Exercise 3-2-1
Solution:
(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 only the
second column, where probabilities sum to (0.15 + 0.25 + 0.10) = 0.5, not one,
so all those probabilities should now be divided by 0.5. Hence
P(Y  20 | X = 2) = 0.25 / 0.5 + 0.1 / 0.5
= 0.35 / 0.5
= 0.7
(c) If X and Y are s.i., then (say) P(Y  20) should be the same as P(Y  20 | X
= 2) = 0.7. However,
P(Y  20) = 0.10 + 0.25 + 0.25 + 0.0 + 0.10 + 0.10
= 0.80  0.7,
hence X and Y are not s.i.
(d) Summing over each row, we obtain the (unconditional) probabilities P(Y = 10)
= 0.20, P(Y = 20) = 0.60, P(Y = 30) = 0.20, hence the marginal PMF of runoff Y
is as follows:
fY(y)
0.6
0.2
0.2
1
2
30
y
(e) Given that X = 2, we use0 the0 probabilities in the X = 2 column, each
multiplied by 2 so that their sum is unity. Hence we have P(Y = 10 | X = 2) =
0.152 = 0.30, P(Y = 20 | X = 2) = 0.252 = 0.50, P(Y = 30 | X = 2) = 0.102 =
0.20, and hence the PMF plot:
fY|X(y
|2)
0.5
0.3
0.2
1
0
2
0
30
y
(f) By summing over each column, we obtain the marginal PMF of X as P(X = 1) =
0.15, P(X = 2) = 0.5, P(X = 3) = 0.35. With these, and results from part (d),
we calculate
E(X) = 0.151 + 0.52 + 0.353 = 2.2,
Var(X) = 0.15(1 – 2.2)2 + 0.5(2 – 2.2)2 + 0.35(3 – 2.2)2 = 0.46,
similarly
E(Y) = 0.210 + 0.620 + 0.230 = 20,
Var(Y) = 0.2(10 – 20)2 + 0.6(20 – 20)2 + 0.2(30 – 20)2 = 40,
Also,
E(XY) =

xyf(x,y)
all
= 1100.05 + 2100.15 + 1200.10 + 2200.25 + 3200.25 + 2300.10 +
3300.10
= 45.5
Hence the correlation coefficient is
E ( XY )  E ( X ) E (Y )
 =
Var( X ) Var(Y )
=
45.5  2.2  20
0.46  40
 0.35
Exercise 3-3-1
Solution:
(a) The mean and median of X are 13.3 lb/ft2 and 11.9 lb/ft2, respectively (as done in Problem 3-3-3).
(b)
The event “roof failure in a given year” means that the annual maximum
snow load exceeds the design value, i.e. X > 30, whose probability is
P(X > 30) = 1 – P(X  30) = 1 – FX(30)
= 1 – [1 – (10/30)4]
= (1/3)4 = 1/81  0.0123  p
Now for the first failure to occur in the 5th year, there must be four years of
non-failure followed by one failure, and the probability of such an event is
(1 – p)4p = [1 - (3/4)4 ]4(1/3)4  0.0117
(b) Among the next 10 years, let Y count the number of years in which failure
occurs. Y follows a binomial distribution with n = 10 and p = 1/81, hence the
desired probability is
P(Y < 2) = P(Y = 0) + P(Y = 1)
= (1 – p)n + n(1 – p)n – 1p
= (80/81)10 + 10(80/81)9(1/81)
 0.994
Exercise 3-3-3
Solution:
(a) Differentiating the CDF gives the PDF,
0
for s  0

 s2
s
f S ( s )  
for 0  s  12
 288 24
0
for s  12

~
The mode s is where fS has a maximum, hence
fS’( ~s ) = 0
 - ~s /144 + 1/24 = 0
 the mode ~s = 6.
The mean of S,

E(S) =

12

sf ( s )ds  (
s  
 (12 )
12

= = - 18 + 24 = 6
4  288 3  24
4
=
(b)
0
 s3 s2
 )ds
288 24
3
Dividing the sample space into two regions R = 10 and R = 13, the total probability of failure is
P(S > R) = P(S > R | R = 10)P(R = 10) + P(S > R)P(R = 13)
= [1 – FS(10)]0.7 + [1 – FS(13)]0.3
= [1 – (-103 / 864 + 102 / 48)]0.7 + [1 – 1]0.3
= 0.0740740740.7  0.0519
Exercise 3-3-5
Solution:
(a)
The only region where fX(x) is non-zero is between x = 0 and x = 20, where
fX(x) = F’X(x) = - 0.005x + 0.1
which is a straight line segment decreasing from y = 0.1 (at x = 0) to y = 0
(at x = 20)
fX(
x)
x
0
xm
2
0
(b) The median divides the triangular area under fX into two equal parts, hence, comparing the two similar triangles
which have area ratio 2:1, one must have
(20 – xm) / 20 = (1 / 2)0.5
 xm = 20(1 – 0.50.5)  5.858
Exercise 3-3-7
Solution:
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)
3
or 1 (for x > 6)
FX is a parabola going from x = 3 to x = 6, followed by a horizontal line when
plotted
6
(b)
E(X) =
24
 x( x
3
) dx = 24(1/3 – 1/6) = 4 (tons)
3
6
(c)
First calculated the variance, Var(X) =
E(X2)
–
[E(X)]2
=
x
2
3
(
24
)dx - 16 =
x3
24(ln 6 – ln 3) = 24 ln 2 - 16

24 ln 2  16

 19.9%

4
When the load X exceeds 5.5 tons, the roof will collapse, hence
P(roof collapse) = P(X > 5.5)
= 1 – P(X  5.5)
= 1 – FX(5.5) = 1 – (4/3 – 12/5.52)  0.063
 C.O.V. =
(d)
Exercise 3-4-1
Solution:
(a) Let F, E, T denote “Floods”, “Earthquakes” and “Tornadoes”, respectively,
and let N denote “Natural hazards”. N also has a Poisson distribution, with the
combined mean rate of occurrence
N = F + E + T
= (1/10 + 1/20 + 1/5) = 7/20 = 0.35 (mean occurrence per
year).
hence for t = 1 year,  = [(7/20) per year](1 year) = 7/20, and P(N = n) = e
n / n!. Thus
P(N = 2) = e0.35 0.352 / 2!  0.043
P(N = 0 in any given year) = e = e–0.35  p, and, adopting an (n = 3, p = e–
0.35) binomial model, where “success” is “no natural hazard in the year”,
P(two out of three years with no natural hazard)
= 3p2(1 – p) = 3 e–0.352(1 – e–0.35)  0.440
(b)
(c) For earthquakes, E = (1/20) per year, therefore, the time T (in years)
between two successive earthquakes is exponentially distributed with a mean
time of <T> = [1/(1/20)] years = 20 years, thus the CDF of T is
FT(t) = 1 – et/20
 P(T  10) = 1 – P(T < 10)
= 1 – FT(10)
= e–10/20 = e–0.5  0.607
Exercise 3-5-1
Solution:
(a) Let X be her cylinder’s strength in kips. To be the second place winner, X must be above 70 but below 100, hence
P(70 < X < 100) = P( 70  80  X   X  100  80 )
20
X
20
= P(- 0.5 < Z < 1)
= 0.191462467 + 0.34134474  0.533
(b) P(X > 100 | X > 90) = P(X > 100 and X > 90) / P(X > 90)
= P(X > 100) / P(X > 90)
= {1 – [(100 – 80)/20]} / {1 – [(90 – 80)/20]}
= [1 – (1)] / [1 – (0.5)]
= 0.15865526 / 0.308537533  0.514
(c) Let Y be the boyfriend’s cylinder strength in kips, which has a mean of 1.0180 = 80.8. Therefore,
Y ~ N(80.8, Y)
Let D = Y – X; D is normally distributed with a mean of
D = Y – X = 80.8 – 80 = 0.8 > 0
This suffices to conclude that the guy’s cylinder is more likely to score higher. Mathematically,
P(D > 0) = P( D   D  0  0.8 )
D
D
= P(Z > a negative number) > 0.5,
hence it is more likely for the guy’s cylinder strength to be higher than the girl’s.
Exercise 3-5-3
Solution:
Let F be the daily flow rate; we're given F ~ N(10,2)
(a) P(excessive flow rate) = P(F > 14)
= P(
F  F
F

14  10
)
2
= P(Z > 2)
= 1 - P(Z  2)
= 1 - 0.97725  0.02275
(b) Let X be the total number of days with excessive flow rate during a threeday period. X follows a binomial distribution with n = 3 and p = 0.02275
(probability of excessive flow on any given day), hence
P(no violation)
= P(zero violations for 3 days)
= P(X = 0)
= (1 - p)3  0.933
(c) Now, with n changed to 5, while p = 0.02275 remains the same,
P(not charged) = P(X = 0 or X = 1)
= P(X = 0) + P(X = 1)
= (1 - p)5 + 5p(1 - p)4  0.995
which is larger than the answer in (b). Since the non-violation probability is
larger, this is a better option.
(d) In this case we work backwards—fix
determine the required parameter values.
We want
P(violation) = 0.01
 P(non-violation) = 0.99
 (1 - p)3 = 0.99
 p = 0.003344507,
the
probability
of
violation,
and
but recall from part (a) that p is obtained by computing P(F > 14)
F   F 14   F
 0.003344507 = P(
)

F
F
14   F
)
2
14   F
 1 – 0.003344507 = P( Z 
)
2
14   F
 0.99666 = P( Z 
)
2
 0.003344507 = P( Z 
Yet we know that (either using a table or typing =NORMSINV(0.99666) in
Excel)
0.99666 = P(Z  2.711967682),
hence, by comparing the two equations above,
14  F
= 2.711967682   F  8.58
2
Exercise 3-6-1
Solution:
(a) Let X be the pile capacity in tons. X is log-normal with parameters
X  X = 0.2,
X  ln 100 – 0.22 / 2
P(X > 100) = 1 – P(X  100)
ln X   X ln 100  (ln 100  0.2 2 / 2)
= 1 – P(

)
0.2
X
= 1 – P(Z  0.1) = 1 – 0.5398  0.460
(b) Let L be the maximum load applied; L is log-normal with parameters
L = 15/50 = 0.3
 L = [ln(1 + 0.32)]1/2 = 0.293560379
 L = ln 50 – 0.2935603792 / 2 = 3.868934157
It is convenient to formulate P(failure) as
P(X < L) = P(X / L < 1)
= P(ln(X/L) < ln 1)
= P(ln X – ln L < 0)
but D = ln X – ln L is the difference of two normals, so it is again normal, with
D = X – L = 0.716236029 and D = (X 2 + L2)1/2 = 0.355215
 P(D < 0) = P[Z < (0 – 2.016345112 0.716236029) / 0.355215]
= (–2.016345112)
 0.0219
(c) P(X > 100 | X > 75) = P(X > 100 and X > 75) / P(X > 75)
= P(X > 100) / P(X > 75)
ln 75  (ln 100  0.2 2 / 2)
= [answer to (a)] / [1 – (
)]
0 .2
= 0.460172104 / [1 – (–1.338410362)]
= 0.460172104 / (1.338410362)
= 0.460172104 / 0.909618587  0.506
(d) P(X > 100 | X > 90) = P(X > 100 and X > 90) / P(X > 90)
= P(X > 100) / P(X > 90)
ln 90  (ln 100  0.2 2 / 2)
= [answer to (a)] / [1 – (
)]
0 .2
= 0.460172104 / [1 – (– 0.426802578)]
= 0.460172104 / (0.426802578)
= 0.460172104 / 0.665238403  0.692
Exercise 3-6-3
Solution:
(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
 P(satisfactory performance at node A)
= P(6 < A < 14)
= P(ln 6 < ln A < ln 14)
= P(
ln 6   A
A

ln A   A
A

ln 14   A
A
)
Substituting the numerical values for A and  A , and since (ln A) ~ N(A,  A ),
the above becomes
P(–2.45412812 < Z < 1.782361183)
 0.962462 – (1 – 0.992857)
 0.955
(b) Let Ni denote the event of pressure at node i being within normal range; i =
A,B. Given:
P(NB) = 0.9  P( N B ) = 0.1;
P( N B | N A ) = 20.1 = 0.2
Hence
P(unsatisfactory water services to the city)
= P( N A  N B )
=
=
=

P( N B | N A )P( N A )
0.2(1 – answer to (a))
0.2(1 – 0.955319)
0.0089
(c) The options are:
(I) to change the c.o.v. of A to 0.15: repeating similar calculations as done
in (a), we get the new values of:
A
A
2.29133509
3
0.15
lower
limit
for Z
–3.3305
upper
limit
for Z
2.31815
P(normal pressure
at A)
0.9893459
hence the probability of unsatisfactory water services,
P( N B | N A )P( N A ) becomes
0.2(1 – 0.9893459)  0.0021
(II) to change P( N B ) to 0.05, and hence P( N B | N A ) = 20.05 = 0.10, thus
P( N B | N A )P( N A ) = 0.10(1 – 0.9555935)  0.0044
 Option I is better since it offers a lower probability of unsatisfactory water
services than II.
Exercise 3-7-1
Consider two random variables X and Y having the following joint probability density function
f(x, y) = (6/5) (x + y2) , 0 < x < 1; 0 < y < 1
Perform the following tasks:
(a) Determine the marginal density function for X, fX(x). (ans. (2/5)(3x + 1))
(b) Compute P(Y > 0.5 | X = 0.5). (ans. 0.65)
(c) Based on the marginal density functions of X and Y, we can derive the following information
E(X) = 3/5 ;
E(Y) = 3/5
E(X2) = 13/30 ;
E(Y2) = 11/25
Determine the correlation coefficient between X and Y. (ans. – 0.131)
Solution:
(a) fX(x) is obtained by “integrating out” the independence on y,
1
6
y3 
6
( x  y 2 )dy =  xy 
fX(x) =

5
5
3 0
0
2
= (3x + 1)
(0 < x < 1)
5
1

(b) fY|X(y|x) =
f X ,Y ( x, y )
f X ( x)
=
x  y2
(6 / 5)( x  y 2 )
=3
3x  1
(2 / 5)(3 x  1)
1
Hence P(Y > 0.5 | X = 0.5) =
f
Y | 0 .5 ( y |
x  0.5)dy
0 .5
1

0 .5  y 2
y3 
=3
dy = (3/2.5) 0.5 y 

1 .5  1
3  0 .5

0 .5
= 0.65
1

1 1
(c) E(XY) =

xyf X ,Y ( x, y )dxdy =
0 0
1
=
6
5
1 1
  (x
2
y  xy 3 )dxdy
0 0
1
2
3 3
ydy 
y dy = 1/5 + 3/20 = 7/20 = 0.35
50
50


 Cov(X,Y) = E(XY) – E(X)E(Y) = 0.35 – (3/5)(3/5) = -0.01,
while
X = {E(X2) – [E(X)]2}1/2 = [(13/30) – (3/5)2]1/2 = 0.27080128
Y = {E(Y2) – [E(Y)]2}1/2 = [(11/25) – (3/5)2]1/2 = 0.282842712,
Hence the correlation coefficient,
XY =
Cov( X , Y )
 XY
=
 0.01
 - 0.131
0.27080128  0.28284271 2
Exercise 4-1-1
Solution:
To have a better physical feel in terms of probability (rather than probability
density), let’s 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
 fY(y) = [FY(y)]’ = 0
But when y  0,
FY(y) = P(Y  y)
1
= P( mX 2  y )
2
= P( 
2y
2y
)
X
m
m
 2y 

2y 
  FX  

 m 


m




= FX 
 2y 

 m 0


= FX 
 2y 

 m 


= FX 
Hence the PDF,
fY(y) =
d
[FY(y)]
dy
 2y  d 2y

 dy m
m


= f X 
8y
=
ma
=
3

exp( 
2y
)
ma 2
1
2my
4 2y
2y
exp( 
)
3
3
a m
ma 2
Hence the answer is
4

f Y ( y)   a 3

0
2y
2y
exp( 
)
3
m
ma 2
y0
y0