P_PracticeExam6_05-1..
P_PracticeExam6_05-1..
P_PracticeExam6_05-1..
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We have:<br />
40<br />
"<br />
( ) !2<br />
1 = C 10 + x<br />
0<br />
dx = !C ( 10 + x)<br />
!1<br />
40<br />
=<br />
0<br />
C C 2<br />
! =<br />
10 50 25 C,<br />
PRACTICE EXAMINATION NO. 6<br />
and therefore C = 25<br />
= 12.5. Now we can calculate the probability over the interval (0, 5):<br />
2<br />
12.5( 10 + x)<br />
!2<br />
5<br />
"<br />
dx = !12.5( 10 + x)<br />
0<br />
!1<br />
5<br />
= 12.5 #<br />
0<br />
1 $ 1 '<br />
!<br />
%<br />
&<br />
10 15(<br />
) = 0.416667.<br />
Answer C.<br />
3. Study Note P-09-<strong>05</strong>, Problem No. 61<br />
An insurer’s annual weather-related loss, X, is a random variable with density function<br />
2.5 ! 200<br />
fX ( x)<br />
=<br />
2.5<br />
x 3.5 "<br />
$ , for x > 200,<br />
#<br />
$<br />
% 0, otherwise.<br />
Calculate the difference between the 25-th and 75-th percentiles of X.<br />
A. 124 B. 148 C. 167 D. 224 E. 298<br />
Solution.<br />
The cumulative distribution function of X is given by<br />
FX ( x)<br />
=<br />
x<br />
"<br />
200<br />
2.5 ! 200 2.5<br />
t 3.5<br />
dt = # 2002.5<br />
t 2.5<br />
x<br />
200<br />
= 1# 2002.5<br />
x 2.5<br />
for x > 200. Therefore, the p-th percentile x p of X is given by<br />
or<br />
p<br />
100 = F x p<br />
( 1 ! 0.01p)<br />
2<br />
( ) = 1 ! 2002.5<br />
2.5<br />
xp ,<br />
5 = 200<br />
,<br />
x p<br />
resulting in<br />
200<br />
xp =<br />
( 1 ! 0.01p)<br />
2 .<br />
5<br />
It follows that<br />
x75 ! x25 = 200<br />
! 200<br />
Answer A.<br />
0.25<br />
2<br />
5<br />
0.75<br />
2<br />
5<br />
= 123.8292.<br />
ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2007 by Krzysztof Ostaszewski - 259 -
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