CE93 Engineering Data Analysis
Joan Walker, Spring 2011
Assignment 3
Due 2/16/2011 11:10 AM
1. There is an active earthquake fault in the middle of the Indian Ocean. Within the next 50 years, there
is a 10% chance that a major earthquake will occur on thi

CE93 Engineering Data Analysis
Prof. Joan Walker, Spring 2011
Assignment #12 Final Exam Review:
th
Due May 4 , 2011 (This assignment will not be graded. Solutions will be posted on due date)
Note: This final exam review can be treated as a practice test i

Graphical
Representation
of Data
Data
There are two type f data:
Discrete (countable)
Continuous (not countable)
Example: SF rainfall data
Seasonal rainfall and rainy days in San Francisco 1960-2003
Season
Rainfall
Rainy
(in.)
Season
days
Rainfall
Rainy
(

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
=
= (0.5 0.2) 2
2
v'
v'
So, v = 50
and, k = 25
Now, v = v +
x

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 the problem, we can get
E '( p ) =
q'
q'r'
= 0.5 Var '( p

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 given in Table 9.1,
0.354 2 3 + 0.62 2.5
=
= 2.629
0.62 +

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 get,
q = 2.4 and r = 9 2.4 = 21.6
From Table 9.1, for a b

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 + 2.15 (2 ) 2
= 2.16km
2 + (2 ) 2
and,
'
L' =
2 (2 )

9.14
(a) The sample mean =
s =
2
1
(32 o 04'+31o 59'+32 o 01'+32 o 05'+31o 57'+32 o 00' ) = 32 o 01'
6
1
cfw_(3' ) 2 + (2' ) 2 + 0 + (4' ) 2 + (4' ) 2 + (1' ) 2 = 9.2
6 1
s = 3.03' = 0.05 o
The actual value of the angle is N (32 o 01' ,
0.05 o
)
6
or,
N(

9.13
(a) T is N( , 10)
Sample mean = t = 65 min.
n=5
So the posterior distribution of ,
f ' ' ( ) = N (t ,
) = N (65,
10
n
) = N (65,4.47) min.
5
(b) We know,
f ( ) = N(65,4.47) min.
L( ) = N (60,
10
) = N(60, 3.16) min.
10
From Equation 8.14 and 8.15,
'

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( , ) = N(0.8, 0.2)
L( ) = N(0.84, 0.08) = N ( x,
)
n
From

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:
P ( X < 6 ) = 1 e 6
Likelihood function is
L( ) = e

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 probability that the tornado hits the town during the next 5

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 upon the observation
(ii) likelihood:
1
h
L( h | x = 4

10.4
n=
ln(1 C )
ln R
Here,
C = 0.95 and R = 0.99
So,
n=
ln(1 0.95)
= 298
ln(0.99)
298 consecutive successful starts would be required to meet the standard.

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 can be calculated by equation 9.34 as
Var (Y | x ) =
6 1

9.24
(1) Prior
Non-informative 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 ) 2
(tn T ) 2
1
1
=
exp
i
exp
ii
exp
2
2
2
2 T 2

4.3
A = volume of air traffic
C = event of overcrowded
(a)
T = total power supply = N ( T , T )
T = 100 + 200 + 400
Where
T = 15 2 + 40 2 + 40 2 = 58.5
(b)
P(Normal weather) = P(W) = 2/3
P(Extreme weather) = P(E) = 1/3
P(Power shortage) = P(S) = P(SW)P(

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
grouted to the constant and can be approximated by the obser

4.2 To have a better physical feel in terms of probability (rather than probability density), lets work with the CDF (which we can later differentiate to get the PDF) of Y: since Y cannot be negative, we know that P(Y < 0) = 0, hence when y < 0: FY(y) = 0

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 of M is:
E '( ) =
k'
k '/ v '2
= 10 '( ) =
= 0 .4
v'
k

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( | = 1/ 5) P '( = 1/ 5) + P( | = 1/10) P '( = 1 /10)
1
2

8.6 (a)
Let Y be the number of years of experience, and M be the measurement error in inches. We have the following data: i 1 2 3 4 5 yi 3 5 10 20 25 63 mi 1.5 0.8 1 0.8 0.5 4.6 yimi 4.5 4 10 16 12.5 47 yi2 9 25 100 400 625 1159
From Eq. 8.4 & 8.3, we hav

8.5 (a) Let X be the car weight in kips; X ~ N(3.33, 1.04). Hence X - 4.5 - 3.33 ) P(X > 4.5) = P( > 1.04 = P(Z > 1.125) = 1 (1.125) = 1 - 0.8697 0.130
(b) Let Y be the gasoline mileage. For linear regression, we assume E(Y | X = x) = + x and seek the bes

8.4 (a)
(b)
Country # Per Capita GNP Per Capita Energy Consumption
Xi
Yi
Xi2
Yi2
XiYi
Yi'=a+bXi
(Yi-Yi')2
Total:
On the basis of calculations in the above table we obtain the respective sample means of X and Y as,
X = 37800/8 = 4725,
and corresponding sam

8.3 (a)
Plot of peak hour traffic vs daily traffic
1.6
1.4
1.2
Peak hour traffic (1000 vehicles)
1
0.8
0.6
0.4
0.2
0 0 1 2 3 4 5 6 7
Daily traffic (1000 vehicles)
(b)
Let X be the daily traffic volume and Y the peak hour traffic volume, both in thousand v

8.2 (a)
(b)
Stopping Distance (m) Yi Xi2 Yi2 XiYi Yi'=a+bXi (Yi-Yi')2
Vehicle No.
Speed (kph) Xi
Total:
On the basis of calculations in the above table we obtain the respective sample means of X and Y as,
X = 679/12 = 56.6 kph,
and corresponding sample va

8.9 (a) The formulas to use are
=
x y
i =1 n i
n
i
- nx y - nx
2
x
i =1
2 i
and
^ = y-x
The various quantities involved are calculated in the following table: n= 3 xi yi xiyi 66.15 168.36 640.56 875.07 xi2 1.1025 3.3489 9.8596 14.311 yi' (yi-yi')2 53.3363