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Chapter 3-Normal distribution
f X ( x) 
2

1
1 x 
exp   
 
2
 2    
  x  
X : N (, )
Example: N (4,2) , P (X < 6.03)
P (5 < X < 6.03)
lognormal distribution
2

1
1  ln x    
exp   


2 x
 
 2  
f X ( x) 
x0
fX(x)
0
2
4
6
x
8
lognormal distribution
1 2
  ln   
2
  ln xm
2



2
  ln 1  2   ln 1   2 
  
 ln a   
P( X  a )   




If X ~LN (, 
lnX ~ N (, 
Example
1. -1(0.95) = 1.645

How about the settlement is Log-normal?
exponential distribution
fX(x)
f X ( x )  e

E( X ) 
1

1
Var ( X )   

2
x
x0
x
 X  100%
Beta distribution
q 1
r 1
( q  r ) ( x  a ) (b  x)
f X ( x) 
q  r 1
( q)( r )
(b  a )
a xb
(x)
fX
0.3
q = 2.0 ; r = 6.0
0.2
0.1
probability
x
0.0
0
2
a = 2.0
4
6
8
10
12
b = 12
Standard Beta distribution
fX(x)
(a = 0, b = 1)
4
q = 1.0 ; r = 4.0
3
q = r = 3.0
q = 4.0 ; r = 2.0
2
q = r = 1.0
1
0
x
0
0.2
0.4
0.6
0.8
1
The difference between Beta and other similar distribution
Review of Bernoulli sequence model

x success in n trials:
binomial

time to first success:
geometric

time to kth success:
negative binomial
n x
n x
 x  p (1  p )
 
t 1
(1  p) p
 t 1  k
t k
 k  1 p  (1  p)


Ex 3.54
Statistics show that 20% of freshman in
engineering school quit in 1 year. What is the
probability that among eight students selected
at random, two of them will quit after 1 year?
Think:
1.
2.
3.
Continuous or discrete?
Students cannot pass or fail “continuously”
Binomial, Geometric or Negative binomial?
Bi: x success in n trials (orderless)
Geo: time to first success (ordered)
Neg: time to kth success (last term ordered)
p = 0.2
8
2
8 2
P( X  2)     0.2  1  0.2   0.293
 2
What is the probability of at least two of them will fail
after 1 year?
Use T.O.T:
P (X ≥ 2)
= 1 – P(X = 0) – P(X = 1)
8
8
0
8 0
1
81
 1     0.2  1  0.2      0.2  1  0.2 
0
1
what is the probability that among eight students selected at
random, two of them will quit within 2 years?
Approach 1: Bayes theorem + TOT
We first consider 1st year scenario:
8
0
80
P(0 student quit in 1st year)     0.2  1  0.2   0.167
0
8
1
8 1
P(1 student quit in 1st year)     0.2  1  0.2   0.335
1
8
2
8 2
P(2 student quit in 1st year)     0.2  1  0.2   0.293
 2
Why not consider X = 3, 4…...8?
For 2nd year:
P
= P(0 student in 1st year) P(2 student in 2nd year)
+ P(1 student in 1st year) P(1 student in 2nd year)
+ P(2 student in 1st year) P(0 student in 2nd year)
8
2
8 2
P (2 student quit in 2nd year)     0.2  1  0.2   0.293
 2
7
1
7 1
P (1 student quit in 2nd year)     0.2  1  0.2   0.367
1
6
0
60
P (0 student quit in 2nd year)     0.2  1  0.2   0.262
0
P = (.167)(.293) + (.335)(.367) + (.293)(.262) = .249
Approach 2: Geometric
Recall geometric is “first time to success”, (1-p)t-1p
Students can quit at 1st and 2nd year.
i.e. t=1, t =2
When t = 2, 1st year pass is defined.
P (t = 1) = 0.2
P (t =2) = (0.8)2-10.2 = 0.16
P (a student quit in 1 or 2 year)
= 0.2 + 0.16 = 0.36
8
2
8 2
P     0.36  1  0.36   0.294
 2