Mohammed Jahanzeb Choudary ipment for finding the wave height and wind speed. The City College of
Figure Anemometer Figure WaveRider Bouy
City University of New York lculation to find the future esti
9.16
Assume the prior distribution of p to be a Beta distribution, then,
E ' ( p) =
q'
= 0.1  (1)
q'+ r '
and
Var ' ( p) =
a' r '
= 0.06 2 (2)
2
(q'+ r ' ) (q '+ r '+1)
From Equations 1 and 2, we ge
9.19
Using conjugate distributions, we assume Gamma distribution as prior distribution of .
From Table 9.1, for an exponential basic random variable,
E ' ( ) =
k'
= 0.5;
v'
Var ' ( ) =
k ' 0 .5
=
= (0
9.20
is the parameter to be updated. Let X denote the crack length.
The probability of crack length larger than 4 is:
P( X > 4) = 1 P ( X < 4) = e 4
The probability of crack length smaller than 6 is
9.21
(a) The occurrence rate of the tornado is estimated from the historical record as 0.1/year.
The probability of x occurrences during time t (in years) is
P( X = x ) =
(0.1t ) x 0.1t
e
x!
The proba
9.22
h is the parameter to be updated.
(a)
10
10
0
0
E ( x ) = E ( x  h ) f ( h )dh = 0.5hi0.003h 2 = 3.75
where
h
h
0
0
E ( x  h ) = xf ( x  h )dx =
x
dh = 0.5h
h
(b)
(i) the range of h is 4<h<10
9.23
(a) Let x denote the fraction of grouted length. Assumptions: the grouted length is normally
distributed; the mean grouted length is normally distributed; the variance of fraction of length
grout
9.24
(1) Prior
Noninformative prior is used as (P.M. Lee. Bayesian Statistics: An Introduction. Edward
Arnold, 1989)
2
f '( T , T )
1
2
T
(2) Likelihood
, tn  T , T )
p(t1 , t2 ,
(t1 T ) 2
(t2 T
9.25
Let x denotes the rated value, and y denotes the actual value.
2
Based on 9.24 to 9.27, = 0.512 , = 0.751 , 2 = 1.732 , s x = 34.167
From equation 9.26, E (Y  x ) = 0.512 + 0.751x
The variance c
Set1
CE26400
Quiz 2
Name: _
1. Suppose an offshore platform is designed against the 100year wave (i.e. a
wave height corresponding to a return period of 100 years) but is intended to
operate for 25
THE CITY COLLEGE OF NEW YORK
Normality Test of Concrete Strength
Distribution
Statistical Analysis of Lognormality and Normal value of
Concrete Strength Distribution
The City College of New York
140t
1. Introduction:
Normal Distribution is the most common and frequently used probability distribution
which is also known as Gaussian distribution Normal distribution is an important
statistical data d
CE264 Term Project Spring 2013

Designing a Bridge over a River
Department of Civil Engineering
The City College of New York
140th Street & Convent Avenue
New York, New York 10031
Contents
Table of C
9.18
(a) The parameter needs to be updated is p. Its suitable to set a conjugate prior for the
problem. Thus, suppose p is Beta distributed with parameter p and q. With the information
presented in th
9.17
is the parameter to be updated with the prior distribution parameter
'
= 3
'
= 0.2 3 = 0.6
(a) From the observation we get
t = N(
2 + 3 0 .5
,
) = N ( 2.5, 0.354)
2
2
From the relationship giv
9.15
Assume, First measurement L=N(2.15, )
Second measurement L=N(2.20, 2)
and Third measurement L=N(2.18, 3)
Applying Eqs. 9.14 and 9.15 and considering First and Second measurements,
L' ' =
2.20 2 +
9.1
(a) Let p = Probability that the structure survive the proof test.
P(p=0.9) = 0.70, P(p=0.5) = 0.25, P(p=0.10) =0.05
P(survival of the structure) = P(S)
= P(Sp=0.9) P(p=0.9) + P(Sp=0.5) P(p=0.5) +
9.3
P(HAHF) = P(HALF) = 0.3, P(LAHF) = P(LALF) = 0.2
(a) P( H A H F ) = P( H A H F HAHF) P(HAHF) + P( H A H F HALF) P(HALF) + P( H A H F LAHF)
P(LAHF) + P( H A H F LALF) P(LALF)
= 0.30.3 + 0.40.3 + 0.
9.7
Since,
where
f ( ) = kL( ) f ( )
L( ) =
e
12
(
12
1!
)1
, f ' ( ) =
0.271
; 0.5 20
= 0, elsewhere
So,
f ( ) = k e
12
12
0.271
;
0.5 20
= 0, elsewhere
To determine the constant k,
20
0.271 12
k
12
9.9
P( =20) = 2/3, P( =15) = 1/3
(a) Let X = Number of occurrences of fire in the next year.
P(X=20) = P(X=20 =15) P( =15) + P(X=20 =20) P( =20)
=
e 15 (15) 20 1 e 20 (20) 20 2
+
20!
3
20!
3
= 0.0139
9.12
Let X = compression index
X is N( , 0.16)
Sample mean x =
1
(0.75 + 0.89 + 0.91 + 0.81) = 0.84
4
(a) From Equation 8.13,
f ( ) = N ( x,
)
n
= N(0.84, 0.16/2) = N(0.84, 0.08)
'
'
(b) f ( ) = N( ,
9.11
M is the parameter to be updated. Its convenient to prescribe a conjugate prior to the Poisson
process. From the information given in the problem, the mean and variance of the gamma
distribution
9.10
is the parameter to be updated with the prior distribution of:
P '( = 1 5) = 1 / 3
P '( = 1 10) = 2 / 3
(a) Let =accidents were reported on days 2 and 5. The probability to observe is
P( ) = P(