This preview shows pages 1–20. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PRACTICE EXAMINATION NO. 5 PRACTICE EXAMINATION NUMBER 5
SOLUTIONS 1. May 1988 Course 110 Examination, Problem No. 3
A fair coin is tossed. If a head occurs, 1 fair die is rolled; if a tail occurs, 2 fair dice are rolled. If Yis the total on the die or dice, then Pr(Y = 6) = A. —l B. 5— C. H D. l E. E
9 36 72 6 36
Solution.
Probability of a head occurring is 0.50. When a head occurs, probability of getting a 6 is 1/6. Probability of a tail is 0.50. When a tail occurs, probability of getting a 6 is g, as there are ﬁve waystogeta6:1+5,2+4,3+3,4+2,5+1.Therefore
Pr(Y=6)=l.l+l.i=i+i=£.
2 6 2 36 12 72 72
AnswerC. 2. May 1988 Course 110 Examination, Problem No. 9
Two independent observations XI and X2 are taken from a distribution in which Pr(X‘. = 0) = 0.4 and Pr(X. = 1) = 0.6 for i = 1,2. What is the probability distribution of the mean, X, of the 2 scores obtained? A. X = 0,1 ,1 with respective probabilities 1,1,1.
2 3 3 3
B. — = 0%,] with respective probabilities 0.16, 0.48, 0.36.
C. 2? = 0,1 with respective probabilities 0.4, 0.6.
D. 2? is normal with mean 0.6 and variance 0.12.
E. i? is normal with mean 0.6 and variance 0.24. Solution.
1 Since X1 and X2 only take on values of 0 and l, the average can only equal 0,; or 1.
Furthermore
Pip—r = o) = Pr({Xl = o} n{X2 = 0}) = Pr(x, = o) Pr(X2 = o) = 0.4  0.4 = 0.16, ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Knysztof Ostnszewski  236  PRACTICE EXAMINATION NO. 5 J: Pr(({X1= 0}n{X2 =1})U({XI=1}0{X2 = =
( Answer B. 3. May 1988 Course 110 Examination, Problem No. 50
Let X and Ybe continuous random variables with joint density function .7
‘3 3 oo oo
f(x’y)={kx e , forl<x< andl<y< 0, otherwise.
Thenk:
l 1 l 1 l
A 1e3 B e3 C. §e3 D 2e3 E e3
Solution. Because the joint density is a product of functions of x and y, separated from each other, on a
rectangular region in the plane, X and Y are independent. We have nektme for 1<x<+oo and ﬁhkhv% for 1< x < +oo, where kl and k2 are constants such that kI k2 = k. ForX we have *°" 1 “’°‘ 1
3dx= (__ 4] =_ ,
1 19x k. 2x 2k,
so that k1 = 2, and for Y
*‘° 2 2 ’"H‘" 1
_[ kze 3dy=[3k2e 3] =3k,e 3,
l )2] I
so that ’9 = %e3. Finally, 1 1 2 1 k=k1k2 =2—e3 =e3. 3 3
Answer C. I ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof 0staszewski  237  PRACTICE EXAMINATION NO. 5
4. Sample Course 1 Examination, Problem No. 16 Micro Insurance Company issued insurance policies to 32 independent risks. For each policy, the probability of a claim is The beneﬁt amount given that there is a claim has probability density function fy(y)={ Calculate the expected value of total beneﬁt paid. 2(1—y), 0<y<l,
0, otherwise. A. E B. § c. 2 D. E E. 2
9 3 9 3 3
Solution. Let k be the index of the 32 policies, k = 1, 2, . . . , 32. Let B, be the random variable representing
the beneﬁt amount, given that there is a claim and let I k be the claim indicator random variable
(1 if there is a claim, and 0 when there is no claim). Then C, = I k B,‘ is the claim random 32
variable, and T = 2C, is the random variable representing total claims. We have E(Ik) = %
k=l
and
‘ 1
E(Bk)=Iy2(1y)dy=§.
0
so that
E(T)§E(C )— 32i—E
kz, " 18 9 '
Answer A. 5. May 1992 Course 110 Examination, Problem No. 1
Let A, B, and C be the events such that A and B are independent, B and C are mutually exclusive, Pr(A)=%, Pr(B)=%, and Pr(C)=%. What is Pr((AnB)cuC)? 11 3 5 23
. — B. — C. — D. — E. l
A 24 4 6 24
Solution. Since B and C are mutually exclusive, C C BC. Therefore, we have: ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004—2008 by Krzysztof Ostaszewski  238  PRACTICE EXAMINATION NO. 5
Pr((AnB)c uC)=Pr(AC UBC UC)= Pr(Ac UBC)= Pr((AnB)c)= = — =  PrB =1—.—=1—=—.
1 Pr(AnB) 1 Pr(A) () 4 6 24 24 Answer D. 6. May 1992 Course 110 Examination, Problem No. 2
LetXbe a random variable such that E(X) = 2, E(X’) = 9, and E((X — 2)’) = 0. What is Var(X)? 1 13 25 49 17
A. — B. — C. — D. — E. —
6 6 6 6 2
Solution.
Note that 0 = E((X — 2)’) = 15(X3 + 3X2 .(—2)+ 3X  (—2)2 +(_2)3) = = E(X3)—6E(X2)+12E(X) 8 = 9 6E(X2)+248 = 25— 6E(X2). Therefore E (X2) = $65 and 25 Var(X)=1E(X2)—(15(X))2 =?_21 =§_4=§_E=l 6 6 6 6'
AnswerA. 7. May 2003 Course 1 Examination, Problem No. 29, also Study Note P0908, Problem No.
72 An investment account earns an annual interest rate R that follows a uniform distribution on the
interval [0 .04, 0.08]. The value of a 10,000 initial investment in this account after one year is given by V =10,000eR. Determine the cumulative distribution function, F (v), of V, for values
ofvthatsatisfy 0 <F(v)<1. V 10.000 _ V _
. 10,000e 10,408 B. ZSeW 41.04 C. v 10,408
425 10,833— 10,408 DE E.251n( v J004
v 10,000 ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostaszewski  239  A Solution. PRACTICE EXAMINATION NO. 5
We are given that R is uniform on the interval [0.04, 0.08] and V = 10,000eR . Therefore, the distribution function of V is given by F(v)=Pr(VSv)=Pr(10,000eRSv)=Pr[eRS v )=Pr RSln(
10000 “(10,300) ln[ v ]—0.04 = I ;dr=k=25[m[;
004 0.08 —0.04 0.08 — 0.04 10,000 AnswerE. 8. May 1992 Course 110 Examination, Problem No. 4
Let X be a continuous random variable with density function 1
—x 1+3x, forl<x<3,
fx(x)= so ( )
0, otherwise.
WhatisE[i]?
X
A.l— 13.1 cis 13.3 EE
12 15 103 20 15
Solution.
Bydeﬁnition
3 3
[i]=j1 ix(1+3x)dx=jlﬂ’—‘dx=i2+i jxdx:
X 1x 1 30 30 10 1
_;+;.[1 2]”; L.[12_12)_
15 10 2 =1 15 10 2 2
1 1 9 l l l 1 2 l 6 7
=—+— —— =—+—4=—+—=—+—=—.
1510221510155151515
AnswerB. 9. May 1992 Course 110 Examination, Problem No. 6
Let X and Y be discrete random variables with joint probability function y
—, forx=1,2,4;y=2,4,8 andey,
fx.r(x’}’)= 24x 0, otherwise. D
tom]. 10,000 .240 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski PRACTICE EXAMINATION NO. 5 What is Pr[X+§SS)? A B.— C.§ D: E.—
8 8 l
'8 24 Solution.
There are only eight pairs of points (x, y) where the joint density is positive:
(1.2).(1,4). (l. 8).(2.2).(2.4). (2.8). (4.4) and (4. 8). Of these, the following (1, 2), (1 , 4), (l, 8), (2, 2), and (2, 4) satisfy the condition x+§ S 5,
while (4, 4), (2, 8) and (4, 8) do not. Therefore,
Pr(X+%SS)=l—Pr[x+§>5]= =1(Pr(X=4,Y=4)+Pr(X=2,Y=8)+Pr(X=4,Y=8)):
4 8 s l 4 2 24—7_17 244 242 244— 24 24 24' 24 24‘
AnswerE. 10. May 1992 Course 110 Examination, Problem No.8
There are 97 men and 3 women in an organization. A committee of 5 people is chosen at
random, and one of these 5 is randomly designated as chairperson. What is the probability that
the committee includes all 3 women and has one of the women as chairperson? I I I I I I I I 3
A. 14.97. B. 1.5.97. C. 35.97. D. 3$5.97. E. 3_
2 100! 2 100! 2 100! 100! 100! Solution.
The number of ways to choose a committee of ﬁve from 100 is [ 5 Then there are ﬁve ways
to choose a chairperson from the ﬁve on the committee, for a total number of possible events of 100
5 5 There is only one way to choose all 3 women for the committee, and then there are 97
[ 2 ) ways to choose the men. Once the ﬁve committee members are chosen, there are three ways to choose the chairperson who is a woman. The probability we are seeking is: _ ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  241  PRACTICE EXAMINATION N 0. 5 3, 97 97!
2 3'—2!95!_3.5!951.97z_§ 41971 5 [100] 5 100! "5 21951 100l'2' 1001' '5195!
Answer A. 11. May 1992 Course 110 Examination, Problem No. 10
Let Y be a continuous random variable with cumulative distribution function 0, fory S a, Fr (3’) ={ AW): 1— e 2 , otherwise,
where a is a constant. What is the 75th percentile of Y? A. F(0.75) B. a— "21n[i) C. a+ "21n[i] D. a—2Jln2 E. a+2~/1n2
a a Solution.
Let p be the 75th percentile of Y. Then leaa)” 0.75=Pr(Ysp)=F,,(p)=1—e 2 , l 2
so that l=e 10’ a) , i.e.,
4
I l 2
1n—=—— — ,
4 200 a) or
21n4=41n2=(p—a)2, and, given that p 2 a, p = a + 2J1n2.
Answer E. 12. May 1992 Course 110 Examination, Problem No. 11
Let X and Y be continuous random variables with joint density function 2x, for0<x<2and0<y<2—x, fx.Y 0‘”) = 4
0, otherwise.
What is Pr(X >1)? ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  242  PRACTICE EXAMINATION N0. 5
3
E. —
4 A 3
B C. — D.
8 l l l
' s ' 4 2
Solution.
The ﬁgure below illustrates the region where the joint density “plays” (i.e., is positive), and the doubleshaded region is where X > 1: Therefore, I 2
“"3 23 3 1 ‘ 1
PrX1= —ddx=—2— dx=—2——3)=—.
( >) {£4xy {4.74 x) [4x 4x 0 2 If you prefer to do this integral in the reverse order, then Pr(X > 1) = iziyEWdy: mix: "2")”: xal Answer D. 13. May 1992 Course 110 Examination, Problem No. 12
Let X and Ybe continuous random variables with joint density function f ( )_ 8xy, forOSxSySl,
x'y x’y — 0, otherwise,
and let W = X Y. What is the density function of W, where nonzero? A. 8wlns/v—v B. 8Wln\/; (3. SW D. 8w(1—J$) E. 8w(1—w) ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Kmysztof Ostaszewski  243  PRACTICE EXAMINATION NO. 5 Solution. In the ﬁgure below, note the region where the density is positive: the triangle bordered by lines x = O, y = 1, and y = x. The shaded region (below the curve xy = w, or, equivalently, y = E ,
x which is a hyperbola), describes the event Y S E , probability of which we will be calculating.
X Note that the line y = x and the hyperbola xy = w intersect at the point (x, y) such that
x = y = 5. Therefore, we have J; x_ l x=_ w l 4W2
= “4"2 _ )dy+ “4ny _y)dy= I4y3dy+ I—dy=
0 x J; J“ o w y
_ 4 F"; 2 "1 _ 2 2 _ _ 2 _ 2
_ (y )L=0 +(4w lny)y=ﬂ _ (w —0)+4w (1n1 Int/5) w 4w lib/J.
Hence,
1 1
w = F' w = 2w— 8wlnsfM74w2——= —8wln~/v—v.
m ) w( ) J; M
Answer A. We could also do this problem by using a bivariate transformation. Let U = Y and
W=XY. Then X=% and Y=U and
a—x a—x w l
— — 1
3(x’y) = det 3“ aw = det u2 u = .
B(u,w) ﬂ a_y 1 0 u
Bu 3w
Notethatthecondition OSxSySI is equivalentto OSKSuSI, or 0 SwSu2 $1. We
u
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004—2008 by Knysztof Ostaszewski  244  PRACTICE EXAMINATION N O. 5
conclude that 8 x,y w 1 8w
f....(u.w)=f...(x(u.w).y(u.w)) =8;u;=7
for 0 vaSu2 51, andO otherwisc.Now note that 0 SwSu2 51 implies that JESu $1. This
gives
+00 1
fw (w) = I f”, (u,w)du = js—wdu = 8w (1n uljj'ﬁ) = —8w1nJ?v'.
... .1; “
Answer A. 14. May 1992 Course 110 Examination, Problem No. 15
Let X and Y be continuous random variables with joint density function 1, for0<y<l—x and 1<x<l,
fx.r(xi}’)={ l 0, otherwise.
What is Var(X)? 1 l 2 ll 2
A. — B. — C — . — E — 18 6 9 18 3
Solution. The joint density is uniform on the region indicated below: The distribution of X is symmetric about the origin, thus E (X) = 0. Furthermore, f).r (x) = l— [x], for 1 S x S 1, because the joint distribution’s density is 1 (the joint distribution is uniform on a
region of area 1). Therefore, again using symmetry, not just of the distribution, but also of the
function x I—> x2, ASM Study Manual for Course Pl] Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  245  PRACTICE EXAMINATION N O. 5 Var(X) = 5(x2)=ix2(1+x)dx+jx2(1—x)dx= ziju—xwx = —l O Answer B. 15. May 1992 Course 110 Examination, Problem No. 16
Let X1 and X2 be random variables with joint moment generating function IMXDX2 (tvtz) = 0.3+ 0.1e'1 + 0.263 + 0.46:,“1 .
What is E(2Xl —Xz)? A. 0.1 B. 0.4 C. 0.8 D. 0.2e+0.4e2 E. 0.3+0.le2" +0.2e“2 +0.482"_" Solution.
Using the properties of the moment generating function: aMx..x2 (’1 J2) aMX 3: (t1 "2 )_ E(2Xl —X2)=2E(X,)—E(X2)= 2 _ = 3" r1 =I2 =0 atz 1. :1; =0
= 2(0.1e' + 0.4e""’) —(0.2e" + 0.4e"+") = 20.5 — 0.6 = 0.4.
rl=rz =0 t.=:, =0
Answer B.
16. May 1992 Course 110 Examination, Problem No. 17
Let X and Y be continuous random variables with joint density function
f ( ) x+y, for0<x<l, 0<y<1,
x'y x’y _ 0, otherwise.
What is the marginal density function for X, where nonzero?
l 5 1
A. + B.2x C.x D.— E.x+—
3’ 2 8 2
Solution.
By deﬁnition,
1 F1
1 2 1
fx(x)=I(x+y)dy= xy+—y =x+—.
0 2 y=0 2
Answer E.
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  246  J PRACTICE EXAMINATION N0. 5
17. May 1992 Course 110 Examination, Problem No. 19
A test for a disease correctly diagnoses a diseased person as having the disease with probability
0.85. The test incorrectly diagnoses someone without the disease as having the disease with
probability 0.10. If 1% of the people in a population have the disease, what is the chance that a
person from this population who tests positive for the disease actually has the disease? A. 0.0085 B. 0.0791 C. 0.1075 D. 0.1500 E. 0.9000 Solution.
Always start by labeling the events. Let D be the event of a person having the disease, and Tbe the event that a person tests positive for the disease. We are given that Pr(TI D) = 0.85, Pr(TlDC) = 0.10, Pr(D) = 0.01, and we are asked to find Pr(DT). This pattern of ﬂip—ﬂop in
probabilities clearly indicates that we should use the Bayes’ Theorem: Pr(TD)Pr(D) ass0.01
=_—____=___—=. 1.
Pr(DlT) Pr(TID).pr(D)+pr(TDC).P1(DC) 0.85.0.01+0.100.99 0079 Answer B. 18. May 1992 Course 110 Examination, Problem No. 21
Let X be a random variable with mean 3 and variance 2, and let Y be a random variable such that for every x, the conditional distribution of Y given X = x has mean x and variance x1. What is the
variance of the marginal distribution of Y? A.2 B.4 C.5 D.11 E.13
Solution.
Recall this key formula:
Var(Y) = E(Var(YX)) +Var(E(YX)).
Therefore, Var(Y)= E(X2)+Var(X)= (Var(X)+(E(X))2)+Var(X) =(2+ 32)+2 =13. Answer E. 19. May 1992 Course 110 Examination, Problem No. 29
The number of serious crimes reported daily in a certain city is a random variable X with mean 2 and variance 4. According to Chebyshev’s inequality, Pr(X 210) is less than or equal to which
of the following? A.O B.— C.— D 1
32 16 'E E l
' 4 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004—2008 by Krzysztof Ostaszewski  247  PRACTICE EXAMINATION NO. 5
Solution. Chebyshev’s Inequality says that if X is a random variable with mean 11 and standard deviation 0' then for any positive real number r, l
Pr(lX—pl 2 rd) Sr—z. In this case, p = 2 and 0'2 = 4. Therefore, h(X2[22r)sri2. Note that we are interested in the probability of X being above the value, whose distance from 2
is 8, and Zr = 8 for r = 4. Thefeore, we choose r = 4. This results in: Pr(X228)si 42'
But
Pr(X—228)=Pr({X228}u{X—2S8})=Pr({X210}u{XS—6})=
=Pr(X210)+Pr(XS6)=Pr(X210)+0=Pr(X210).
Therefore,
1 1
PrX_>.10 ——=——.
( )<41 16
AnswerC. 20. May 1992 Course 110 Examination, Problem No. 31 Bowl I contains 5 red marbles and 5 blue marbles, while Bowl II contains 8 red marbles and 2
blue marbles. We draw one marble at random from Bowl I and place it in Bowl 11. We then
draw one marble at random from Bowl H. Let X be the number of red marbles drawn from Bowl I and let Y be the number of red marbles drawn from Bowl II. What is Pr(X S Y)? B.£ CH D: E 10 A, l ._
2 11 22 10 11 Solution.
Since we only draw one marble from either bowl, X and Y can only assume two values: 0 and 1. The event {X S Y} means that either a blue marble was drawn from Bowl I, or a red marble was drawn from both. Its complement {X S Y}c is the event of drawing a red marble from Bowl I 2 1 . . .
and a blue marble from Bowl H. Therefore, Pr({X S Y}C) = %  H = ThlS implies that
l 10
Pr xsr =lPr xsr C =1—=—.
({ 1) (1 1) I, 1,
Answer E.
ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 20044008 by Krzysztof Ostaszewski  248  PRACTICE EXAMINATION N0. 5
21. May 1992 Course 110 Examination, Problem No. 32
Let X be a continuous random variable with density function x fx(x)= 5’ 0, otherwise.
What is E([X—E(X))? forOSxSZ, 2 4
. .— C.— D.— E.—
AO B 9 3 Solution.
First we note that E(X)=Ix§dx=[%x3] Therefore, 4
4 3 4 x 2 4 x _
E(IX—E(X)I)=EUX3—D—{(—JHEJE +£[x—Ejidx—
‘ 3
x3 it2 “3 x3 x2 m2_32
"("6_+?Lo {73L ‘5 Answer C. 22. May 1992 Course 110 Examination, Problem No. 33
Let X be a continuous random variable with density function 1 x, for —1<x<1,
fx(x)={ " 0, otherwise.
What is the density function of Y = X 2, where nonzero? l 1
A.—l B.2 — C.2 D.1— y E.— Solution. This is an unusual case, as on the domain where X has positive density, Y = X2 does not have an
inverse function. We cannot use the standard approach, and it is best to play carefully with the cumulative distribution function, noticing that Y is positive almost surely, so that we only need to
concern ourselves with positive values of it: ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  249  PRACTICE EXAMINATION NO. 5 F,(y)=Pr(YSy)=Pr(X2sy)=Pr({—J§sx<o}u{osxsﬁ})=
0 J; o J;
=£_(lx)ix+‘£(lxl)dx= jf(1+x)¢x+£(1—x)dx=
1 , ° 1 2 J; 1 1
= y+[Ex JJ;+ y—[Ex )0 =2J§+(Ey](Ey]=2J§—y.
Hence,
fy(y)=Fy’(y)=%1
AnswerA. 23. May 1992 Course 110 Examination, Problem No. 34 If four dice are rolled, what is the probability of obtaining two identical odd numbers and two
identical even numbers? A. i B. i C. ~1 D. i E. l
144 72 36 24 6
Solution. The total number of possible outcomes is 6‘. There are three ways to choose the ﬁrst odd
number, and three places to put the second one among the remaining three dice throws after the
ﬁrst one. Then there are three ways to pick the even number for the remaining two dice. But the
first number may be even or odd, and this multiplies the total number of desired outcomes by 2.
Thus the probability desired is: 2 3  3  3 _ 3 ~ 3 1 1 6‘ 666‘226=Z'
4
You can also count the number of favorable outcomes this way: there are (2) ways to choose the two dice that will have even numbers on them, and then the other two will need to have odd
numbers on them. There are 3 ways to choose the even number that will appear on the dice with
even numbers, and 3 ways to choose the odd number that will appear on the dice with odd
numbers on them. Therefore, the probability sought is 4
.36
[2] _2333_ 3.3 1 __1_ 64 ' 6‘ 666 2.26 2
AnswerD. o ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004—2008 by Knysztof Ostaszewski  250  é...) PRACTICE EXAMINATION NO. 5
24. May 1992 Course 110 Examination, Problem No. 35
Ten percent of all new businesses fail within the ﬁrst year. The records of new businesses are
examined until a business that failed within the ﬁrst year is found. Let X be the total number of
businesses examined prior to ﬁnding a business that failed within the ﬁrst year. What is the probability function for X? A. 0.10.9‘, forx=0,l,2,3,...
B. 0.90.1‘, forx= 0,1,2,3,...
C. 0.1x0.9", forx=1,2,3,...
D. 0.9x0.l‘, forx=1,2,3,...
E. 0.1(x—l)0.9", forx=2,3,4,... Solution.
Let a failure of a business be a success in a Bernoulli Trial, and a success of a business be a failure in the same Bernoulli Trial. Then X has the geometric distribution with p = 0.1, and
therefore fx (x) = 0.10.9‘ forx = 0,1, 2, 3, .
Answer A. 25. May 1992 Course 110 Examination, Problem No. 38
Let Xl ,...,Xwo and Y1 ,...Yloo be independent random samples from uniform distributions on the intervals [IOs/3JOJ3] and [—30,30],respectively. According to the Central Limit Theorem, what is the approximate value of Pr(i7  J? <1)? A. 0.31 B. 0.34 C. 0.66 D. 0.69 E. 0.72 Solution. (20%)2 The mean of each of the X ’s is 0, and the variance of each of them is = 100. The mean 2
of each of the Y’s is 0 and the variance of each of them is 61% = 560 = 300. Therefore, the mean of I? is O and the variance of it is 1, while the mean of i7 is 0 and its variance is 3. The random variable )7 — X can be expressed as a sum of independent, identically distributed random
variables in the following way: }7_X'=Y1—X1+Y2—X2 Y — X
+ . . . + —100 '00 ,
100 100 100
and therefore, by the Central Limit Theorem, Y — J? can be approximated by a normal random variable Wwith the same mean and variance as )7 )? , i.e., W ~ N(0,4). Hence, if we denote a ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  251  PRACTICE EXAMINATION NO. 5
standard normal random variable by Z, we have: Pr(?—X<1)=Pr(w <l)=Pr[W2_O <§J=n(2<%)=0.6915. Answer D. 26. May 1992 Course 110 Examination, Problem No. 39
A box contains 5 balls numbered 1, 2, 3, 4, and 5. Three balls are drawn at random and without replacement from the box. If X is the median of the numbers on the 3 chosen balls, then what is
the probability function for X, where nonzero? A. g(x)=0.2 forx= l,2,3,4,5 B. g(x)=% forx= 2,3,4
E forx=1,5
125
31 0.3 forx=2,4
C. = — f =2,4 D. =
gm 125 0” gm {0.4 forx=3
22 forx=3
125
0.1 forx=2%,%,4
E'3(x)= 8 10
0.2 forx=,3,—
3 3 Solution.
0f the ﬁve numbers, neither 1 nor 5 can ever be the median, 2 is the median for the following samples: {1,2,3}, {1,2,4}, and {1,2,5}, 3 is the median for the following samples {1,3,4},
{2,3,4}, {1,3,5}, and {2,3,5}, while 4 is the median for the following samples {1,4,5},
{2,4,5}, and {3,4,5}. The total number of way to choose an unordered threesample from ﬁve 5 .
elements is [3] = 52—5 = 10. Therefore, the probability function of the median is: 3
— for x = 2
10
4 0.3 forx=2,4,
x = — for x = 3 =
g( ) 10 0.4 forx=3.
3
— for x = 4
1
Answer D.
ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  252  PRACTICE EXAMINATION NO. 5
27. May 1992 Course 110 Examination, Problem No. 40 X and Y be discrete random variables with joint probability function given by the following table: What is the variance of Y — X? A.i B.1—6 Gag D.2 25 25 25 4 5
Solution.
From the joint distribution, we obtain:
—2 with probability %,
. .. 2
—1 With probability —,
Y—X= f
0 with probability 3,
l with probability Therefore,
1 2 1 l 2 2 1 3
E Y—X = 2 —+ 1 +0 —+1 —=————+—=—
( ) ( )5 < ) 5 5 5 5
Finally,
Var(Y—X)=
2 2 2 2
3 1 3 2 3 1 3 l
= 2— —— —+ —l— —— —+ 0— — —+ 1— —— —=
[(5)151 (51H (51H Ms
(7)” (2)” (3)” (8)”
= —— —+ —— —+ — ~—+ — o—=
5 5 5 5 5 5 5
49142 91641130 26
=——.—+—.—+—.—+—.—=—=—.
255 255 255 255125 25
AnswerC. ASM Study Manual for Course Pl] Actuarial Examination. © Copyright 2004~2008 by Krzysztof Ostaszewski  253  PRACTICE EXAMINATION N0. 5 28. May 1992 Course 110 Examination, Problem No. 42
Let X1,X2 ,X3 be random variables satisfying the constraints Pr(Xl > X2) = 0.7 and Pr(X2 > X3) = 0.6. What is the minimum possible value of Pr(X1 > X2 > X3)? A. 0.12 B. 0.30 C. 0.58 D. 0.70 E. 0.88
Solution.
We have: Pr(Xl > X2 > X,) = Pr({Xl > X;.'}r\{X2 > X3})= = Pr(X, > X2)+Pr(X2 > X,)— Pr({xl > X2} up:2 > X3}).
Now we see that the minimum will be reached when
1>r({Xl > X2} o{x2 > X3})
is the largest possible. Since this is a probability, its largest possible value is 1. Therefore, the
minimum possible value of Pr(X, > X2 > X3) is
1>r(Xl > X2)+ Pr(X2 > X3)— 1 = 0.7 +0.6 —1 = 0.3.
Answer B . 29. May 1992 Course 110 Examination, Problem No. 44
Let X and Y be discrete random variables with joint probability function (x +1)(y + 2)
———, for =0,l,2 and =0,l,2,
fx.r (xsy) = 54 x y 0, otherwise.
What is E(YX =1)? ll 11 y + 2 y2 + 2y
A. — B. 1 C. — D. —— E. 27 9 9 9
Solution. First we see that 4 6 8 13 1
fx(l)=fx.r(1'0)+fx.r(1,1)+fx.y(1,2)=gz+§+§=32:3.
Therefore,
E(YX=1)=0.fy(olx=1)+1.fy(1Xz1)+2,ﬁ(2X=l)= fx.r(1’l)+2.fX.Y(1’2)=0+1.3.£+2.3._§. 11 =°“‘f.u) no) 54 54 9' Answer C. ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  254  PRACTICE EXAMINATION NO. 5
30. May 1992 Course 110 Examination, Problem No. 45
Let X and Y be continuous random variables with joint density function ( ) 6x, for0<x<y<l,
f” x’y — 0, otherwise. Note that E(X) = and E(Y) = What is Cov(X,Y)? A. i B. 3 c. 3 D. 1 E. E
40 5 8 8 Solution. The graph below shows where the joint density is nonzero. We have: E(XY) = 6xdydx = jj6x2ydydx =
0 x 0 x l 1
=I3x2(y2H)dx=J‘3x2(lx2)dx=[x3—2x5] =2.
0 y=x 0 5 x: 5
Therefore
2 l 3 16 15 l
Cov X,Y :5 XY E X E Y = ———— —=———=—.
()()()()524404040
AnswerA.
ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostaszewski  255  ...
View
Full
Document
This note was uploaded on 10/27/2010 for the course PSTAT 172a taught by Professor Staff during the Winter '08 term at UCSB.
 Winter '08
 Staff

Click to edit the document details