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

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

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

4. a

5. Page 380 Table of Standard Normal Probability

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

8. fX(x) x  a b

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

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

11. Lognormal distribution
Parameter l 
fX(x) x

12. Parameters  
for   0.3,

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

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

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

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

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

18. 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%

19. 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.

20. 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

21. 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

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

23. Exponential distribution
fX(x)
x  0

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

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

26. Shifted exponential distribution
Lower bound not necessarily 0
x  a

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

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

29. 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?

30. 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

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

32. 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

33. 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)

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

35. 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) =
= 0.919

36. P (flooding during 5 years)
= P (X  1)
= 1 – P( X = 0) =
= 0.41

37. 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

38. P (first flood in 3rd year) = ?

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

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

41. 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

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

43. 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

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

45. 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)
= 0.005 = 0.003

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

47. 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

48. 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

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

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

51. 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 =

52.  T = year
VD = mph

53. 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 =

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

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

56. 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) min divided into 30 second interval
 No. of interval = 12

57. (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

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

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

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

61. 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.

62. 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)

63. 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

64. 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%

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

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

67. 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)

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

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

70. (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

71. Total = 0.301 x P( S| X = x ) P( X = x ) P( S| X = x ) P( X = x ) 11
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
Total = 0.301

72. 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)

73. 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

74. 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

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

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

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

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

79. 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

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

81. 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)

82. 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

83. 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

84. 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) = = 0.093

85. 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) = = 0.432

86. Conditional PMF P(Y=70 | X=8) = PY|X(70| 8) PY|X (y|8) y 0.5 0.41750 70 90 y
0.5
0.417
0.083

87. If X and Y are s.i. or

88. 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)

89. a) Calculate probability

90. b) Derive marginal distribution

91. c) Conditional distribution

92. Correlation coefficient
a measure of correlation between X and Y

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

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

95. Estimation of r from data

96. 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

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

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