

 ...  1 
1! 2! 3!

 lim 
h0
h
2
h h

h3
e x  

 ... 
1! 2! 3!

 lim 
h0
h

h h2
x1

 lim e   
 ... 
h0
 1! 2! 3!



 ex
符號介紹：
dy
,
dx
y,

y,

y,

f x  ,
f x  ,
dy
dy
,
dx x 3 dx 3,6 


f 3 ,
d 2y
dx 2
y 
f x 
f 7 
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Maths (M1)
D. Rule of Differentiation 微分法則
Rule 1
d

c   0
dx

d
du
cu   c
dx
dx

d
x n   nx n1
dx

d x
e  ex
dx

d
1
ln x 
dx
x

d x
a  a x ln a
dx

d
1
loga x 
dx
x ln a
Rule 2
d
du dv


u  v  
dx
dx dx

d
dv
du
v
uv   u
dx
dx
dx

d u 
 
dx  v 
log a x 
v
(Product rule 積法則)
du
dv
u
dx
dx (Quotient rule 商法則)
2
v
ln x
ln a
d
1 1
log a x 

dx
ln a x
let y  a x
ln y  x ln a
1 dy
 ln a
y dx
dy
 y ln a  a x ln a
dx
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Maths (M1)
D. Rule of Differentiation 微分法則
Rule 3 (Chain Rule 鏈式法則)
dy dy du


dx du dx
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Maths (M1)
D. Rule of Differentiation 微分法則
Find 求
dy
dx
(a) y  x 3  7
1
(b) y  x x  23
1
(c) y  x 2x  12
(d) y  x 2x 2  1
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Maths (M1)
D. Rule of Differentiation 微分法則
Find 求
dy
dx
3
(a) y  2x  82  3x 2
(b) y 
x
x 2
(c) y  x  2 3x  6  8x
(d) y 
9  4x 2
6  2x 2
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Maths (M1)
D. Rule of Differentiation 微分法則
6x
, find 求 f x  .
x 3
dr
2
(b) If 若 r  3 
, find 求
.
dt
2t
dV
4
(c) If 若 V  πr 3 , find 求
.
dr
3
dV
(d) If 若 V  243πr  πr 3 , find 求
.
dr
5π 2
dA
r , find 求
(e) If 若 A 
.
3
dr
(a) If 若 f x  
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Maths (M1)
D. Rule of Differentiation 微分法則
(a) If 若 y  x 2 h  x where 其中 h  constant 常數.
dy
Find 求
.
dx
dp
2.1
(b) If 若 p  8 
, find 求
.
dt
t4
3k
(c) If 若 x  4  λt
where 其中 k, λ  constant 常數.
2 k
dx
Find 求
.
dt
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Maths (M1)
D. Rule of Differentiation 微分法則
(a) If 若 g u   e u u 2  2u  2 , find 求
d
g u  .
du
dx
.
du
4
(c) If 若 y  1  ax   3e2x where 其中 a  constant 常數.
dy
Find 求
.
dx
(b) If 若 x  2u , find 求
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Maths (M1)
D. Rule of Differentiation 微分法則
(a) If 若 y 
dy
340
,
find
求
.
2  e t  2e 2t
dt
(b) Let h be a constant 設 h 為㇐常數
y  x 2 h  x , find 求
(c) If 若 g x   x 
dy
.
dx
5
 ln x 4 , find 求 g x  .
x
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Maths (M1)
D. Rule of Differentiation 微分法則
dy
ex
(a) If 若 y  3
, find 求
.
dx
x x 2
(b) Let a be a constant 設 a 為㇐常數
f x   ax 8  152x 5  4320x 2 , find 求 f x  .
(c) If 若 u  e 62t , find 求
du
.
dt
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Maths (M1)
D. Rule of Differentiation 微分法則
(a) If 若 N  Ae u , where A is a constant 其中 A 為常數,
dN
find 求
.
du
dy
(b) If 若 y  1  7 x 4  3e 2x , find 求
.
dx
(c) If 若 x  4 
3k
dx
, where 其中 k, λ =constant 常數, find 求
.
2 k
dt
λt
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Maths (M1)
D. Rule of Differentiation 微分法則
2
ln x 

(a) If 若 f x  
, find 求 f x  .
x
dV
d 2V
 t  1 3  t  1 , find 求
(b) If 若
.
dt
dt 2
t
t




5
10






(c) If 若 R t  4 4  e   9 2  e  , find 求 R t  .




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Maths (M1)
D. Rule of Differentiation 微分法則
(a) If 若 g x  
(b) If 若 N 
x3
, find 求 g x  and 及 g x  .
1  x6
40
, find 求
d 2N
dN
and 及
.
dt
dt 2
3 1  0. 2t  1
5
(c) If 若 g x   x   4 ln x , find 求 g x  and 及 g x  .
x
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Maths (M1)
D. Rule of Differentiation 微分法則
2
3
1
(a) If 若 f x   x x  23 , find 求 f x  and 及 f x  .
2
t
33 15
te , find 求 g t  and 及 g t  .
(b) If 若 g t  
10
(c) If 若 y t  
15 lnt 2  100
, find 求 y t  and 及 y t  .
16
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Maths (M1)
D. Rule of Differentiation 微分法則
1 2t
e , find 求 f t  and 及 f t  .
(a) If 若 f t  
t
(b) If 若 f t   lne t  t  , find 求 f t  and 及 f t  .
(c) If 若 f x  
e 0.1x
, find 求 f x  and 及 f x  .
x
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Maths (M1)
D. Rule of Differentiation 微分法則


(a) If 若 f x   ln x 2  16  ln3x  20 , find 求 f x  and 及 f x  .
(b) If 若 f x  
(c) If 若 f x  
ln x
, find 求 f x  and 及 f x  .
x2
x
, find 求 f x  and 及 f x  .
2x
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Maths (M1)
D. Rule of Differentiation 微分法則
(a) If 若 f x   e x ln x , find 求 f x  and 及 f x  .


(b) If 若 At   ln t 2  8t  95 , find 求 At  and 及 At  .


(c) If 若 At   5 ln t 2  8t  20 , find 求 At  and 及 At  .
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Maths (M1)
D. Rule of Differentiation 微分法則
(a) If 若 At  
601  10t 
, find 求 At  and 及 At  .
e 2t
(b) If 若 pt   2t ln t 2  4  , find 求 p t  and 及 p t  .
(c) If 若 S 
200
1  42
1
t
2
, find 求
dS
d 2S
and 及
.
dt
dt 2
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Maths (M1)
D. Rule of Differentiation 微分法則
27
dN
d 2N
,
find
求
and
及
.
2  0.03te 0.1t
dt
dt 2
40
dN
d 2N
(b) If 若 N  10.2t
, find 求
and 及
.
3
1
dt
dt 2
(a) If 若 N 
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Maths (M1)
D. Rule of Differentiation 微分法則
64
(a) If 若 V 
e
1
t
2
4
, find 求
dV
d 2V
and 及
.
dt
dt 2
32
dP
d 2P
(b) If 若 P  54t
, find 求
and 及
.
dt
dt 2
2
8
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Maths (M1)
Content
Tangent
切線
A.
Properties of Tangent
切線的特性
B.
Type 1 : Given Internal Point
已知內點
C.
Type 2 : Given Slope
已知斜率
D.
Type 3 : Given External Point
已知外點
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Maths (M1)
A. Properties of Tangent 切線的特性
Tangent Properties
切線特徵:
Each point on the curve has one tangent only.

每㇐點在曲線上只有㇐條切線。
A line can only touch the curve at one point.

㇐直線只能相切於曲線上的㇐點。
  0
B. Type 1: Given Internal Point 已知內點
Given Point Presentation 已知點演繹:
(3, 4)
(a, a)
when x = 3 / when y = 3
x-coordinate = 3 / y-coordinate = 2
cut y-axis / cut x-axis







Point – Given
Slope – f '  x1 
Use point-slope form to find the tangent
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Maths (M1)
C. Type 2: Given Slope 已知斜率
Given that Slope of required tangent  α
已知所需切線的斜率  α
D. Type 3: Given External Point 已知外點
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Maths (M1)
Content
Rate of change
改變率
A.
Relation Chart
關係圖
B.
Translate
翻譯
C.
3 Steps
三個步驟
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Maths (M1)
A. Relation Chart 關係圖
(a) g u   e
(b) r  3 
u
1
3
u  2u  2 , u  x .
2
2
4
, V  πr 3 .
2t
3
4 3
5π 2 dV
πr , A 
r ,
 2π .
9
3
dt
2.1
(d) p  8 
, C  2p .
t4
(c) V 
dg u 
.
dx
dV
Find 求
.
dt r 2.5
Find 求
dA
.
dt r 3
dC
Find 求
.
dt t 5
Find 求
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Maths (M1)
A. Relation Chart 關係圖
(a) V 
4 3 dV
πr ,
 100 .
3
dt
Find 求
dr
.
dt r 10
(b) u  e 62t , N  Ae u , where A is a constant 其中 A 為常數
dN
dN
Find
in terms of u. 求以 u 表
。
dt
dt
dy
(c) y  1  7 x 4  3e 2x , x  2 u .
Find 求
.
du u  0
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Maths (M1)
B. Translate 翻譯

f(x) is increasing
f(x) 遞增

f(x) is decreasing
f(x) 遞減

Rate of change of A
A 的變率

Leaking at a constant rate of 100 cm3/s
以每秒 100 cm3 的恆速漏氣
C. 3 Steps 三個步驟
Translate question
Connect relationship
Find the answer
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Maths (M1)
Content
Maximum, Minimum, Greatest, Least
極大、極小、最大、最小
A.
Concept
概念
B.
Vocabulary
生字
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Maths (M1)
A. Concept 概念
a
b
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Maths (M1)
B. Vocabulary 生字
y  x2
Q1: Find the minimum value?
求極小值？
Q2: Find the minimum point?
求極小點？
Q3: y attain minimum when x  x 0 , find x 0 .
y 達致極小當 x  x 0 ，求 x 0 。
Find the maximum slope.
求極大斜率？
Rate of change of volume attain maximum when t  1 .
當 t  1 時，體積的改變率達致極大。
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Maths (M1)
Content
Integration
積分
A.
Concept of Integration
積分概念
B.
Basic Integration Rule
基本積分技巧
C.
Integration by substitution
代換積分法
D.
Integration Training
積分訓練
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Maths (M1)
A. Concept of Integration 積分概念
Integration ~ Reverse of Differentiation
d
x 2  3x  9  2x  3
dx
d

x 2  3x  5   2x  3
dx
d

x 2  3x  k   2x  3
dx
 2x  3 dx  x  3x  C
2

where C is a constant
其中 C 為常數
Indefinite Integral 不定積分 :
 2x  3 dx  x  3x  C
2


Definite Integral 定積分 :  2x  3 dx  x 2  3x a  b 2  3b   a 2  3a 
b
a
b
B. Basic Integration Rule 基本積分技巧
x n 1
C
n 1

n
 x dx 

0
 1 dx   x dx 





x 0 1
C  x C
0 1
 f x   g x dx   f x dx   g x dx
 kf x dx  k  f x dx
 e dx  e  C
x
x
ax
 a dx  ln a  C
1
 x dx  ln x  C
x
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Maths (M1)
C. Integration by substitution 代換積分法
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Maths (M1)
D. Integration Training 積分訓練



 x  7   12x  9dx
 e dx
 e  6e  8dt
3
2x
t
0.5t
3

1

  2x  x  dx
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Maths (M1)
D. Integration Training 積分訓練
t

 1  t dt



501  10t 
dt
1  2t
6x
 x  3dx
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Maths (M1)
D. Integration Training 積分訓練
t

 4t  1 dt

 t  3dt


t 8
61t
5
t  12
dt
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Maths (M1)
D. Integration Training 積分訓練




t 9  t2
 3 dt
t 2
 t 2  4t  11dt
t 2  3t  9
 t 2  4t  11dt
12x  48
 3x  24x  49 dx
2
2
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Maths (M1)
D. Integration Training 積分訓練

ln x
 x dx
ln x 2 dx



 x ln x dx

1

x
e
ln x 2  3ln x  2 dx
x
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Maths (M1)
D. Integration Training 積分訓練

x
1
2
 3 x dx

23 x
 1  23x dx

 t  1 3  t  1 dt
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Maths (M1)
D. Integration Training 積分訓練

230  t   10
 t  35  25dt
2
t
5

  4e

3

32t 2
 1  32t dt
2x
 9e
t
10
 2dt
 10  3x  9dx
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Maths (M1)
D. Integration Training 積分訓練



d
x ln x  , find 求  ln xdx .
dx
d
ve v , find 求  ve v dv .
By considering 考慮
dv
d
xemx  , find 求  xe mx dx .
By considering 考慮
dx
By considering 考慮
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Maths (M1)
D. Integration Training 積分訓練


d
x  7 x  , find 求  x  7 x dx .
dx
d
xe x ln x  , find 求  x  1e x ln x  1 dx .
By considering 考慮
dx
x
By considering 考慮
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Maths (M1)
D. Integration Training 積分訓練


d
x 6  1lnx 2  1 , find 求  x 5 lnx 2  1dx .
dx
1
1
t 
t
d  20
20
 te  , find 求  te dt .
By considering 考慮


dt 

By considering 考慮
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Maths (M1)
D. Integration Training 積分訓練
dN
225  t 
,
 0.04 t
dt
e
 4t
By considering v  1  4te 0.04t and
考慮 v  1  4te 0.04t 及
dv
, find N in terms of t.
dt
dv
，求以 t 表 N。
dt
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Maths (M1)
D. Integration Training 積分訓練
du
t 2  2t  1
and
.
2
dt
t  2t  1
t2  1
Find  2
dt .
t  2t  1 t 2  2t  1
By considering u 



du
t 2  2t  1
及
。
2
dt
t  2t  1
t2  1
求  2
dt 。
t  2t  1 t 2  2t  1
考慮 u 



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Maths (M1)
Content
Area
面積
A.
Concept of Area
面積概念
B.
Area Training
面積訓練
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Maths (M1)
A. Concept of Area 面積概念
y
y  f x 
a
x
b
b
Shaded Area 陰影面積   f x dx
a
a
 no area is included 沒有面積
 f x dx  0
 f x dx    f x dx
 kf x dx  k  f x dx
 f x   gx dx   f x dx   g x dx  vertical add-up 垂直增加
 2 parts of area 兩部份的面積
 f x dx   f x dx   f x dx

a
b

a
a
b

b
b
a
b


a
b
b
a
b
c
a
b
a
a
a
c
Remarks 備註:
y
+ve
+ve
–ve
x
–ve
If the curve is above the x-axis, the value is positive, or else is negative.
若曲線在 x 軸上，它的值為正數，否則為負數。
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Maths (M1)
B. Area Training 面積訓練
y
5
3
4
x
1
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Maths (M1)
Content
Trapezoidal Rule
梯形法則
A.
Concept of Trapezoidal Rule
梯形法則的概念
B.
Over-estimate, under-estimate
高估、低估
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Maths (M1)
A. Concept of Trapezoidal Rule 梯形法則的概念
b a
 f x dx  2n f  f  2f  f  ...  f 
b
a
0
n
1
2
n 1
B. Over-estimate, under-estimate 高估、低估
f x   0
for a  x  b

under-estimate 低估
f x   0
for a  x  b

over-estimate
高估
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Maths (M1)
Content
Curve  Straight line
曲線直線
A.
Demo
示範
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Maths (M1)
A. Demo 示範
Demo 1:
Curve : y  ae bx  8 where a, b  integers
曲線 : y  ae bx  8 其中 a, b  整數
x
y
1
30
Find a, b.
求 a、b。
4
8951
7
3607821
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Maths (M1)
A. Demo 示範
Demo 2:
400
where b, k  integers
ke bt  3
400
曲線 : y  bt
其中 b, k  整數
ke  3
Curve : y 
t
y
0.2
19.8
Find b, k.
求 b、k。
0.4
5.4
0.6
1.6
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Maths (M1)
Content
Bayes’ Theorem
貝葉斯定理
A.
Concept of Bayes’ Theorem
貝葉斯定理概念
B.
Formula of Bayes’ Theorem
貝葉斯公式
C.
Reverse Tree Diagram
反樹形圖
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Maths (M1)
A. Concept of Bayes’ Theorem 貝葉斯定理概念
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Maths (M1)
B. Formula of Bayes’ Theorem 貝葉斯公式

P A  B  P B  A

P B 
P B 
P A  B 
P B A 
P A
 P A  B   P B   P A B   P A  P B A
P A B  

P A  P A  1

P A  P A  B   P A  B 

P A  B   P A  P B   P A  B 

Independent 獨立
P A  B   P A  P B 
P A B   P A B   P A

Mutually Exclusive 互斥
P A  B   0
P A  B   P A  P B 
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Maths (M1)
C. Reverse Tree Diagram 反樹形圖
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Maths (M1)
Content
Normal Distribution 正態分佈
A.
Properties of normal distribution
正態分佈特性
B.
Normal Distribution Table
正態分佈表
C.
Training
練習
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Maths (M1)
A. Properties of Normal Distribution 正態分佈特性

Symmetry 對稱

Mean 平均 = Mode 眾數 = Median 中位數

Tend to infinity and negative infinity 兩邊無限伸延

x   , f x   0

 f x dx  1

Bell-shape curve 鐘形曲線

X ~ N μ, σ 2 


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Maths (M1)
正態分佈 02
B. Normal Distribution Normal
TableDistribution
正態分佈表
f x  
1
2
e
1  x 
 

2  
2
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.0
.0000
.0040
.0080
.0120
.0160
.0199
.0239
.0279
.0319
.0359
0.1
.0398
.0438
.0478
.0517
.0557
.0596
.0636
.0675
.0714
.0753
0.2
.0793
.0832
.0871
.0910
.0948
.0987
.1026
.1064
.1103
.1141
0.3
.1179
.1217
.1255
.1293
.1331
.1368
.1406
.1443
.1480
.1517
0.4
.1554
.1591
.1628
.1664
.1700
.1736
.1772
.1808
.1844
.1879
0.5
.1915
.1950
.1985
.2019
.2054
.2088
.2123
.2157
.2190
.2224
0.6
.2257
.2291
.2324
.2357
.2389
.2422
.2454
.2486
.2517
.2549
0.7
.2580
.2611
.2642
.2673
.2704
.2734
.2764
.2794
.2823
.2852
0.8
.2881
.2910
.2939
.2967
.2995
.3023
.3051
.3078
.3106
.3133
0.9
.3159
.3186
.3212
.3238
.3264
.3289
.3315
.3340
.3365
.3389
1.0
.3413
.3438
.3461
.3485
.3508
.3531
.3554
.3577
.3599
.3621
1.1
.3643
.3665
.3686
.3708
.3729
.3749
.3770
.3790
.3810
.3830
1.2
.3849
.3869
.3888
.3907
.3925
.3944
.3962
.3980
.3997
.4015
1.3
.4032
.4049
.4066
.4082
.4099
.4115
.4131
.4147
.4162
.4177
1.4
.4192
.4207
.4222
.4236
.4251
.4265
.4279
.4292
.4306
.4319
1.5
.4332
.4345
.4357
.4370
.4382
.4394
.4406
.4418
.4429
.4441
1.6
.4452
.4463
.4474
.4484
.4495
.4505
.4515
.4525
.4535
.4545
1.7
.4554
.4564
.4573
.4582
.4591
.4599
.4608
.4616
.4625
.4633
1.8
.4641
.4649
.4656
.4664
.4671
.4678
.4686
.4693
.4699
.4706
1.9
.4713
.4719
.4726
.4732
.4738
.4744
.4750
.4756
.4761
.4767
2.0
.4772
.4778
.4783
.4788
.4793
.4798
.4803
.4808
.4812
.4817
2.1
.4821
.4826
.4830
.4834
.4838
.4842
.4846
.4850
.4854
.4857
2.2
.4861
.4864
.4868
.4871
.4875
.4878
.4881
.4884
.4887
.4890
2.3
.4893
.4896
.4898
.4901
.4904
.4906
.4909
.4911
.4913
.4916
2.4
.4918
.4920
.4922
.4925
.4927
.4929
.4931
.4932
.4934
.4936
2.5
.4938
.4940
.4941
.4943
.4945
.4946
.4948
.4949
.4951
.4952
2.6
.4953
.4955
.4956
.4957
.4959
.4960
.4961
.4962
.4963
.4964
2.7
.4965
.4966
.4967
.4968
.4969
.4970
.4971
.4972
.4973
.4974
2.8
.4974
.4975
.4976
.4977
.4977
.4978
.4979
.4979
.4980
.4981
2.9
.4981
.4982
.4982
.4983
.4984
.4984
.4985
.4985
.4986
.4986
3.0
.4987
.4987
.4987
.4988
.4988
.4989
.4989
.4989
.4990
.4990
3.1
.4990
.4991
.4991
.4991
.4992
.4992
.4992
.4992
.4993
.4993
3.2
.4993
.4993
.4994
.4994
.4994
.4994
.4994
.4995
.4995
.4995
3.3
.4995
.4995
.4995
.4996
.4996
.4996
.4996
.4996
.4996
.4997
3.4
.4997
.4997
.4997
.4997
.4997
.4997
.4997
.4997
.4997
.4998
3.5
.4998
.4998
.4998
.4998
.4998
.4998
.4998
.4998
.4998
.4998
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Maths (M1)
C. Training 練習

P 0  Z  2

P 0  Z  0.63

P 0  Z  4 

P  1.5  Z  0 

P Z  2.46

P Z  1.4 

P Z  2.16

P Z  0.78
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Maths (M1)
C. Training 練習

P 1.3  Z  1.16

P 1.3  Z  1.16

P  2.1  Z  1.15

P 0  Z  a   0.3438

P a  Z  0   0.3264

P Z  a   0.0606

P Z  a   0.3974

P b  Z  b   0.663
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Maths (M1)
C. Normal Distribution Basic Training 正態分佈特訓
X ~ N 100,400 

P 100  X  150 

P X  120 

P 55  X  65

P 105  X  85
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Maths (M1)
C. Normal Distribution Basic Training 正態分佈特訓
X ~ N 20,4 

P X  a   0.6026

P X  a   0.9066

P X  a   0.1587

P 20  a  X  20  a   0.668
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Maths (M1)
C. Normal Distribution Basic Training 正態分佈特訓
X ~ N μ, σ 2 


P 35  X  65  0.5468 &
P 18  X  24   0.2746 &
P X  65  0.2266
P X  18  0.119
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Maths (M1)
Content
Binomial Distribution 二項分布
A.
Continuous vs. Discrete
連續 vs. 離散
B.
Binomial vs. Poisson Distribution
二項 vs. 泊松分佈
C.
Origin of Binomial Distribution
二項分佈由來
D.
Formula of Binomial Distribution
二項分佈公式
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Maths (M1)
A. Continuous vs Discrete 連續 vs 離散
Continuous: Lifetime, concentration, Intelligent quantity level, weight,
length…
Discrete:
number of passenger, number of accident, population,
number of complaint…
連續︰ 壽命、濃度、智商、體重、⾧度 …
離散︰ 乘客人數、意外數字、人口、投訴個案 …
B. Binomial vs. Poisson Distribution 二項 vs.泊松分佈
with upper limit:
Binomial Distribution
e.g. number of even numbers outcome out of 10 dice
drawings
without upper limit: Poisson Distribution
e.g. number of typhoons in the next year
有上限 :
二項分佈
例: 擲 10 次骰子，出現雙數的次數
無上限 :
泊松分佈
例: 明年會出現的颱風數目
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Maths (M1)
C. Origin of Binomial Distribution 二項分佈由來
A game winning probability is p.
遊戲勝出的概率為 p
A game losing probability is q.
遊戲落敗的概率為 q
(p+q=1)
Play 2 times 玩 2 次
p  q 2  p 2  2pq  q 2
win 2
win 1
win 0
Play n times 玩 n 次
p  q n  C 0n p n q 0  C1n p n 1q 1  ...  Cnn p n n q n
N X  0   C 0n

P X  0   C 0n p n 0 q 0
N X  1  C1n

P X  1  C1n p n 1 q 1
N X  2  C2n

P X  2  C2n p n 2 q 2

P X  r   Crn p n r q r

P X  n   Cnn p n n q n
…
N X  r   Crn
…
N X  n   Cnn
D. Formula of Binomial Distribution 二項分佈公式
P X  r   Crn p n r q r
Expected Value 期望值︰
E X   np
Variance
Var X   np 1  p 
方差︰
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Maths (M1)
Content
Poisson Distribution 泊松分佈
A.
Conditions of Poisson Distribution 泊松分佈條件
B.
Formula of Poisson Distribution
泊松分佈公式
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Maths (M1)
A. Conditions of Poisson Distribution 泊松分佈條件
Satisfy 3 conditions:
滿足 3 大要求︰
Binomial Distribution  Poisson Distribution
二項分佈  泊松分佈
(a) n  
(b) p  0
(c) λ  np
e.g. number of typhoons, number of telephone calls,
number people entering lift
例: 颱風數目、電話接聽數目、入升降機人數
B. Formula of Poisson Distribution 泊松分佈公式
For x  0, 1, 2, 3, ...
對於 x  0, 1, 2, 3, ...
e λ
P X  x  
x!
where λ = mean
e  λ λx
P X  x  
x!
其中 λ = 平均值
λ x
Expected Value 期望值︰
E X   λ
Variance
Var X   λ
方差︰
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Maths (M1)
Content
E(X), Var(X)
A.
Formula of E(X), Var(X)
E(X), Var(X) 公式
B.
Demo
示範
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Maths (M1)
A. Formula of E(X), Var(X)
E(X), Var(X) 公式
n

x
E X   k 1
n
k
n

 x  E X 
2
k
Var X   k 1
n
n
x
Var X   k 1
k
2
n
k 1
n
 xk 2
Var X   k 1
n
 2 xk E X    E X 
n
n
n
 xk E X 
 2 k 1
n
Var X   E X 2   2E X E X  
2
k 1
n
 E X 
2
 k 1
n E X 
n
2
Var X   E X 2   2E X   E X 
2
n
n


xk 


 E X   k 1 

n 




2
Var X   E X 2   E X 
2

Var X   E X 2   E X 

E aX  b   aE X   b

E X 2   E X 

Var aX  b   a 2Var X 

Var X 2   Var X 2
2
2
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Maths (M1)
B. Demo 示範
The table below shows the probability distribution of a discrete random
variable X , where a and b are constants.
下表顯示㇐離散隨機變量 X 的概率分佈，其中 a 及 b 均為常數。
x
P X  x 
1
0.3
2
a
5
b
It is given that E X   2.9 .
(a) Find a , b and Var X  .
(b) Find E 5  3X  and Var 5  3X  .
(c) Find E 5  3X 3  and Var 5  3X 3  .
已知 E X   2.9 。
(a) 求 a 、b 及 Var X  。
(b) 求 E 5  3X  及 Var 5  3X  。
(c) 求 E 5  3X 3  及 Var 5  3X 3  。
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Maths (M1)
Content
Central limit theorem
中心極限定理
A.
Central limit theorem
中心極限定理
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Maths (M1)
A. Central limit theorem 中心極限定理
Any Distributions
任何分佈
A random sample of size n
隨機抽取㇐個容量為 n 的樣本
Normal Distribution 正態分佈
X  Sample Mean 樣本平均
Var X 
Var X 
n
 
[ X  the mean of n independent random observation of X]
[ X  n 個 X 的獨立隨機觀察的平均值]
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Maths (M1)
Content
Confidence interval 置信區間
A.
Confidence interval
置信區間
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Maths (M1)
A. Confidence interval 置信區間
Population
Mean
μ
-
σ
Standard Deviation
總體
-
平均
μ
-
標準差
σ
A random sample of n
隨機抽 n 個樣本
Sample
- Sample Mean  E X  unbiased estimate for μ
-
 
  σn
Var X 
2
樣本
 
-
樣本平均  E X  μ 的無偏估計值
-
Var X 
  σn
2
95% Confidence interval for μ
μ 的 95% 置信區間
σ
σ 

  E X  1.96 
, E X  1.96 

n
n

 
 
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Maths (M1)