P_PracticeExam6_05-1..

The survival function for the Weibull distribution is sX ( x)
= e
)
# x &
!
$
%
" '
(
PRACTICE EXAMINATION NO. 6
for x > 0, and from the
information given in the problem we have sX ( 3)
= e !1 and sX ( 6)
= e !4 . Therefore, e
as well as e
" 3 %
#
$
! &
'
(
that sX 4
Answer D.
)
# 6 &
!
$
%
" '
(
= e !4 . We are looking for sX ( 4)
= e
= 1, or ! = 3. Furthermore, e
( ) = e
)
# 4 &
!
$
%
" '
(
= e
2
# 4 &
!
$
%
3'
(
)
# 6 &
!
$
%
" '
(
= e !2)
= e !16
9 * 0.1690133.
)
# 4 &
!
$
%
" '
(
. But e
18. You are given that the hazard rate for a random variable X is
! X ( x)
= 1
x" 1
2
2
for x > 0, and zero otherwise. Find the mean of X.
A. 1 B. 2 C. 2.5 D. 3 E. 3.5
Solution.
The hard way is to calculate the survival function of X via:
! 1
x 1
!
t 2 dt
e
" 2
0
! (
= e
x
t ) 0
for x > 0. Therefore,
( ) = e ! x +"
# dx
0 ! # " $#
Substitution of
z= x ,2zdz=dx
E X
= e
( ! x + 0 ) ! x
= e
+"
+"
! z
= # 2ze dx = 2
! z
# ze dx = 2.
0
0 ! # " $#
Mean of EXP(1)
The easy way is to notice that this is the hazard rate
! X ( x)
= "
# $ x
" +1
% (
&
'
# )
* = "
# " $ x" +1
for the Weibull distribution with ! = 1
and ! = 1, so that
2
E( X)
= !" 1+ 1 $ '
%
&
# (
) = 1* " 3 ( ) = 2! = 2.
Answer B.
)
# 3 &
!
$
%
" '
(
)
# 3 &
!
$
%
" '
(
= e !1 implies that
= e !1 ,
= e !4 implies that ! = 2. Therefore, we conclude
ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2007 by Krzysztof Ostaszewski - 275 -