3.55 (a) T = trouble-free operational time follows a gamma distribution with mean of 35 days and c.o.v. of 0.25 k/ = 35,
1 k
= 0.25
Hence, k = 16, = 0.457 P(T>40) = 1-Iu(0.457x40, 16) = Iu(18.3, 16) =
3.12
x
(a) For 3 x 6, FX(x) =
24
t
3
dt = 4 / 3 12 / x 2 ; elsewhere FX is either 0 (for x < 3) or 1 (for x >
3
6)
FX is a parabola going from x = 3 to x = 6, followed by a horizontal line when plotte
3.13
(a)
Since the return period, , is 200 years, this means the yearly probability of exceedance (i.e.
encountering waves exceeding the design value) is
p = 1 / = 1 / (200 years) = 0.005 (probability
2.48
P(E) = 0.001, P(L) = 0.002, P(T) = 0.0015
P(LE) = 0.1
T is statistically independent of E or L
(a)
P(ELT) = P(LE)P(E)P(T)
= 0.1x0.001x0.0015
= 1.5x10-6
(b)
P(ELT) = 1-P( E L T )
= 1-P( T )P( E L
2.49
(a) Let E1, E2 denote the respective events of using 100 and 200 units, and S denote shortage of
material.
P(S) = P(S | E1)P(E1) + P(S | E2)P(E2)
= 0.10.6 + 0.30.4
= 0.06 + 0.12 = 0.18
(b) Using
3.16
(a) Let C be the number of microscopic cracks along a 20-feet beam. C has a Poisson
distribution with mean rate C = 1/40 (number per foot), and length of observation t =
20 feet, hence the parame
2.50
Let H and S denote Hard and Soft ground, respectively, and let L denote a successful landing.
Given probabilities:
P(L | H) = 0.9; P(L | S) = 0.5;
P(H) = 3P(S) , but since ground is either hard o
3.17 (a) Let N be the number of poor air quality periods during the next 4.5 months; N follows a Poisson process with mean value (1/month)(4.5 months) = 4.5, hence -4.5 2 P(N 2) = e (1 + 4.5 + 4.5 /2!
3.18 (a) Let E and T denote the number of earthquakes and tornadoes in one year, respectively. They are both Poisson random variables with respective means 1 E = Et = 1 year = 0.1; T = 0.3 10 years Al
2.52
Let A and B be water supply from source A and B are below normal respectively
P(A) = 0.3, P(B) = 0.15
P(BA) = 0.3
P(SA B ) = 0.2, P(S A B) = 0.25, P(S A B ) = 0,
P(SAB) = 0.8
Where S denotes even
3.19
Mean rate of accident = 1/3 per year
(a)
P(N=0 in 5 years) = e-1/3x5 = 0.189
(b)
Mean rate of fatal accident = 1/3x0.05 = 0.0167 per year
P(failure between inspection)
= 1-P(no fatal accident wit
2.53
Let W be weather favorable
F be field work completed on schedule
C be computation completed
T1 be computer 1 available
T2 be computer 2 available
Given: P(FW) = 0.9,
P(F W ) = 0.5
P( W ) = 0.6
P(
3.20
(a)
Let X be the number of accidents along the 20 miles on a given blizzard day. X has a
1
20 miles = 0.4, hence
Poisson distribution with X =
50 miles
0.4
P(X 1) = 1 P(X = 0) = 1 e
= 1 0 6703200
2.46
A, B, C denote branch A, B, C will be profitable respectively
E denotes bonus received
P(A) = 0.7, P(B) = 0.7, P(C) = 0.6
P(AB) = 0.9 = P(BA)
C is statistically independent of A or B
(a)
P(Exactl
3.56
Seismic capacity C is lognormal with median of 6.5 and standard deviation of 1.5
= ln 6.5
For lognormal distribution, > x m
/ < / x m = 1.5 / 6.5 = 0.23
and
Since =ln 6.5 = ln (1.5 / ) 2
6.7;
2.39 (a) P(shipment accepted on a given day) = P(at most 1 defective panel) = 0.2 + 0.5 = 0.7 P(exactly one shipment will be rejected in 5 days) = 5xP(0.7)4(0.3) = 0.36 P(acceptance of shipment on a g
3.57 (a) Let A and B denote the pressure at nodes A and B, respectively. Since A is log-normal with mean = 10 and c.o.v. = 0.2 (small), we have A 0.2, and A = ln(A)
A2
2
ln(10) 0.02 = 2.282585093
2.40
The possible scenario of the working condition and respective probabilities are as follows:
(a)
P(completion on schedule) = P(E)
= P(EG) P(G) + P(EN) P(N) + P(EBC) P(BC) + P(EB C ) P(B C )
= 1x0.
3.58
(a) fX(x) is obtained by integrating out the independence on y,
1
y3
6
6
fX(x) =
( x + y 2 )dy = xy +
5
5
3 0
0
2
(0 < x < 1)
= (3x + 1)
5
1
(b) fY|X(y|x) =
f X ,Y ( x, y )
=
f X ( x)
(6 / 5)( x
2.41
Let L, N, H denote the event of low, normal and high demand respectively; also O and G
denote oil and gas supply is low respectively.
(a)
Given normal energy demand, probability of energy crisis
3.59 (a) Summing over the last row, 2nd & 3rd columns of the given joint PMF table, we obtain P(X 2 and Y > 20) = 0.1 + 0.1 = 0.2 (b) Given that X = 2, the new, reduced sample space corresponds to onl
2.42
Given:
P(H=1) = 0.2, P(H=2) = 0.05, P(H=0) = 0.75
Where H = number of hurricanes in a year
P(J=H=1) = P(DH=1) = 0.99
P(J=H=2) = P(DH=2) = 0.80
Where J and D denote survival of jacket and deck sub
3.1 Total time T = A + B, which ranges from (3 + 4 = 7) to (5 + 6 = 11). Divide the sample space into A = 3, A = 4, and A = 5 (m.e. & c.e. events) P(T = 7) = =
P(T = 7 | A = n)P(A = n) P(B = 7 - n)P(A
2.43
(a) Sample space = cfw_FA, HA, EA, FB, HB, EB
in which FA, FB denote fully loaded truck in lane A, B respectively
HA, HB denote half loaded truck in lane A, B respectively
EA, EB denote empty tru
3.2
Let X be the profit (in $1000) from the construction job.
(a) P(lose money) = P(X < 0)
= Area under the PDF where x is negative
= 0.0210 = 0.2
(b) Given event is X > 0 (i.e. money was made), hence
University of California at Berkeley
Department of Civil Engineering
Instructor: P.J.M. Monteiro
HOMEWORK
Question 4
Explain the creep mechanism in cement pastes.
Creep mechanisms in cement pastes cou