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