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Chapter 3-Normal distribution
x
Example: N (4,2) , P (X < 6.03)
P (5 < X < 6.03)
X : N (, )
lognormal distribution
0 2 4 6 8
fX(x)
x
lognormal distribution
21ln
2
2
2 22
ln 1 ln 1
ln mx
ln( )
aP X a
If X ~LN (
lnX ~ N (
Example
1. -1(0.95) = 1.645 How about the settlement is Log-normal?
exponential distribution
x
fX(x)
( ) xXf x e x 0
1( )E X
21
( )Var X
100%X
Beta distribution1 1
1
( ) ( ) ( )( )
( ) ( ) ( )
q r
X q r
q r x a b xf x
q r b a
a x b
0.0
0.1
0.2
0.3
0 2 4 6 8 10 12x
fX(x)q = 2.0 ; r = 6.0
a = 2.0 b = 12
probability
Standard Beta distribution
0
1
2
3
4
0 0.2 0.4 0.6 0.8 1
q = 1.0 ; r = 4.0
q = r = 3.0 q = 4.0 ; r = 2.0
q = r = 1.0
x
fX(x)
The difference between Beta and other similar distribution
(a = 0, b = 1)
Review of Bernoulli sequence model
x success in n trials:
binomial
time to first success:
geometric
time to kth success:
negative binomial
1(1 )tp p
(1 )x n xnp p
x
1(1 )
1k t ktp p
k
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. Continuous or discrete?
Students cannot pass or fail “continuously”
2. 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)
3. p = 0.2
2 8 28( 2) 0.2 1 0.2 0.293
2
P X
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)
0 8 0 1 8 18 81 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:
Why not consider X = 3, 4…...8?
0 8 0
1 8 1
2 8 2
8(0 student quit in 1st year) 0.2 1 0.2 0.167
0
8(1 student quit in 1st year) 0.2 1 0.2 0.335
1
8(2 student quit in 1st year) 0.2 1 0.2 0.293
2
P
P
P
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)
P = (.167)(.293) + (.335)(.367) + (.293)(.262) = .249
2 8 2
1 7 1
0 6 0
8(2 student quit in 2nd year) 0.2 1 0.2 0.293
2
7(1 student quit in 2nd year) 0.2 1 0.2 0.367
1
6(0 student quit in 2nd year) 0.2 1 0.2 0.262
0
P
P
P
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
2 8 280.36 1 0.36 0.294
2
P