3.9
(a) Differentiating the CDF gives the PDF,
0
for s 0
s2
s
+
for 0 < s 12
f S ( s ) = 288 24
0
for s > 12
s
The mode ~ is where fS has a maximum, hence setting its derivative to
s
fS( ~ ) = 0
~ /144 + 1/24 = 0
- s
the mode ~ = 6.
s
The mean of S,
E(S

3.7
(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 >

3.8
(a) Note that we are given the CDF, FX(x) = P(X x) and it is a continuous function. Hence
P(2 X 8) = FX(8) FX(2) = 0.8 0.4 = 0.4
(b) The median is defined by the particular x value where F(x) = 0.5, which occurs somewhere
along x = 2 and x = 6; precis

3.4
(a) The median of X, xm, is obtained by solving the equation that defines xm,
P(X xm) = 0.5
FX(xm) = 0.5
1- (10 xm)4 = 0.5
xm 11.9 (inches)
(b) First, obtain the PDF of X:
fX(x) = dFX/dx
= 4104x-5 for x 10, and zero elsewhere.
Hence the expected am

3.3
(a) Applying the normalization condition
f
X
( x)dx = 1
6
x2 x3
x2
c( x )dx = 1 c = 1
6
2 18 0
0
18
= 1/6
c=
9 36 6 3
6
(b) To avoid repeating integration, lets work with the CDF of X, which is
1 x2 x3
FX(x) =
6 2 18
2
3
= (9x x )/108
(for x b

3.5
Let F be the final cost (a random variable), and C be the estimated cost (a constant), hence
X=F/C
is a random variable.
(a) To satisfy the normalization condition,
a
1
a
3
3
3
dx = = 3 = 1 ,
2
a
x
x 1
hence
a = 3/2 =1.5
(b) The given event is F exc

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 the conditional probability,
P (X > 4 0 | X > 0)
= P(

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 = n)
n = 3, 4 ,5
=
n = 3, 4 ,5
= P(B =4)P(A = 3) = 0.20