Exercises
Exercise 4-2-1
Solution:
P(T<30)=P(T<30|N=0)P(N=0)+P(T<30|N=1)P(N=1)+P(T<30|N=2)P(N=2)
30  30
30  35
30  40
 (
)  0.2  (
)  0.5  (
)  0.3
2
5
25  3
25  2  32
 0.5  0.2  (0.857)  0.5  (1.525)  0.3
=0.217
Exercise 4-2-3
Solution:
(a) X 5  X 1  X 2  X 3
 5  1   2   3  10  15  20  45
 52   12   22  2 1,2 1 2  32  32  2  0.6  3  3  28.8
 5  5.37
(b)
V1  60 X 1
V2  60 X 2
 P(V2  V1  400)  P(Z  400)
  Z  60 2  601  60 15  60 10  300
 Z2  60 2  22  60 2  12  2 1, 2  60 2  1 2  25920
 Z  161
 P(V2  V1  400)  1   (
400  300
)  1   (.621)  .267
161
(c) X 5  X 1  X 2  X 3
 5  1   2   3  10  15  20  3n  45  3n
 52   12   22  2 1,2 1 2  32  32  2  0.6  3  3  28.8
 5  5.37
 P( X 5  70)  .05
 P( X 5  70)  .95
 70  45  3n 

  .95
5.37


 25  3n  5.37 1 (.95)  5.37 *1.645  8.839
 n  5.4
Exercise 4-2-5
Solution:
(a)
E (Y | X  0.4)  1.12  0.4  0.05  0.498
 (Y | X )  0.0025  0.05
0.3  0.498
)  1  (3.96)  0.999963
0.05
E (Y | X  0.35)  1.12  0.35  0.05  0.442
0.3  0.442
 P(Y  0.3 | X  0.35)  1  (
)  0.9977
0.05
P(Y>0.3)=P(Y>0.3|X=0.35)P(X=0.35)+P(Y>0.3|X=0.4)P(X=0.4)
=0.9977*0.2+0.999963*0.8
=0.9995
YA=N(0.442,0.05)
YB=N(0.498,0.05)
P(YA>YB)=P(YB-YA<0)=P(Z<0)
 Z  0.498  0.442  0.056
 P(Y  0.3 | X  0.4)  1  (
(b)
(c)
 Z  0.05 2  0.05 2  0.0707
P(Z<0)=  (
0  0.056
)   (0.7921)  1  0.786  0.214
0.0707
Exercise 4-2-7
Solution:
(a)
(b)
(c)
P(X2)=1-P(X=0)-P(X=1)
5
 5
 1   0.6 0 0.4 5   0.610.4 4
0
1
 1  0.4 5  5  0.6  0.4 4
 0.046
P(H=1B=0)=P(H=1)P(B=0)
 3
 2
=  0.610.4 2  0.6 0 0.4 2
1
 0
=0.046
T=H1+H2+B
T in Normal distribution with
T  100  100  80  280
 T  40 2  40 2  20 2  60
300  280
)  1  (0.333)  0.2695
60
T  100  100  80  280
 P(T  300)  1  (
(d)
 T  40 2  40 2  20 2  2  0.8  40  40  78.5
 P(T  300)  1  (
300  280
)  1  (0.255)  0.4
78.5
Exercise 4-2-9
Solution:
(a)
C=F+B
 C   F   B  20  30  50
 C  (0.2  20) 2  (0.3  30) 2
(b)
 9.85
T=C1+C2=2  C =0.197
 T  9.85 2  9.85 2  2  0.8  9.85 2  18.69
  T  18.69
(c)
 0.187
100
P(failure)=P(T<L)=P(T-L<0)
0  Z
 (
)
Z
 (
 (100  50)
18.69 2  (.3  50) 2
 50
)
23.95
 0.0184
 (
Exercise 4-3-1
Solution:
Let Xi be the net amount won (in dollars) on the i-th play. Then X1, X2, … are independent, identically
distributed (i.i.d.) RV's with such a distribution:
f(–1) = 20/ 38, f(+1) = 18/38
Hence
E(Xi)   = (–1)(20/38) + (1)(18/38) = –1/19;
2
Var(Xi)   = (–1 + 1/19)2(20/38) + (+1 + 1/19)2(18/38) = 360/361   = 360 / 19
It is important to realize that, after n plays, the net winnings is
Sn = X1 + X2 + … + Xn
which is the sum of a large number of i.i.d. RV's whose individual  and  we have found, which permits using
the central limit theorem: Sn must be approximately normal with mean n and standard deviation n , hence
P(being ahead or just breaking even) = P(Sn  0)
= P(
S n  n
n
 P(Z 

0  n
)
n
 (  n )  19
)
19 360
= P(Z  n / 360 ) = 1 – ( n / 360 )
which is tabulated below for increasing values of n:
Number of plays, n
365
1000
5000
10000
( n / 360 )
0.843013501
0.95220967
0.999903
0.999999932
Probability of not losing money
0.156986499
0.04779033
9.69995E-05
6.81784E-08
Since  approaches 1 as its argument goes to infinity, the probability of winning diminishes to zero as n  .
Moral of the story: don't gamble, but if you must, stop as early as you can.
Exercise 4-3-2
Solution:
(a) Let C denote car weight and T denote truck weight. The total vehicle weight (in kips) on the bridge is
V = C1 + C2 + … +C100 + T1 + T2 + … + T30
Since V is a linear combination of independent random variables, it mean and variance can be found as
E(V) = 100E(C) + 30E(T) = 1005 + 3020 = 1100 (kips);
Var(V) = 100Var(C) + 30Var(T) = 10022 + 3052 = 1150 (kip2),
Hence the c.o.v.
V = 11500.5 / 1100  0.031
(b) Assuming that the distribution of V approaches normal due to CLT,
V  V 1200  1100
P(V > 1200) = P(
)

V
1150
= 1 – (2.948839)  0.0016
(c)
(i)
Let D denote the total dead load in kips, where D ~ N(1200, 120)
The difference S = V – D is again normal,
S ~ N(1100 – 1200, (1150 + 1202)1/2) = N( –100, 155500.5), hence
P(V > D) = P(V – D > 0) = P(S > 0)
S   S 0  (100)
= P(
)

S
15550
= 1 – (0.80192694)  0.211
(ii)
Let T denote the total (dead + vehicle) weight, T = V + D. T ~ N(1100 + 1200, 155500.5), hence
P(V + D > 2500) = P(T > 2500)
T   T 2500  2300
= P(
)

T
15550
= 1 – (1.60385388)  0.054
Exercise 4-3-3
Solution:
(a) Let B be the total weight of one batch. B = 402.5 kg = 100 kg
(b) Since n > 30, we may apply the Central Limit Theorem:
B is approximately normal with B = 100 kg and B =
P(penalty) = P(B > 101) = P(Z >
40  0.1 kg, hence
101  100
)
40  0.1)
= 1 – P(Z  1.5811)
= 1 – 0.943076848
 5.69%
101  100
(c) In this case, P(penalty) = P(B > 101) = P(Z >
)
40  1)
= 1 – P(Z  0.158113883)
= 1 – 0.562816481
 43.7%
This penalty probability is much higher, caused by the large standard deviation which makes the
PDF occupy more area in the tails. Hence a large standard deviation (i.e. uncertainty) is
undesirable.
Exercise 4-3-5
Solution:
(a), (b)
First let's consider the outcome of any particular step Xi, where i = 1,2,…,N. It has two possibilities a and –a,
with respective probabilities p and q = 1 – p, hence
E(Xi) = pa + q(–a) = (p – q) a, while
Var(Xi) = p[a – (p – q)a]2 + q[–a – (p – q)a]2
= [p(2q)2 + q(–2p)2 ]a2
= 4pq(q + p) a2 = 4pq a2
Now, since X is a large sum of these identically distributed Xi's, by CLT, it must be normal with
E(X) = NE(Xi)
= N(p – q) a,
and
Var(X) = NVar(Xi)
= 4N pq a2
(c) Since the normal distribution peaks at its mean, the most likely location is x* = N(p – q)a = (2p – 1)(Na) =
(2p – 1)L, hence (i) x* = (20.5 – 1)L = 0 (i.e. at the origin); (ii) x* = (20.2 – 1)L = – 0.6L; (iii) x* =
(20.99 – 1)L = 0.98L
Exercise 4-4-1
Solution:
(a) E(Qc)=0.463E(n)-1E(D)2.67E(S)0.5=0.463*0.015-1*3.02.67*0.0050.5=41.0ft3/sec
Var(Qc)= 
=
Qc
n

 Q 
 Q 
Var (n)   c  Var ( D)   c  Var ( S )
 D  
 S  
2
2
 0.463D
 0.463n
2.67
1
S 0.5 D 2

2

2


2
Var (n)  0.463n 1  2.67 D1.67 S 0.5  Var ( D)

2
D 2.67  0.5S 0.5  Var ( S )
=22.665(ft3/sec)2
(b) The percentage of contribution of each random variable to the total uncertainty:
Qc
 2 Var(n) / Var(Qc )  74.2%
n
2
 Qc 
 D  Var ( D) / Var (Qc )  21.2%



 Qc 
 S  Var ( S ) / Var (Qc )  4.6%


2
(c) QC follows the log-normal distribution with:
 Q  41.0
C

2
QC
 22.665

22.665

 0.116

41
  ln   12  2  3.707
 
 P(QC  30)  1  (
ln 30  3.707
)  0.9958
0.116
Exercise 4-4-3
Solution:
(a)
 F  1,  F  0.3
 X  5,  x  121 (10) 2  2.89
M  F  X  5
 M2  (
M 2
M 2
) 0 Var ( F )  (
) 0 Var ( X )   2 X  F2   F2  X2  10.583
F
X
20
(b)
T  Mi
i 1
(c)
(d)
 T  20 M  100
 T2  20 M2  211.66
 T  14.55
120  100
P(T  120)  1  (
)  1  (1.375)  0.0847
211.66
P( F )  P( F | C  100) P(C  100)  P( F | C  120) P(C  120)
 P( F | C  140) P(C  140)
 P(T  100)  0.2  P(T  120)  0.3  P(T  140)  0.5
100  100 
120  100 
140  100 



 1  (
)  0.2  1   (
)  0.3  1   (
)  0.5
14.55 
14.55 
14.55 



 0.5  0.2  0.0847  0.3  0.003  0.5
 0.1269