Normal distribution X : N (, ) fX(x) x  = 5 N (5, 2)

Effect of varying parameters ( & ) fX(x)   for C   for B B C A x

Standard normal distribution S: N (0,1) fX(x) x

a

Page 380 Table of Standard Normal Probability

Given probability (a) = p,  a = -1(p) = ?

fX(x) x  a b

Example: retaining wall Suppose X = N(200,30) x F

If the retaining wall is designed such that the reliability against sliding is 99%, How much friction should be provided? 2.33

Lognormal distribution Parameter l  fX(x) x

Parameters   for   0.3,

Probability for Log-normal distribution If a is xm, then  is not needed.

P 3.19 Project completion time T a) Given information: T is normal  = 30 P (T<40) = 0.9   = 7.81

P (T < 50) b) P ( T < 0 ) Normal distribution ok? Yes

c) If assume Log-normal distribution for T, with same value of m and s.  = 7.81/30 = .26 P (T<50)

Construction job that has a fleet of similar equipments In order to insure satisfactory operation, you require at least 90% equipments available.

Lognormal with mean 6 months, c.o.v. 25% Each equipment has a breakdown time T: time until break down Lognormal with mean 6 months, c.o.v. 25%

0) Suppose scheduled time period for maintenance is 5 months P (an equipment will break down before 5 months) = P (T<5) Expect 27% of equipment will not be operative ahead of the next scheduled maintenance.

P (breakdown of an equipment) - Suppose need at least 90% equipment available at any time - What should be the scheduled maintenance period to? P (breakdown of an equipment) = P (T < to)  0.1  to = 4.22 months

P (will go for at least another month | has survived 4.22 months) = ? b) If to = 4.22 months P (will go for at least another month | has survived 4.22 months) = ? = P (T > 5.22 | T > 4.22) = 0.6 = 0.749

p.224-225: table of common distribution Other distributions Exponential distribution Triangular distribution Uniform distribution Rayleigh distribution p.224-225: table of common distribution

Exponential distribution fX(x) x  0

Example of application Quake magnitude Gap between cars Time of toll booth operative

Given: mean earthquake magnitude Example: Given: mean earthquake magnitude = 5 in Richter scale P (next quake > 7)

Shifted exponential distribution Lower bound not necessarily 0 x  a

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

Standard beta PDF (a = 0, b = 1) fX(x) x q = 1.0 ; r = 4.0 q = r = 3.0

Bernoulli sequence and binomial distribution Consider the bulldozer example If probability of operation = p and start out with 3 bulldozers, what is the probability of a given number of bulldozers operative?

ppp 3pp(1-p) 3p(1-p)(1-p) (1-p)3 Let X = no. of bulldozers operative ppp GGG GGB GBG BGG BBG BGB GBB BBB X = 3 X = 2 3pp(1-p) X = 1 3p(1-p)(1-p) (1-p)3 X = 0 binomial coefficients

Suppose start out with 10 bulldozers P (8 operative) = P( X=8 ) If p = 0.9, then P (8 operative)

Bernoulli sequence Discrete repeated trials 2 outcomes for each trial p = probability of a success Discrete repeated trials 2 outcomes for each trial s.i. between trials Probability of occurrence same for all trials

P ( x success in n trials) Binomial distribution S F x = number of success p = probability of a success P ( x success in n trials) = P ( X = x | n, p)

Examples Number of flooded years Number of failed specimens Number of polluted days

Example: Given: probability of flood each year = 0.1 Over a 5 year period P ( at most 1 flood year) = P (X =0) + P(X=1) = 0.95 + 0.328 = 0.919

P (flooding during 5 years) = P (X  1) = 1 – P( X = 0) = 1- 0.95 = 0.41

E (no. of success ) = E(X) = np For binomial distribution E (no. of success ) = E(X) = np Over 10 years, expected number of years with floods E (X) = 10  0.1 = 1

P (first flood in 3rd year) = ?

Geometric distribution In general, T = time to first success P (T = t) = (1-p)t-1p t=1, 2, … geometric distribution Return period

P (2nd flood in 3rd year) = P (1 flood in first 2 years) P (flood in 3rd year)

P (kth flood in tth year) In general P (kth flood in tth year) = P (k-1 floods in t-1 year) P (flood in tth year) negative binomial distribution

Review of Bernoulli sequence No. of success  binomial distribution Time to first success  geometric distribution E(T) =1/p = return period

Significance of return period in design Service life Suppose 中銀 expected to last 100 years and if it is designed against 100 year-wind of 68.6 m/s design return period P (exceedence of 68.6 m/s each year) = 1/100 = 0.01 P (exceedence of 68.6 m/s in 100th year) = 0.01

P (1st exceedence of 68.6 m/s in 100th year) = 0.99990.01 = 0.0037 P (no exceedence of 68.6 m/s within a service life of 100 years) = 0.99100 = 0.366 P (no exceedence of 68.6 m/s within the return period of design) = 0.366

If it is designed against a 200 year-wind of 70.6 m/s P (exceedence of 70.6 m/s each year) = 1/200 = 0.005 P (1st exceedence of 70.6 m/s in 100th year) = 0.995990.005 = 0.003

P (no exceedence of 70.6 m/s within a service life of 100 years) = 0.995100 = 0.606 > 0.366 P (no exceedence of 70.6 m/s within return period of design) = 0.995200 = 0.367

How to determine the design wind speed for a given return period? Get histogram of annual max. wind velocity Fit probability model Calculate wind speed for a design return period

Frequency Example N (72,8) 0.01 V100 Annual max wind velocity Design for return period of 100 years: p = 1/100 = 0.01 V100 = 90.6 mph

Alternative design criteria 1 Suppose we design it for 100 mph, what is the corresponding return period? T  4300 years

Pf = P (exceedence within 100 years) Probability of failure Pf = P (exceedence within 100 years) = 1- P (no exceedence within 100 years) =1- (1-0.000233)100 = 0.023

Let T = design return period Alternative design criteria 2 If P (exceedence within the life time of the building, i.e., 100 years) = 0.01 Q: What should be the design wind velocity? Let T = design return period P (annual exceedence) = 1/T P (no exceedence in 100 year) =(1-1/T)100 = 1- 0.01

 T = 10000 year VD = 101.76 mph

Hence, 1- P(no exceedence in 25 years) = 0.3 A preliminary planning study on the design of a bridge over a river recommended an acceptable probability of 30% of the bridge being inundated by flood in the next 25 years a) p : probability of exceedence in one year ? P (exceedence of design flood within 25 years) = 0.3 Hence, 1- P(no exceedence in 25 years) = 0.3  p = 0.0142

Return period of design flood b ) what is the return period for the design flood? Return period of design flood T = 1/p = 1/0.0142 = 70.4 year

Review of Bernoulli sequence model x success in n trials: binomial time to first success: geometric time to kth success: negative binomial

Mean number of occurrence in 6 min = 9 Suppose: average rate of left turns is n = 1.5 /min Q: P (8 LT’s in 6 min) = ? Mean number of occurrence in 6 min = 9 (a) 6 min divided into 30 second interval  No. of interval = 12

(b) 6 min divided into 10 second interval  No. of interval = 36 (c) In general, P ( 8 occurrences in n trials) = No. of occurrences in time interval = nt

P ( x occurrences in n trials) = x = 0, 1, 2, … Poisson distribution

e.g. x = 8, t = 6 min; n = 1.5 per min P (2 LT’s in 30 sec) = P(X = 2)

P (at least 2 LT’s in 1 min) = P(X2) = 1- P(X=0)-P(X=1)

Poisson Process An event can occur at random and at any time or any point in the space Occurrence of an event in a given interval is independent of any other nonoverlapping intervals.

Example: Mean rate of rainstorm is 4 per year P (2 rainstorms in next 6 months) P (at least 2 rainstorms in next 6 months) = P(X2) = 1- P(X=0)-P(X=1)

Design of length of left-turn bay Suppose LT’s follow a Poisson process 100 LT’s per hour How long should LT bay be? Assume: all cars have the same length Let the bay be measured in terms of no. of car length k Suppose traffic signal cycle = 1 min Cars waiting for LT will be clear at each cycle

If k = 0 ,ok 19% of time If k = 1 , ok 50% If k = 2 , ok 76% If k = 3 , ok 91% If k = 4 , ok 97%

If criteria changes, k changes Suppose criteria is adequate 96% of time  k = 4 In general, k = ? If criteria changes, k changes

Variance of Poisson r.v.: nt Mean of Poisson r.v.: nt Variance of Poisson r.v.: nt c.o.v. =

P 3.42 Service stations along highway are located according to a Poisson process Average of 1 station in 10 miles  n = 0.1 /mile P(no gasoline available in a service station)

No. of service stations (a) P( X  1 in 15 miles ) = ?

binomial (b) P( none of the next 3 stations have gasoline) No. of stations with gasoline binomial

(c) A driver noticed the fuel gauge reads empty; he can go another 15 miles from experience. P (stranded on highway without gasoline) = ? No. of station in 15 miles P (S) binomial Poisson

Total = 0.301 x P( S| X = x ) P( X = x ) P( S| X = x ) P( X = x ) 1 1 e-1.5 = 0.223 0.223 0.2 1.5 e-1.5 = 0.335 0.067 2 0.22 1.52/2! e-1.5 = 0.251 0.010 3 0.23 1.53/3! e-1.5 = 0.126 0.001 4 0.24 1.54/4! e-1.5 = 0.047 0.00007 Total = 0.301

P (S) = P ( no wet station in 15 mile) Alternative approach Mean rate of service station = 0.1 per mile Probability of gas at a station = 0.8  Mean rate of “wet” station = 0.10.8 = 0.08 per mile Occurrence of “wet” station is also Poisson P (S) = P ( no wet station in 15 mile)

P 3.48 Consider reliability of a tower over next 20 years Time Quake magnitude 5 6 7 1921 1931 1941 1951 1961 1971 50-year Record of Earthquake

The tower can withstand an earthquake whose magnitude is 5 or lower Damaging earthquake  magnitude > 5 n, no. of damaging earthquakes Probability of failure 0.5 1.0 1 2 3 4 5 0.2 0.8

a) P (tower subjected to less than 3 damaging earthquakes during its lifetime) = ? from record

Lift-time reliability b) P ( tower will not be destroyed by earthquakes within its useful life) = ?

c) tower also subjected to tornadoes If a tornado hits tower, the tower will be destoryed. = tower damaged by tornadoes

P (tower damaged by natural hazards) s.i.

Time to next occurrence in Poisson process Time to next occurrence = T is a continuous r.v. = P (X = 0 in time t) Recall for an exponential distribution

T follows an exponential distribution with parameter l = n  E(T) =1/n If n = 0.1 per year, E(T) = 10 years

Example: P ( next storm occurs between 6 to 9 months from now) Storms occurs according to Poisson process with n = 4 per year =1/3 per month P ( next storm occurs between 6 to 9 months from now)

Comparison of two families of occurrence models Bernoulli Sequence Poisson Process Interval Discrete Continuous No. of occurrence Binomial Poisson Time to next occurrence Geometric Exponential Time to kth occurrence Negative binomial Gamma

Duration and productivity (x,y) Multiple Random Variables E 3.24 Duration and productivity (x,y) No. of observations Relative frequencies 6, 50 2 0.014 6, 70 5 0.036 6, 90 10 0.072 8, 50 8, 70 30 0.216 8, 90 25 0.180 10, 50 8 0.058 10, 70 10, 90 11 0.079 12, 50 12, 70 6 0.043 12, 90 Total = 139

Joint PMF PX,Y (x,y) y x PX,Y (x,y) PX,Y(6,50) = 0.014 0.079 0.4 0.014 90 0.3 70 0.2 0.014 50 0.1 x 0.0 6 8 10 12 PX,Y(6,50) = 0.014 P(X>8,Y>70) = 0.079+0.014 = 0.093

Marginal PMF PY(y), PX(x) 0.180 0.475 0.345 PY(y) PX,Y (x,y) y 0.180 0.216 0.432 0.4 90 0.317 0.3 70 0.036 0.2 50 0.122 0.129 0.1 PX(x) x 0.0 6 8 10 12 P(X=8) =0.036+0.216+0.180 = 0.432

Conditional PMF P(Y=70 | X=8) = PY|X(70| 8) PY|X (y|8) y 0.5 0.417 50 70 90 y 0.5 0.417 0.083

If X and Y are s.i. or

Joint and marginal PDF of continuous R.V.s marginal PDF fX (x) marginal PDF fY (y) x=a fX (a) = Area fX,Y (x, y=b) Surface = fX,Y (x,y) y =b Joint PDF fY (b) = Area fX,Y (x=a, y)

a) Calculate probability

b) Derive marginal distribution

c) Conditional distribution

Correlation coefficient a measure of correlation between X and Y

Significance of correlation coefficient y: elongation x: Length Steel y: strength x: Length Glass r = +1.0 r = -1.0

y: ID No x: height y: weight x: height r = 0 0< r <1.0

Estimation of r from data

Review of Chapter 3 Random variables Main descriptors discrete – PMF, CDF continuous – PDF, CDF Main descriptors central values: mean, median, mode dispersion: variance, s, c.o.v. skewness Expected value of function

Common continuous distribution Occurrence models normal, lognormal, exponential Occurrence models Bernoulli sequence – binomial, geometric, negative binomial Poisson process – Possion, exponential, gamma

Multiple random variables Discrete: joint PMF, CDF, marginal PMF, conditional PMF Continuous: joint PDF, marginal PDF, conditional PDF Correlation coefficient