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Unformatted text preview: SECTION 14
PRACTICE EXAMINATION NUMBER 10 SOLUTIONS 1. An automobile insurance company has a block of oneyear car insurance policies. The policies
are divided into three classes: A, B, and C. A randomly chosen policy has 40% chance of being
in class A, 10% in class B, and 50% in class C. The probability that a policy will produce a claim
is 20% in class A, 10% in class B and 5% in class C. A class of policies (i.e., either class A, or
class B , or class C) is chosen at random, with probability of being chosen proportional to the
random chance of a policy being chosen from class (i .e., 40% for class A, 10% for class B, and
50% for class C) and ﬁve policies are selected at random from that class. It turns out that exactly
one of the ﬁve policies produced a claim. What is the probability that these policies are from class A? A. 0.287 B. 0.339 C. 0.458 D. 0549 E. 0.699 Solution.
Let pA = 0.20 be the probability of producing a claim for class A, Pa = 0.10 be the same probability for class B, and pc = 0.05 be the corresponding probability for class C. Let us write
p for any of these three probabilities. Then the probability of producing exactly one claim among 5
ﬁve policies is [1) pl (1— p)4 . By the Bayes’ Theorem,
Pr(5 policies came from A One claim exactly among 5 policies) = 5 02010804 040
1 ' ' ~—'r—I
“W _——J Pr(0ne claim exactly among 5 policiesIS policies came from A) 5 l 4 5 1 4 5 l 4
1 .020 0.80 o.4o+ 1 o.10 o.90 0.10+ 1 .0.05 .095 0.50 = 054895444.
Answer D. 2. You are given that the joint density of random variables X and Y is f“, (x, y) = x + y for
0 < x < 1 and 0 < y < l, and zero otherwise. What is the coefﬁcient of variation of Z = X + Y ? A. 0.1889 B. 1.3611 C. 1.55 D.0.50 E.0.3194 Solution.
Note that the function f“, is a density because ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  381 . PRACTICE EXAMINATION NO. 10 y=l 11mmIez+yx):;]dy=randym Observe also that 0 < Z < 2 with probability 1. Using convolution we obtain fz(z)=fx+r(z)= Ifxy(xz" )xdx= I zdx= I mix: . 05x51 0.1nzl.z
OSz—x51 [ 1 l ] PO 1
Izdx, 0<le,
__ o _{ZZ, 0<z<1}_{z, 0<le,
“1 " 2— ,1<zS2._2—2,1<52.
Izdx, 1<z52. z( Z) z z Z 2—1 Therefore,
1 2 I 2 1 z=2
EZ=z3d+ 22—3d=—+(—3_4] =
( ) I Z I( Z Z) 2 4 32 42 z:
1 [16 2 I) l 14 3 28 24 7
=— _" "—+— =—+—4=—+———=,
4 3 3 4 2 3 6 6 6 Hence, 3 49 54 49 5
Var<z>=E(zz)— (5(2))2=§‘§g=§"%=3—6 Finally, the coefﬁcient of variation is .I_v(z \)_I_’5‘
”(Z 36 ‘5 O..31943828 E—(Z) = —"
6
Answer B. We could also use the formulas E(Z) = E(X + Y) = E(X)+ E(Y),
Var(Z) = Var(X+ r) = Var(X)+ VaI(Y) + 2Cov(X,Y),
and calculate all required items directly. We have y=l
fx(x)= I()x+y dy= x+[1y2 ]=x+l’
2 F0 2 l
for 0 < x <1, zero otherwise, and because of symmetry f, (x) = y + 2 for 0 < y < 1, zero otherwise. Therefore ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004—2008 by Krzysztof Ostaszewski 382 SECTION 14 ‘ 1 13 12 "“' 4 3 7
E(X)—£x[x+5}1a—(§x Z JenE 55,
7
EY=—,
H 12
14 7
EZ=EX EY=—=—,
() ( )+ () 12 6
l 2:]
5(x2)=jx2[x+l)¢‘=[lx4 1x3] =3 i=3,
0 2 4 6 no 12 12 12
5
EY2 =—,
( ) 12 l and 11 ll 2 5
V Z=V X Y=V X V Y 2C X,Y =—+——=.
ar<>ar<+>ar<1+ar<>+ ov< )144144144 36 Therefore, the coefﬁcient of variation of Z is 5 ﬂ=i=£=031943828 13(2) Z 7 ' '
6 Answer E, again. 3. An insurance policy has a deductible of 10. Losses follow a probability distribution with
density fx (x) = xe" for x > 0 and fx (x) = 0 otherwise. Find the expected payment. A. e"0 B. 2e"° C. lOe'l0 D. 12e'lo E. lOOe'l0 Solution.
Let X be the random variable describing losses, and let Y be the amount of payment. Then ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  383  PRACTICE EXAMINATION N0. 10 Y={o, xslo, J X — 10, X > 10.
We can calculate the expected value of Y directly u=x2 v=e" _
du=2xdx dv=e"dx _ \_—_—I
Integration by pans forth: ﬁrs integral E(Y) = T(x —10)xe“dx = sze"dx T10xe"dx =
lo 10 lo XDw (xzex) x=lO + T2xe"dx  Tlee'Vx = 100e"° — T 8xe"dx =
10 IO l0 u = 8x v = e" _ _ xw “” _
= _x =100e1°+(8xe ‘) —8 [e ’dx =
du = de dv = e dx “1° ,0
1n ‘ b Ifortheﬁrst‘ s 'vnlfuncul‘ r
te I te um ono
gnaw ypam n gal exponential distribution Widthnmdmeevnlualed
atx=l0 = 1002'lo — 80e"° — 8e"° =12e"°.
Answer D. We could also do this problem by applying the Darth Vader Rule. We have sy (y): Pr(Y > y): {P4X >10). ify=0,} =PrX 10= '* ,=
Pr(X10>y). ify>0, ( >y+ ) I” it 10+);
.3 +ua
u = "X V = e _ xtw _
= = xe ’ + I e ’dx =
dx = dx dv = —e"‘ x=w+y [0+ .)
y _ ——'v———4 __‘,_a lawman by pm: Survival function ofex neutinl with and tale 1
evaluated at lO+y = (10 + y)e"°" + (“H =(11+ y)e"°" for any nonnegative y. As the payment random variable is nonnegative almost surely, the
expected payment is «a «a +oo
    o  lO
I(ll+y)e'° ydy=lle'° Ie’dy +e‘  Iye’dy =12e .
0 0 0
Thisequals one because This equals one
e" for y>0 isadensity WSNhFﬂtlBWd
ofanexpomu’al random “Mdemnyln ““5
variable with hazard rate 1 problem. Answer D. 4. You are given that the joint density of X and Y is f,” (x,y) = 3(1x)y2 for —1 < x <1 and
0 < y <1, zero otherwise. Find E(X2Y2). D E A.— B.— cl .1 .l
12 10 6 4 3 ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004—2008 by Knysztof Ostaszewski  384  SECTION 14
Solution.
From the joint density form we see that the random variables and Yare independent and fxa' (753’) = (1 _ le) ' a)”: i W
t
422‘.“ 6292:“ and zero otherwise. Therefore, Answer B . 5. May 1982 Course 110 Examination, Problem No. 1
If a normal distribution with mean u and variance 0'2 > 0 has 46th percentile equal to 200',
then u is equal to A. 18.250 B. 19.900 C. 20.106 D. 21.730‘ E. Cannot be determined Solution. X  l1
0' distribution. Let us write x0“ for the 46th percentile of X. We have 0,46=PI(X<xo"6)=Pr[X—ll<xo.46‘/~‘]=Pr[X,u<200'ﬂ). has the standard normal Let X be the random variable described in the problem. Then O' 0' 0" O'
From the table, <1:(0.1) = 0.5398 and 0(011): 0.5438. Therefore, c1>(—0.1)= 0.4602 and <I>(—0.1 1) = 0.4562. The probability desired is 0.46, and that is
0.4602 0.46 _ 0.0002 _ 1 0.4602 — 0.4562 ' 0.0040 — E
of the distance between 0.4562 and 0.4602, down from 0.4602. This means that the 46th percentile is approximately 2—10 of the distance between —0.1 and —O.1 1, down from —0.1, i.e., we can get the desired percentile of the standard normal distribution via the following linear
interpolation approximation: 1 0.01
=_0.1__. _o_1 _ _o,11 =—0.1—=—0.1005. Therefore,
ASM Study Manual for Course PI] Actuarial Examination. © Copyright 2004—2008 by Krzysztof Ostaszewski  385  PRACTICE EXAMINATION NO. 10 200"“ =—o.1005,
0' so that u z 20.10050'.
Answer C. 6. May 1982 Course 110 Examination, Problem No. 2
Suppose that U and V are independent random variables, each uniformly distributed on the interval [100,200]. What is the number 2‘ for which the probability that at least one of U and
V exceeds t is 0.25? 2
A. 20050~/§ B. 150 c. 175 D. 100+50J§ E. zoo—G) 100 Solution.
We have Pr({U >t}U{V >t})=Pr({U sr}c U{VSt}C)=Pr(({U St}ﬂ{VSt})c)=
=1—Pr({U St}n{VSt})=1—Pr(USt)Pr(VSt)= _1_ t—lOO t—lOO _1_[t—100]2
200—100 200100 100 ‘
Bysetting
2
Pr({U>t}u{V>t})=l—(t101§0] =o.25, we obtain this equation
2
[“100] =0.75=3,
4 100
leading to
t— 100 _ _J_§
100 _ 2 '
and t= 100 +50J3.
Answer D. 7. May 1982 Course 110 Examination, Problem No. 5
3
Suppose the joint density function of X and Yis given by f“, (x, y) = Z for 0 < y2 < x < 1. What is the marginal density function of X for 0 < x < l? ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  386  ”Q; SECTION 14 3 l 3 3 1 3 3
A. 5x2 B. 1(1—x2) C. 2x2 D. 5(lx2) E. Z
Solution. The condition 0 < y2 (x <1 is equivalent to —l <  x < y < J: <1. Therefore,
4.. J; 3 3 3
= , d = —d =—2J‘=—J§.
fx(x) ifxxh J’) y “£4 y 4 2
AnswerA. 8. May 1982 Course 110 Examination, Problem No. 7
Let X be a random variable with cumulative distribution function F( )_ 0 forxSO,
X x ' le" forx>0.
What is Pr(0 se" s4)? A. e’4 3% 0% D; E. le" Solution.
First note that X has the exponential distribution with mean 1. Also, because the exponential
function f (x) = e" is strictly increasing and always positive, we have 1 Pr(05e"s4)=Pr(e"s4)=Pr(X51n4)=1e""“‘ =1Z_%. Answer D. 9. May 1982 Course 110 Examination, Problem No. 8
Suppose Q and S are independent events such that the probability that at least one of them occurs is g and the probability that Q occurs but S does not occur is 5. What is Pr(S)? A. i B. l c. E D. l E. l
9 3 9 7 9
Solution. We are given that Pr(Q u S) = %, Pr(Q 0 SC) = %, and that Q and S are independent (this last piece of information turns out to be irrelevant). Note that ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  387  PRACTICE EXAMINATION N0. 10 QUS= (QnSC)US. _,_/
Union of mutually exclusive events
This implies that
Pr(S)=Pr(QUS)Pr(QnSC)=%—%=g. Answer C. 10. May 1982 Course 110 Examination, Problem No. 9
Suppose Y is a random variable with mean u > 0 and variance 0'2 > 0. For what value of a, 2
where a>0, is E[[aY—1] ] minimized?
a 1 1 1 l l A. E B. M 2 C. 0' 2 D. (02 +;12)_4 E. p(a’ +u2)'4
Solution.
2
Deﬁne a function g(a) = E [(aY — l) ]. Then by the wellknown result of the great Henri
a Lebesgue, you can move the derivative inside the expected value, i.e., the derivative of the
expected value is an expected value of the derivative: w=%E[(w—2ﬂ=Estaif}Emma(ma Y Y 1 l 2 2
=2E Y2+—————— =2E Y2—— =2 EYZ ——=2 02+ 2 ——.
[a a a a3) (a as] a( ) a3 a( p) :23
Setting this derivative equal to zero gives
2
20(02+ﬂ2)—?=0,
and
l l
1 4 2 2
a= = 0' + 4.
[0.24112] ( .u)
Notealsothat g"(a) = 2(02 + p2)+ % 1
and this is positive at the point a = (0'2 + #2) 4 , verifying that this is where a local minimum of the function g(a) occurs.
Answer D. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Osmszewski  388  SECTION 14
11. May 1982 Course 110 Examination, Problem No. 10 A rectangle is to be constructed having dimensions X by 2X, where X is a random variable with density function fx (x) =% for 0 < x < 2. What is the expected area of the rectangle? A.2 B.4 C.8 D. 3—32 E. 16
Solution.
The area equals X . 2X = 2X 2 , so that the expected area is 1 “2 1 2 2
E(2X2)=12x2 5 =j'x3dx =—x‘ = —. 16: 4.
o 2 o 4 no 4 Answer B . 12. May 1982 Course 110 Examination, Problem No. 12
A random variable X has the density function 0.4, 0 < x S 1,
fx(x)= 0.6, 1<xSZ,
0, otherwise.
Which of the following statements about the random variable X are true? A. It is continuous with Pr(X = 2) = 0.6 B. It is continuous with Pr(l < X < 2) = 0.6 C. It is discrete with Pr(X = 2) = 0.6 D. It is discrete with Pr(l < X S 2) = 0.6 E. It has a distribution that is a mixture of continuous and discrete Solution.
For this random variable to be discrete or a mixture, it would have to have clearly indicated
point—masses, and it does not. Thus this is a continuous distribution. For a continuous distribution, Pr(X = 2) = 0. So the answer is B, but just to check 2
Pr(1< X < 2) = IO.6dx = 0.6.
1
Answer B. 13. May 1982 Course 110 Examination, Problem No. 14
Suppose the random variable X has moment generating function M x (t) = (1 — ﬁt)'k , provided ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  389  PRACTICE EXAMINATION NO. 10
1 .
1‘ <3. What IS 500)? A. a“ ‘12!)(kti) B. B"ﬁ)(k+i) c. nn‘ﬁ(k+i) D. (5)" gun) E. (argon)
Solution.
We have
5(X")=j,.((1—ﬁt)'*) =%.—.((k)(—m(1/st)"") =...
= (—k)(—m  (k  1)  (13) ...(k  (n 1))(p)(1— W“ =
Note that the number of minus signs is even. because each one in from of ﬂ is paired with one in from of k = B” :i:§(k + i). Answer B. 14. May 1982 Course 110 Examination, Problem No. 15
Suppose that X1, X2, and X3 are random variables with E(Xi) = 0, Var(X,.) =1 for i = 1,2,3 3
and Cov(X,.,X,.) = 0.5 for i¢ j. What is the variance of Y = ):i . x. ?
i=1 A.2.0 B. 3.0 C.6.0 D. 8.5 E. 14.0 Solution.
We have 3
Var(zi.x,.)= Var(X1+2X2 + 3X3):
i=l = Var(X,)+4Var(X2)+9Var(X3)+4Cov(X,,X2)+ 6Cov(X,,X3)+12COv(X2,X3)= =1+4+9—23—6=3.
AnswerB. 15. May 1982 Course 110 Examination, Problem No. 16
A fair coin is tossed until a head appears until a head appears. Given that the ﬁrst head appeared
on an even numbered toss, what is the conditional probability that the head appeared on the fourth toss? ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  390  SECTION 14 A.l 13.1 oi 13.1 E
16 8 l6 3 16
Solution.
Wehave
1 3 1 1 ‘ 1
Pr(Theﬁrstheadappearsonthefourthtoss)= — —= — =—,
2 2 2 16
and Pr(The ﬁrst head appears on an evennumbered toss) = =(%)‘%+(%J’é+(%)’—%+=(%)‘+(%)‘+GJ°+=PG]: The probability we are looking for is the ratio of the ﬁrst of the above probabilities to the second oneie i
“"16' l
3. Answer C. 16. May 1982 Course 110 Examination, Problem No. 18 Two balls are dropped in such a way that each ball is equally likely to fall into any one of four
holes. Both balls may fall into the same hole. Let X denote the number of unoccupied holes at the
end of the experiment. What is the moment generating function of X? A.Zlt ift=2 or 3, otherwise B. 3+3:
4 2 4 8
1 2r 3: l 21 31 1 31 L
C. —(3e +e ) D. —(e +3e ) E. — e4 +3e4
4 4 4
Solution. Each ball had four possible holes to fall into, so that the total number of ways for the balls to fall
is 44 =16. There can be only two or three holes unoccupied. If three holes are unoccupied, there are four ways to put two balls in one occupied hole. Hence, Pr(X = 3) = % = l and 4
Pr(X=2)=l—Pr(X=3)=%. This gives us the following moment generating function
_ 3 2 1 31 _ 1 2 3:
Mx(t)—Ze '+Ze —Z(3e ' +e ).
Answer C. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  391  PRACTICE EXAMINATION N0. 10
17. May 1982 Course 110 Examination, Problem No. 20 The number of potholes on a certain highway has a Poisson distribution with an average of 3
potholes per mile. What is the probability that a given 2mile portion of this highway has no
more than 3 potholes? A. 25e'6 B. 6Ie" C. —e'3 D. 13.24 E. l—25e'6 N0 Solution.
If there are 3 potholes per mile, this implies that there are 6 potholes per 2mile stretch. Let X be
a random number of potholes in a 2mile stretch. Then the probability we are looking for is Pr(XS3)=Pr(X=0)+Pr(X=l)+Pr(X=2)+Pr(X=3)=
62e"S 63e'6
3! =e'°+6e“+ =6le". Answer B . 18. May 1982 Course 110 Examination, Problem No. 24 A card is drawn at random from an ordinary deck of 52 cards and replaced. This is done a total
of 5 independent times. What is the conditional probability of drawing the ace of spades exactly
4 times, given that this ace is drawn at least 4 times. 12 C 13 D 60 E 255 A. l B. — .— .— . —
2 13 14 61 256 Solution.
Probability that exactly 4 aces of spades are drawn is (Ella.1; and the probability that exactly 5 aces of spades are drawn is
5 1 5
so that the probability sought is
5 1 4 51
[415] 5—2 _ £5
[511T 51 {51 1 )5 — 256'
4 52 52 5 52 Answer E. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004—2008 by Knysztof Ostaszewski  392  t; SECTION l4
19. May 1982 Course 110 Examination, Problem No. 26 2
Suppose that X is a random variable with density function fx (x) = 3% for 0 < x < 2 and zero otherwise. Let Y = mX2 , where m is a positive number. What is the density function of Y, where
nonzero? A.32for0<y<§ﬂ Rihfor0<y<4m C.Lﬂfor0<y<4m
8m 3 16( ’m) 2m 3
D.l—“8y for0<y<2 E.—3for0<y<§
2 3 2 8 3
Solution. As 0 < x < 2, and Y = mX‘, we have 0 < y < 4m. Also,x is positive, so that the transformation is onetoone for 0 < x < 2, and the inverse is x =41. The derivative of the inverse is
m E: i._1_=_1_
dy m 25 2Jrr—Iy'
Therefore, 3[ .31]: J—
L’x _’" _1 3.L_3. y
My)_f"(x(y»'ldy 8 '2 myl6 me_y—16 mJZ’ 20. May 1982 Course 110 Examination, Problem No. 27
Let X and Y have joint density function 2e"’e‘y for 0 < x S y < co,
x, =
fx'Y( y) {0 otherwise.
What is the marginal density of X where nonzero? A. ﬁe" for0<x<+°o B.e"‘ for 0<x<+oo C. 2e"(1e") for 0<x<+°°
D. 2e'2" for 0<x<+oo E. 252* for0<x<y<+oo Solution.
Note that only the values of y meeting the condition 0 < x S y < oo will be used in the calculation of marginal density, because for other points f“ (x, y) = 0. Therefore, for x > 0, ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  393  PRACTICE EXAMINATION N0. 10 ym fx (x) = foy (x.y)dy = TZe"e"dy = 2e" 'Te"dy = 2e" (e" )= 2e'z". Otherwise, the density is zero.
Answer D. )‘=‘ 21. May 1982 Course 110 Examination, Problem No. 30
Suppose there are (n +1) slips of paper, numbered 1, 2, 3, ..., (n — 2), (n  l), n, n. One slip of paper is randomly selected, its number recorded, and then replaced. This is done a total of n
independent times. What is the probability that the numbers recorded, in any order, are l, 2, 3, ..., (112), (nl), n? l l l v I
A. n. n B. n. n C. (n+1);l D. 2(n.)n E. n. n+1
(n+1) 2(n+l) (n+1) (n+1) (n+1)
Solution.
Consider ﬁrst the probability of getting 1, 2, 3, ..., (n — 2), (n — l), n in order. Probability of
getting 1 in the first selection is %. Since slips are selected and replaced, probability of
n
getting 2 in the second selection is also %. So is the probability of getting any number up to
n
n — 1. For n, the probability is %. The probability of getting 1, 2, 3, ..., (n — 2), (n— 1), n in
n order is therefore
1 1 1 2 2 n+l'n+l.m n+1 n+l=(n+1)"' 2(n!) (n+l)" ' There are n! ways to order these n slips, so the probability desired is Answer D. 22. May 1982 Course 110 Examination, Problem No. 31
Let X1 ,X2 ,. . . ,X” be a random sample from a continuous uniform distribution on the interval (0,5). Let T denote the smallest observation in the sample. What is the cumulative distribution
function, F, (t), for t in the interval (0,5)? t 17 t I? t 17
A. (E) 13.1—(52)l7 c.1—(1—5t)l7 D.1—[§] 13.1—[1—3) Solution.
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  394  SECTION 14
We have T = min(T, ,T2,. ..,T,,), so that the survival function of Tis 17 5—; '7
s,(t)=Pr(T>t)=1'[Pr(X,>t)= 5— .
(:1
Therefore,
5—: I7 t 17
F}(t)=ls,(t)=l—[—5) =1‘[“§] .
AnswerE. 23. May 1982 Course 110 Examination, Problem No. 32
Let X1,X2,...,Xw0 be a random sample from the uniform distribution with mean u and variance 0'2 > 0. If (I) is the cumulative distribution function of the standard normal distribution, then It!)
Pr(2 X, S an) is closest to: A.¢[(a—l)u] 30¢[(a—1)#] C_¢[(a100)y] a 100' 0'
D. (p (a100)[1 E. (I) (a—100)p
100' 1000'
Solution.
By the Central Limit Theorem
I00 all
2X " —#
10° = ‘ all — an X'# 100
Pr XS =Pr LS— =Pr XS— =Pr S =
(E. ' a”) 100 100 [ 100) a o
JIOO 100
ﬂ_ ﬂ
zq, 100 ’1 =d, 100 ’1 =¢[aﬂ100#)=¢ (a100)p
g 100' 100'
100 10
AnswerD. 24. May 1982 Course 110 Examination, Problem No. 35 Suppose an experiment consists of tossing a fair coin until three heads occur. What is the
probability that the experiment ends after exactly six ﬂips of the coin with a head on the ﬁfth toss
as well as on the sixth? A. i B. l c. i D. l E. E
16 8 32 4 32 ASM Study Ma...
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