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Unformatted text preview: PRACTICE EXAMINATION NO. 1 PRACTICE EXAMINATION NUMBER 1
SOLUTIONS 1. May 2001 Course 1 Examination, Problem No. 4, also Study Note P0908, Problem No.
66 A company agrees to accept the highest of four sealed bids on a property. The four bids are
regarded as four independent random variables with common cumulative distribution function F(x)=%(l+sin7rx) for 3 S x S ‘2‘. Which of the following represents the expected value of the accepted bid? . 1—16 (1+sinnx)‘dx C. —1 x(1+sin7tx)4dx xcosztx dx B
16 A.:r Nu0—.uu
Nu'—.NUt
NIu'—.Muu 11: xcos 7rx(1 + sin 7tx)3dx cos 7tx(1 + sin nx)3dx E Din .1
4 4 Nlu'——.Nl'h
Nlu‘——.NIM Solution.
Let X1 ,X2 ,X3 and X 4 denote the four independent bids with common distribution function F. Then if we define
Y = max(Xl,X2,X3,X4),
the distribution function F} of Y is
F}(y)=Pr(Y Sy)=Pr(Xl sy)n(X2 Sy)n(X3 Sy)n(X4 Sy)= 4
9 = Pr(Xl S y)Pr(X2 S y)Pr(X3 S y)Pr(X4 S y) = (F(y))4 = %(1+ sin ﬂy) 3 5
for a S y S 5. It then follows that the density function g of Y is given by
l
f), (y) = 5307) = Z(1+ sin ivy)3 oncosn'y = %cos ny(l + sin my)3 ,
3 5
for — S S —. Final] ,
2 y 2 y
s 5
5 5 ﬂ 3
150’) = 1% Mb = Izycowyﬂ + sin 70’) dy
3 3
5 5
Answer E.
ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  118  J ‘ PRACTICE EXAMINATION NO. 1 / ‘ 2. May 2001 Course 1 Examination, Problem No. 5, also Study Note P0908, Problem No.
W 117 A company is reviewing tornado damage claims under a farm insurance policy. Let X be the
portion of a claim representing damage to the house and let Ybe the portion of the same claim
representing damage to the rest of the property. The joint density function of X and Yis 61— x+ forx>0, >0,x+ <1,
f(x.y)={0( ( y» . y y
, otherwrse. Determine the probability that the portion of a claim representing damage to the house is less
than 0.2. A. 0.360 B. 0.480 C. 0.488 D. 0.512 E. 0.520 Solution. Always start problems such as this one by drawing a graph of the area where the joint density is
positive, and where events considered happen. The area where the joint density of X and Y is
positive is shown in the ﬁgure below as the triangle bounded by the axes and the line y = l — x. The shaded region is the portion of the domain of the joint density over which X < 0.2. Pr(X <0.2)=TT6(1—(x+y))dydx= 60f[y_w_ly2] x=0 2 ' =0.83+1=0.488. Answer C. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  119  PRACTICE EXAMINATION N0. 1
3. May 2001 Course 1 Examination, Problem No. 6, also Study Note P0908, Problem No.
20
An insurance company issues life insurance policies in three separate categories: standard,
preferred, and ultrapreferred. Of the company’s policyholders, 50% are standard, 40% are
preferred, and 10% are ultrapreferred. Each standard policyholder has probability 0.010 of '
dying in the next year, each preferred policyholder has probability 0.005 of dying in the next
year, and each ultrapreferred policyholder has probability 0.001 of dying in the next year. A
policyholder dies in the next year. What is the probability that the deceased policyholder was
ultrapreferred? A. 0.0001 B. 0.0010 C. 0.0071 D. 0.0141 E. 0.2817 Solution. Start by labeling the events: let S be the event of a policyholder having a standard policy, P be
the event of a policyholder having a preferred policy, U be the event of a policyholder having an
ultrapreferred policy, and D be the event that a policyholder dies. Then by the Bayes’ Theorem: _ Pr(DU)Pr(U) _
MUID) _ Pr(DS)Pr(S)+Pr(DP)Pr(P)+Pr(DU)P(U) —
0.001.o.1o = ——————— = 0.0141.
0.01  0.50 + 0.005 .0.40 + 0.001  0.10 Answer D. 4. May 2001 Course 1 Examination, Problem No. 7, also Study Note P0908, Problem No.
104
A joint density function is given by
f( ) kx, for0<x<l, O<y<1,
x. = .
y 0, otherwrse, where k is a constant. What is Cov(X,Y)? A. — B.0 C .l D 13.3
6 9 3 l
. 6
Solution.
The joint density is a product of functions of x, fx (x) = k, x, and y, f, (y) = k2 l, where kl k2 = k, on a rectangular region, thus X and Y are independent, and their covariance must be zero. This is an important shortcut to remember: a necessary condition for independence of two
random variables is that the region where their joint density is positive is a rectangle with sides
parallel to the axes (this could be the whole plane, or a quadrant of it, or in a discrete case a
collection of points positioned in a rectangular shape, with points present in each position along
each vertical or horizontal line) and if also that joint density is a product of a function of the ﬁrst ASM Study Manual for Course PI] Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostaszewski ~ 120  \ PRACTICE EXAMINATION N0. 1
variable only and a function of the second variable only, then the two variables are independent. Answer B. 5. May 2001 Course 1 Examination, Problem No. 9, also Study Note P0908, Problem No.
13 An actuary is studying the prevalence of three health risk factors, denoted by A, B , and C, within
a population of women. For each of the three factors, the probability is 0.1 that a woman in the
population has only this risk factor (and no others). For any two of the three factors, the
probability is 0.12 that she has exactly these two risk factors (but not the other). The probability that a woman has all three risk factors, given that she has A and B, is 31);. What is the probability that a woman has none of the three risk factors, given that she does not have risk factor A?
A. 0.280 B. 0.311 C. 0.467 D. 0.484 E. 0.700 Solutiong"? .
Let A, B,‘ and C be the events of having the three health risk factors A, B , and C, respectively.
The Venn diagram below summarizes the unconditional probabilities of all mutually exclusive pieces (e.g., Pr(An BC n CC) = 0.10, Pr(A n B nCC) = 0.12, etc.) given in the problem. S 0.10 Furthermore, we are told that l:Pr(AanCAnB)=——Pr(AanC)= x .
3. Pr(AnB) x+0.12
Itfollowsthat
x=l(x+0.12)=lx+0.04.
3 3 Therefore, %x = 0.04, and x = 0.06. The probability we are looking for is ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  121  PRACTICE EXAMINATION NO. 1 We add u probabilities of mutually exclusive
pieces in mated in the Venn diagram above r—’—
l—Pr(AUBuC) 1— (30.10+30.12+0.06) 0.28
=—————=——=—=0.4667.
l—Pr(A) 1— (0.10+20.12+0.06) 0.60 w
We add u probabilities of mutually exclusive
pieces in icated in the Venn diagram above Answer C. 6. May 2001 Course 1 Examination, Problem No. 10, also Study Note P0908, Problem No.
113 Two life insurance policies, each with a death beneﬁt of 10,000 and a onetime premium of 500,
are sold to a couple, one for each person. The policies will expire at the end of the tenth year.
The probability that only the wife will survive at least ten years is 0.025, the probability that only
the husband will survive at least ten years is 0.01 , and the probability that both of them will
survive at least ten years is 0.96. What is the expected excess of premiums over claims, given
that the husband survives at least ten years? A. 350 B. 385 C. 397 D. 870 E. 897 Solution. Start by labeling the events. Let W be the event that wife survives at least 10 years, and H be the
event that husband survives at least 10 years. Let us also denote by B the amount of beneﬁt paid,
and H be the proﬁt (i.e., excess of premiums over claims) from selling policies. We have: Pr(H)= Pr(Hnw)+ Pr(H nWC) = 0.96+0.01= 0.97. Furthermore:
Pr H 0Wc 0.01
Pr(W CIH)=%=6§7=00103.
It follows that: Elan)= E((1000B)H)=1000E(BH)=
=1000(0Pr(WH)+10000PI(W‘ H))=1000—100000.0103 = 897. Answer E. 7. May 2001 Course 1 Examination, Problem No. 12, also Study. Note P0908, Problem N0.
3 You are given Pr(AUB)=0.7 and Pr(AUBC)=0.9. Determine Pr(A). ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  122  PRACTICE EXAMINATION No.1
A. 0.2 B. 0.3 C. 0.4 D. 0.6 D. 0.8 Solution.
Since Pr(A U B) = 0.7, we see that Pr((A U B)C) = 0.3. From Pr(A U BC) = 0.9 we get Pr((AUBC)C)=0.l. But (AUB)c =AC nBC and (AUBC)C =Ac OB, so that these two events, (A U B)c and (A U BC )c are mutually exclusive and their union is Ac. This gives Pr(A‘) =0.3+o.1 = 0.4 and Pr(A)= 0.6.
Answer D. 8. May 2001 Course 1 Examination, Problem No. 13, also Study Note P0908, Problem No. 41 A study is being conducted in which the health of two independent groups of ten policyholders is
being monitored over a oneyear period of time. Individual participants in the study drop out
before the end of the study with probability 0.2 (independently of the other participants). What is
the probability that at least 9 participants complete the study in one of the two groups, but not in both groups?
A. 0.096 B. 0.192 C. 0.235 D. 0.376 E. 0.469 Solution.
Deﬁne the following random variables:
X = number of group 1 participants that complete the study.
Y = number of group 2 participants that complete the study.
We are given that X and Yare independent. Therefore, Pr(({X 2 9} n {Y < 9}) U ({X < 9} n{Y .>. 9})) =
=Pr({X 29}n{Y < 9})+Pr({X<9}n{Y 29}).
Because of symmetry and independence between X and Y the above equals:
2Pr({X 2 9} n {Y < 9}) = 2Pr(X 2 9)Pr(Y < 9). Note that probabilities concerning X are same as those concerning Y. Observe that X counts the
number of successes in ten Bernoulli trials, where success is deﬁned as completion of the study
by an individual participant, so that p = 0.8, and n = 10. Based on this, the above is equal to: 2m(x 29)Pr(X< 9)=2Pr(X 29)(1Pr(X 29)): =2[[ 1: W£ :3)03'°1[1[( ‘9" Wi 10°31} = 2  0.3758  (1 — 0.3758) = 0.4692.
Answer E. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  123  ~ PRACTICE EXAMINATION NO. 1
9. May 2001 Course 1 Examination, Problem No. 14, also Study Note P0908, Problem No.
115
The stock prices of two companies at the end of any given year are modeled with random
variables X and Y that follow a distribution with joint density function 2x, for0<x<1,x<y<x+1,
f(x,y) = .
0, otherW1se.
What is the conditional variance of Y given that X = x? A.— B.1 C.x+l D. xz—l E.x2+x+1
12 6 2 6 3 Solution. 
Let fx (x) be the marginal density function of X. Then x+l fx (x) = I 2xdy = 23:34::+1 = 2x(x + 1) — 2x2 = 2x
for 0 < x < l, and‘ fx (x) = 0 otherwise. Consequently,
fx Y (x’y) {
f y X = x = —‘——— =
A I ) fx (x)
We see that the random variable (Y! X = x) is uniform on the interval (x,x +1), and therefore its . . . . l .
mean 18 x + 2’ and its variance is 1—2. You could calculate those parameters usmg calculus, but 1, ifx<y<x+1,
0, otherwise. you should not. You should know key properties of the uniform distribution.
Answer A. 10. May 2001 Course 1 Examination, Problem No. 17, also Study Note P0908, Problem
An auto insurance company insures an automobile worth 15000 for one year under a policy 'with
a 1000 deductible. During the policy year there is a 0.04 chance of partial damage to the car and
a 0.02 chance of a total loss of the car. If there is partial damage to the car, the amount X of
damage (in thousands) follows a distribution with a density function a
f(x)={0.5003e , for0<x<15, 0, otherwise.
What is the expected claim payment? A. 320 B. 328 C. 352 D. 380 E. 540 Solution. ‘. ' ., . “
Let Ydenote the claim payment made by the insurance company, in thousands. As'the‘ﬁiilicy
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof OstaszeWSki  124 . PRACTICE EXAMINATION NO. 1
has a deductible of 1 (thousand), the claim payment is 0, when X <1,with probability 0.94, . .
Y = X—l, when l S X <15,note that Pr(l S X < 15) =0.04, WX.
14, when X 2 15,with probability 0.02.
Therefore, the expected value ofs this random variable is calculated as follows: E(Y)= 0.94 o+o.04 “iIOHm x1)e 2ax+o.0214= — __ '3‘
=0020012'[:1f Ixegdx_ TedeJ‘I'OZS: u—x v—~ 28x M—J _
du=dx dv=e 2die
=0.04 05003
ﬁ—w———l
Integration by pans in the ﬁrst integral ”0020012 —2xe ‘2‘ :s+2je Zarx— J'e ;dx]+0.28= —30e 75 + 2605+ +[2e2] % =0 “020012 = 0.020012 (—30e'75 + 2e'05 — 2e 75 + 2e’05 ) + o .28~ 0 32819738. Since this is E (Y) expressed m thousands, the expected claim payment is approximately 328.
Answer B. = 0.020012 [—30e'” + 2e'05 + je" 2 dx] +0 28— _ x x=l5
' J+028= x=1 11. May 2001 Course 1 Examination, Problem No. 19, also Study Note P0908, Problem
No. 83 A company manufactures a brand of light bulb with a lifetime in months that is normally
distributed with mean 3 and variance 1. A consumer buys a number of these bulbs with the
intention of replacing them successively as they burn out. The light bulbs have independent
lifetimes. What is the smallest number of bulbs to be purchased so that the succession of light
bulbs produces light for at least 40 months with probability at least 0.9772? A. 14 B. 16 C. 20 D. 40 D. 55 Solution.
Let X,,...,X,, denote the life‘ spans of the n light bulbs purchased. Since these random variables are independent and normally disuibuted with mean 3 and variance 1, the random variable (3,. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostaszewski  125  PRACTICE EXAMINATION NO. 1 .
S = XI + ...+ X” is also normally distributed with mean it = 3n and standard deviation J 0' = Jn = J5. We want to choose the smallest value for n such that 0.9772 s Pr(S > 40) =Pr[s_ 3" 40" 3"]. T>Jz We know that 53—3" is standard normal, because S — N (3n,n). Let Z denote a standard normal
n
random variable. Since Pr(Z > —2) = 0.9772, n should satisfy the following inequality:
40 — 3n < _2 ‘
J; 0
To ﬁnd such an n, we solve the corresponding equation for n: 40f“ = —2,resulting in n = 16.
n Answer B. ~ 12. May 2001 Course 1 Examination, Problem No. 20, also Study Note P0908, Problem
A No. 46 ‘ A device that continuously measures and records seismic activity is placed in a remote region.
The time, T, to failure of this device is exponentially distributed with mean 3 years. Since the
device will not be monitored during its first two years of service, the time to discovery of its failure is X = max(T,2). Determine E(X). 2 4 g
A. 2+:e'6 B.2—2e3+5e3 C.3 D.2+3e3 E.5
Solution. _1 I The survival function of Tis s, (t): e 3 for 0 < t < co, while its PDF IS s, (t)  3 —e 3 for 0 < t < co, and T is nonnegative with probability one. Using the Darth Vader Rule for
E(max(T,2)), we obtain +oo w : = I 2
E(max(r,2))= 2+ js,(:)dt = 2+ je‘idt = 2+ 3I§e_3dt = 2+ 3e—3.
2 2 2 =Pf(T >2) AnswerD. E(7MM7'Z)>; E(T1V’)+Ei°1/7<>>
:Ltw 362751?” 5? PM”) 13. May 2001 Course 1 Examination, Problem No. 22, also Study Note P09 08, Problem
No.88 The waiting time for the ﬁrst claim from a good driver and the waiting time for the ﬁrst claim
from a bad driver are independent and follow exponential distributions with means 6 years and 3 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  126  3") PRACTICE EXAMINATION NO. 1
years, respectively. What is the probability that the ﬁrst claim from a good driver will be ﬁled within 3 years and the ﬁrst claim from a bad driver will be ﬁled within 2 years? 1 1 1 1 1 1 2 1 l
A.—l—e3—e2+e‘ B.—e° C.1e3—e2+e6
18 18
2 1 1 _1 1 1 1
D.le3—e3+e3 E.l——e3e2+—e6
6 18
Solution. Let X denote the waiting time for a ﬁrst claim from a good driver, and let Y denote the waiting time for a ﬁrst claim from a bad driver. The respective cumulative distribution functions for X
x y and Yare Fx(x)=1—e'3 for x>0, and my)=1e'3 for y>0. Therefore,
Pr({X s 3}n{Y s 2}) = Pr(X s 3)Pr(Y s 2) = Fx(3)F,,(2) = 1 2 2 1 1
=[le 2][l—e 3)=le 3e 2+e 5. 14. May 2001 Course 1 Examination, Problem No. 23‘, also Study Note P0908, Problem
No. 28 A hospital receives 1/5 of its ﬂu vaccine shipments from Company X and the remainder of its
shipments from other companies. Each shipment contains a very large number of vaccine vials.
For Company X’s shipments, 10% of the vials are ineffective. For every other company, 2% of Answer C. the vials are ineffective. The hospital tests 30 randomly selected vials from a shipment and ﬁnds ; . that one vial is ineffective. What is the probability that this shipment came from Company X?
A. 0.10 B.O.14 C.0.37 D.0.63 E.0.86 Solution. As always, we start by labeling the events. Let X be the event that shipment came from Company
X, and F be the event that exactly one of the vaccine vials tested is ineffective. By applying the
Bayes Theorem we get: Pr(FX)Pr(X) MXIF): Pr(FX)Pr(X)+Pr(FXC)Pr(XC)' Now we use the data given in the problem. We have Pr(X) = g, and therefore Pr(Xc) = %. Furthermore
Pr(FX)=[ 310 J0.100.9029=0.141 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  127  PRACTICE EXAMINATION NO. 1
and Pr(FIXC)=[ 31° ].o.020.98”=o.334. Substituting these values into the Bayes formula, we get 0.141l Pr(XF) = —15—4 = 0.096.
0'141‘§+0'334'§ Answer A. 15. May 2001 Course 1 Examination, Problem No. 24, also Study Note P0908, Problem No. 79
A device contains two components. The device fails if either component fails. The joint density function of the lifetimes of the components, measured in hours, is f (s,t) , where 0 < s < 1 and
0 < t < 1. What is the probability that the device fails during the first half hour of operation? 0505 10.5 I l
A. I Jf(s,t)ds dt B. ij(s,t)ds dt C. I Jf(s,t)ds dt
0 0 0 0 0.50.5
o. fIf(s,t)ds dt+ Iff(s,t)ds dt E. ff f(s,t)ds dt+j°ff(s,t)ds a:
0 D 0 0 0 0.5 0 0
Solution. Always begin by drawing a graph indicating the areas where the density is positive. In this case
that region where the density is positive is the unit square. The shaded region is the portion of the
domain of the density over which the device fails sometime during the ﬁrst half hour. We
divided it into two subregions: A and B, as indicated in the ﬁgure. l S ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20044008 by Krzysztof Ostaszewski  128  J PRACTICE EXAMINATION NO. 1 The probability sought is l l
l '2' E l
Pr[{SS ﬂu {T < 3): Hm.) t )(dsdHHf s, r )dsdt,
o o o 1
2
where the ﬁrst integral is over the subregion indicated 1n the ﬁgure as A and the second integral is over the subregion indicated in the ﬁgure as B.
Answer E. 16. May 2001 Course 1 Examination, Problem No. 26, also Study Note P0908, Problem No. 109 A company offers earthquake insurance. Annual premiums are modeled by an exponential
random variable with mean 2. Annual claims are modeled by an exponential random variable
with mean 1. Premiums and claims are independent. Let X denote the ratio of claims to
premiums. What is the density function of X? A. l B. #2 c. e" D. 29'” E. xe"
211: +1 (2x + 1)
Solution. Let U be the annual claims and let Vbe the annual premiums (also random), and let fay (u,v) be the joint density of them. Furthermore, let fx be the density of X and let Fx be its cumulative
distribution function. We are given that U and V are independent, and hence 1 1 1 fu'v(u,v)=e'" 3e 2 =Ee'“e 2 ‘ for 0 < u < no.0 < v < co. Also, noting the graph below, we have: ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  129  on V1 Fx(x)=Pr(XSx)=Pr[%—Sx)= "EV! ‘5‘ 1e "e 2dudv=I
2 1w
0 2 ll
°'—.3 14:0 0 ll
°'—»3 Finally, x(x)= F (10’ (2x+ 1) Answer B. This problem can also be done with the use of bivariate transformation. We will now
give an alternative solution using that approach. Consider the following transformation 22:9, y=v.
V Then the inverse transformation is
U = XY, V = Y.
We know that f,,.v(u.v)=§eue‘E for u > 0, v > 0. It follows that
3(u, v) fx.y (LY) = fay ("(x’Y)’v(xvY)) a(x’y)
for xy > 0 and y > 0, o...
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