Chapter 3-Normal distribution

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Presentation transcript:

Chapter 3-Normal distribution X : N (, ) Example: N (4,2) , P (X < 6.03) P (5 < X < 6.03)

lognormal distribution fX(x) x

lognormal distribution If X ~LN (l, z) lnX ~ N (l, z)

Example 1. F-1(0.95) = 1.645 How about the settlement is Log-normal?

exponential distribution fX(x) x  0

Beta distribution x fX(x) q = 2.0 ; r = 6.0 probability a = 2.0 b = 12

Standard Beta distribution fX(x) (a = 0, b = 1) q = 1.0 ; r = 4.0 q = r = 3.0 q = 4.0 ; r = 2.0 q = r = 1.0 x 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

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” 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

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)

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?

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

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