CIVL 281
Solutions to Ang & Tang Ch. 4 problems
4.11
Let R and P denote reservoir and pumped water supply, respectively. Given:
R ~ N(30,3),
P ~ N(15,4)
Then the total supply is S = R + P, where
S ~ N(30 + 15, 32  4 2 ), i.e.
S ~ N(45, 5)
(a) Let D be the water demand with the PMF as shown in Fig. P4.11. By the theorem
of total probability,
P(insufficient water supply)
= P(S < D)
= P(S < 20)P(D = 20) + P(S < 30)P(D = 30) + P(S < 40)P(D = 40)
20  45
30  45
40  45
)  0.3 + (
)  0.6 + (
)  0.1
= (
5
5
5
 0.0167
(b) Suppose now D is also normal, with
D = 0.3*20 + 0.6*30 + 0.1*40 = 28, and
D =
0.3  (20  28) 2  0.6  (30  28) 2  01
.  (40  28) 2 = 6, hence
if we let F denote the difference S - D, then F ~ N(45 - 28, 52  6 2 ), i.e.
F ~ N(17, 61 ).
Then
P(insufficient water supply)
= P(S < D)
= P(F < 0)
0  17
)
= (
61
 0.0148
4.15
(a) For people who go to the airport (A) via route II-A, the probability of taking more
than 1 hour is
60  50
) = 0.1586553, whereas
1 - (
10
for people going along route I-A, that probability is only
60  40
) = 0.0227501
1 - (
10
Hence the total probability of taking more than 1 hour is
0.30.1586553 + 0.70.0227501
 0.0635
(b) Let T be the total travel time (in minutes), where
T~ N(40 + 15 + 50,
10 2  52  10 2 ), i.e.
T ~ N(105, 15). Hence
120  105
) = (1)  0.841
P(T < 120) = (
15
Now let U be the total travel time for two round trips, we have
U ~ N(105 + 105, 152  152 ), i.e.
U ~ N(210, 15 2 ).
Hence the required probability is
240  210
P(U < 240) = (
)  0.921
15 2
which is different from P(T < 120). This is expected since not everything inside  got
scaled by a factor of 2 (namely, the standard deviation only went up by a factor of
2)
(c) Let R = minutes spent after 7:50am for the friend to arrive at the airport; R ~
N(50,10). Given fact: R > 60, and we want
P(R < 70 | R > 60)
P( R  70 and R  60)
=
P( R  60)
70  50
60  50
(
)  (
)
10
10
=
60  50
1  (
)
10
 0.857
4.19
(a) No, since A, B, D all depend on R, they depend on each other, and are therefore
not s.i.
(b)
T=A+B+D
= 0.38R + 0.7, hence
E(T) = 0.38E(R) + 0.7 = 0.3815 + 0.7 = 6.4;
Var(T) = 0.382 Var(R) = 0.38222  0.578
T is normally distributed.
(c)
S = 30 + T - I - M - E, hence
E(S) = 30 +E(T) - E(I) - E(M) - E(E)
= 30 + 6.4 - 1.5 - 1 - 2.5
= 31.4;
Var(S) = Vat(T) + Var(I) + Var(M) + Var(E)
= 0.5776 + 0.09 + 0.01 + 0.16
 0.8376
S is normally distributed.
(d)
P(S > 30)
= 1 - P(S < 30)
= 1 - (
30  314
.
)
0.8376
 0.937
4.23
(a) Given parameters:
Median xm = 4, x = 0.2
 x = ln(xm) = ln 4; x  x = 0.2; also
Median fm = 0.15, F = 0.1
 F = ln(fm) = ln 0.15; F  F = 0.1
Note that F is the fraction of pollutant remaining untreated, i.e.
Y
F  , hence
X
Y = FX
being the product of log-normal random variables, it must also be log-normal, with
parameters
Y = F + X = ln 0.15 + ln 4
 - 0.51
Y  F2  X2  01
. 2  0.2 2
 0.22
(b)
P(Y > 1) = 1 - P(Y < 1)
ln 1  Y
)
= 1 - (
Y
ln 1  0.51
)
= 1 - (
0.22
= 1 - (2.28448)
 0.0112
(c) On a B(ad day), the median of X becomes 5 instead of 4, while other parameters
remain the same, hence the parameters of Y change to
Y = F + X = ln 0.15 + ln 5;
Y  0.22 (as before)
Hence the probability of exceedence, given a bad day, is
P(Y > 1 | B)
ln 1  (ln 5  ln 015
. )
)
= 1 - (
0.22
= 1 - (1.287)
 0.099
Hence on any given day (Good or Bad),
P(exceedence)
= P(Y > 1 | G)P(G) + P(Y > 1 | B)P(B)
= 0.01120.9 + 0.0990.1
 0.020
4.26
Given:
Arrival of planes: Poisson, with mean rate A = 20 (planes/hr);
Departure of planes: Poisson, with mean rate D = 30 (planes/hr)
(a) Let X = total number of arrivals and/or departures, whose mean occurrence rate is
X = A + D = 50 planes/hr,
hence, over a 6-minute period, i.e.
t = (6/60) hr = 0.1 hr,
  = t = 500.1 = 5, hence
e   2
P(X = 2 | t = 0.1) =
 0.084
2!
(b)
(i) Let Ni be the number of passengers on the i-th plane that arrived, and T be the total
number of them in the last hour. Clearly,
T = N1 + N2 + … + N25, hence
E(T) = E(N1) + E(N2) + … + E(N25)
= 100 + 100 + … +100 = 25100 = 2500, while
Var(T) = Var(N1) + Var(N2) + … + Var(N25)
(s.i. assumed)
= 25(0.4100)2 = 40000
(ii) Assuming s.i. between Ni , and that their distributions are well-behaved functions
that allow application of the central limit theorem, we conclude that T is
approximately normal (with T = 2500 and T = 200), hence
P(T > 3000) = 1 - P(T < 3000)
3000  2500
)
= 1 - (
200
 1 - 0.99379032
 0.0062
4.32
(a) Given:
IA = 1000  IA = IA IA = 0.11000 = 100
IB = 2000  IB = IB IB = 0.12000 = 200
The noise intensity at C,
IA
IB
IC =
= 0.01 IA + 0.0025 IB,
2 
(9  1)
(19  1) 2
hence
E(IC) = 0.01 E(IA) + 0.0025 E(IB)
= 0.011000 + 0.00252000 = 15 units, while
Var(IC) = 0.012 Var(IA) + 0.00252 Var(IB) = 1.25 (sq. units)
(b) In number of decibels, the intensity at C is
DC = 40 ln(2IC),
where IC is a RV with mean = 15 and variance = 1.25. We can approximate the mean
and variance of DC by recalling the formulas (for Y = g(X))
E(Y)  g(X),


2
Var(Y)  g ' (  X ) Var ( X )
therefore, since
E(IC) = 15, and
d
d
40
( DC ) 
(40 ln(2 I C )) 
,
dI C
dI C
IC
we have
E(DC)  40 ln(215)  136.0, and
2
 40 
Var(DC)     125
.  8.89
 15 