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Unformatted text preview: PRACTICE EXAMINATION NO. 2
PRACTICE EXAMINATION NUMBER 2
SOLUTIONS 1. November 2000 Course 1 Examination, Problem No. 27, also Study Note P0908, Problem No. 98
Let Xl ,X2 ,X3 be a random sample from a discrete distribution with probability function , forx=0, , forx=1, , otherwise. Determine the moment generating function M (t) of Y = X,X2X3. 19 s 1 2 3 1 8 3, 1 2,,
.—+—' B.l+2’ c. —+—' D.—+—e E.—+—e
27 27e e [3 39] 27 27 3
Solution. Let f, be the probability function for Y = X1X2X3. Note that Y = 1 if and only if
XI = X2 = X3 =1. Otherwise, Y: 0. We also know that Pr(Y= 1)= Pr({xl =1}n{X2 =1}n{x3 =1})= 2 3 8
= Pr(Xl =1)Pr(X2 =1)1>r(x3 = 1) = [3] = .27,
We conclude that
%, fory=0,
—, fory=1,
27
and is zero otherwise. Y is a Bernoulli Trial random variable and
19 8
Mt =M 1=E ‘Y =—+— ',
() r() (e ) 27 273
AnswerA. 2. November 2000 Course 1 Examination, Problem No. 28, also Study Note P0908, Problem No. 12 A doctor is studying the relationship between blood pressure and heartbeat abnormalities in her patients. She tests a random sample of her patients and notes their blood pressures (high, low, or
(WV normal) and their heartbeats (regular or irregular). She finds that: ASM Study Manual for Course PI] Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  149  PRACTICE EXAMINATION N0. 2
(i) 14% have high blood pressure. (ii) 22% have low blood pressure. '
(iii) 15% have an irregular heartbeat.
(iv) Of those with an irregular heartbeat, onethird have high blood pressure.
(v) Of those with normal blood pressure, oneeighth have an irregular heartbeat.
What portion of the patients selected have a regular heartbeat and low blood pressure? A. 2% B. 5% C. 8% D. 9% E. 20%
Solution.
We are directly given that Pr(Irregular heartbeat) = 0.15,
Pr(High blood pressure) = 0.14, and
Pr(Low blood pressure) = 0.22. We calculate other relevant probabilities and place all probabilities in a table (below):
Pr(Normal blood pressure) = l— 0.14 — 0.22 = 0.64, Pr ({High blood pressure} A {Irregular heartbeat}) = = Pr(High blood pressure] Irregular heartbeat)  Pr(lrregular heartbeat) = g  0.15 = 0.05, Pr({Irregular heartbeat} m {Normal blood pressure}) = = Pr(Irregular heartbeat Normal blood pressure) o Pr(Normal blood pressure) = = 10.64 =0.08,
8 Pr (Regular heartbeat) = 1— 0.15 = 0.85,
Pr({High blood pressure} n {Regular heartbeat}) = 0.14 — 0.05 = 0.09, Pr({Normal blood pressure} 0 {Regular heartbeat}) = 0.64  0.08 = 0.56,
Pr({Low blood pressure} n {Irregular heartbeat}) = 0.15 — 0.05  0.08 = 0.02,
Pr({Low blood pressure} n {Regular heartbeat}) = 0.22 — 0.02 = 0.20, m We see from this table or from the last calculation that 20% of patients have a regular heartbeat
and low blood pressure.
Answer E. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20044008 by Krzysztof Ostaszewski  150  PRACTICE EXAMINATION N0. 2
3. November 2000 Course 1 Examination, Problem No. 30, also Study Note P0908, Problem No. 11
An actuary studying the insurance preferences of automobile owners makes the following conclusions:
(i) An automobile owner is twice as likely to purchase collision coverage as disability coverage.
(ii) The event that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage.
(iii) The probability that an automobile owner purchases both collision and disability coverages is 0.15 .
What is the probability that an automobile owner purchases neither collision nor disability coverage?
A. 0.18 B. 0.33 C. 0.48 D. 0.67 E. 0.82 Solution.
We begin by labeling the events. Let C be the event that a policyholder buys collision coverage, and let D be the event that a policyholder buys disability coverage. We are given that
Pr(C) = 2Pr(D) and Pr(C n D) = 0.15. Using independence of C and D, we see that 0.15 = Pr(C n D) = Pr(C)Pr(D) = 2Pr(D)Pr(D) = 2(1’r(D))2
and therefore (Pr(D))2 = g = 0.075. From that Pr(D) = , and Pr(C) = 2%. The problem asks for
Pr((C u D)C) = Pr(CC 0 DC).
Independence of C and D also implies the independence of CC and DC. As a result, we have
Pr(Cc nDc) = Pr(CC)  19(06): (l—Pr(C))(1 — Pr(D)) =
=(1—2W)(1W)=0.3284. We could also calculate the answer as
Pr((Cu D)C) =1Pr(CU D)=1—(Pr(C)+Pr(D)— Pr(CnD)) = = 1 (240.075 + J0.075 — 0.15) = 1.15  340.075 = 0.3284. Answer B. 4. November 2000 Course 1 Examination, Problem No. 32, also Study Note P0908,
Problem No. 75 The monthly proﬁt of Company I can be modeled by a continuous random variable with density
function f. Company 11 has a monthly proﬁt that is twice that of Company I. Detemiine the
probability density function of the monthly profit of Company 11. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  151  PRACTICE EXAMINATION N O. 2 A. B.f(§] C. 24%] D. 2f(x) E.2f(2x) Solution.
Let X and Y be the monthly proﬁts of Company I and Company H, respectively. Let us write fx
for the PDF of X, instead of just f. Then Y = 2X, so that X = éY. Therefore
dx 1 l
= x  — =  .
my) f.( (a) dy f.[2y] 2 You can, of course, also use the CDF approach. In this case
5(y)= Pr(Y Sy)=Pr(2XSy)=Pr[XS%)= Fx(%], and f.(y)=F;(y)=in[Z)=lF;[l]=§f.[z). dy 2 2 2 2
Answer A. 5. November 2000 Course 1 Examination, Problem No. 34, also Study Note P0908,
Problem No. 29 The number of days that elapse between the beginning of a calendar year and the moment a high
risk driver is involved in an accident is exponentially distributed. An insurance company expects
that 30% of highrisk drivers will be involved in an accident during the ﬁrst 50 days of a
calendar year. What portion of highrisk drivers are expected to be involved in an accident
during the first 80 days of a calendar year? A. 0.15 B. 0.34 C. 0.43 D.0.57 E. 0.66 Solution.
Let The the number of days that elapse before a highrisk driver is involved in an accident. We
know that T is exponentially distributed with unknown parameter A. We are also given that 0.3 = Pr(T s 50)=1—e"°‘. ln0.7 Therefore, e"°" = 0.7 and 2. = — . It follows that Q Pr(T s 80) = 1 e’m =1— e50'1"°'7 =1— 0.7‘6 = 0.4349.
Answer C. 6. November 2000 Course 1 Examination, Problem No. 36, also Study Note P0908, Problem No. 91
An insurance company insures a large number of drivers. Let X be the random variable ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  152  PRACTICE EXAMINATION N0. 2
representing the company’s losses under collision insurance, and let Yrepresent the company’s
losses under liability insurance. X and Y have joint density function 2x+2—y
f000= , for0<x<land0<y<2, 0. otherwise.
What is the probability that the total loss is at least 1? A. 0.33 B. 0.38 C.0.4l D.0.71 E. 0.75 Solution.
We want to ﬁnd Pr(X + Y > 1). We start by drawing a graph of the area where the density of this random vector “plays.” The probability is calculated as an integral of the joint density over the
region indicated with dotted lines in the ﬁgure below. Therefore ‘2 2x+2— ‘ 1 1 1 '
Pr(X+Y>1)=IJl7lldydx=llliw+5y7y2l
01: l
=J[x+l—l—lx(l—x)—%(l—x)+l(lx)2] =
0
l x=l
=“if+3x+ljdx=(ix3+2x2+lx) =1 2 l=ﬂ
o 8 8 24 8 =0 24 8 8 24 Answer D. 7. November 2000 Course 1 Examination, Problem No. 38, also Study Note P0908,
Problem No. 101
The proﬁt for a new product is given by Z = 3X  Y — 5. X and Y are independent random variables with Var(X) = l and Var(Y) = 2. What is the variance of Z? ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  153  PRACTICE EXAMINATION NO. 2
A.l B.5 C.7 D.ll E. 16 Solution.
Due to independence of X and Y Var(Z) = Var(3X—Y —5) = Var(3X)+Var(Y) = 32Var(X)+Var(Y) = 91+ 2 =11.
AnswerD. 8. November 2000 Course 1 Examination, Problem No. 40, also Study Note P0908,
Problem No. 122 A device contains two circuits. The second circuit is a backup for the ﬁrst, so the second is used
only when the ﬁrst has failed. The device fails when and only when the second circuit fails. Let
X and Y be the times at which the ﬁrst and second circuits fail, respectively. X and Y have joint
probability density ﬁmction
6 " '2’ , < < < oo,
f(x’y)={ e e forO .x y
0, othermse. What is the expected time at which the device fails? A. 0.33 B. 0.50 C. 0.67 D. 0.83 E. 1.50 Solution.
Note that the joint density is positive only when 0 < x < y. Therefore, the marginal density of Y is calculated as
y 3’
fr (3’) = Ilsexe'z’dx = 6e'2’ Ie‘QLx = 6e'2’e" + 6e"2y = 6e'2’ — 6e'”
0 0
for 0 < y < co. Given this, E(Y) = Tyfr (y)dy = T(6ye2y " 6ye"’)dy = 6 °° 6 °° 6 1 6 1 3 2 5
=_ ,2 2yd __ .3 3.vd = _____ ._=___=_=0.83.
2 {y e y 3 {y e y 2 2 3 3 2 3 6 ﬁr—J ‘—w—’
Mean of an ex nemial Mean of an ex nential
random vnriab c with random varinb c with
hazard late 2 hazard rate 3 Answer D. 9. May 2000 Course 1 Examination, Problem No. 1, also Study Note P0908, Problem No. 2 The probability that a visit to a primary care physician’s (PCP) ofﬁce results in neither lab work
nor referral to a specialist is 35%. Of those coming to a PCP’s ofﬁce, 30% are referred to
specialists and 40% require lab work. Determine the probability that a visit to a PCP’s ofﬁce
results in both lab work and referral to a specialist. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostaszewski  154  PRACTICE EXAMINATION NO. 2
A. 0.05 B. 0.12 C. 0.18 D. 0.25 E. 0.35 Solution.
We deﬁne L as the event of lab work needed, and R as the event of a referral to a specialist needed. We are given that Pr(R) = 0.3, Pr(L) = 0.4, and
Pr((Lu 1a)“) = Pr(LC n R‘) = 0.35. It follows that
Pr(Lu R) = 1 P((LUR)C)= 0.65. Finally, ﬁom the formula
Pr(L u R) = Pr(L)+Pr(R)— Pr(L n R), we get
Pr(LnR)= Pr(L)+Pr(R)—Pr(LuR)= 0.3+ 0.40.65 =0.05. Answer A. 10. May 2000 Course 1 Examination, Problem No. 2, also Study Note P0908, Problem No. 27
A study of automobile accidents roduced the followin ; data: I odel year Proportion of all Probability of involvement
vehicles in an accident 1997 m
1998 m [EE—
m_ An automobile from one of the model years 1997, 1998, and 1999 was involved in an accident.
Determine the probability that the model year of this automobile is 1997. A. 0.22 B. 0.30 C. 0.33 D. 0.45 E. 0.50 Solution.
Let us start by labeling the events:
E97  The model year is 1997, 1!?93  The model year is 1998,
E99  The model year is 1999,
E0  The model year is other than 1997, 1998 or 1999, A  The car is involved in an accident.
We are given Pr(E9.,)=0.l6, Pr(59,)=0.18. Pr(E..,)=o.20, Pr(Eo)=O.46,
Pr(A[ 159,) = 0.05, Pr(A E93) = 0.02, Pr(AE99) = 0.03, Pr(AEo) = 0.04.
It might seem that this is a Bayes’ Theorem problem, but note that the accident happened to a ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Krzysztol' Ostaszewski  155  PRACTICE EXAMINATION N0. 2
1997, 1998, or a 1999 model, but not one of the other years’ model, so that not all of the pieces of the partition {E97 ,E93 ,E99 ,Eo} are considered, and we can’t use Bayes directly. We are i actually asked to ﬁnd this conditional probability:
Pr(E97lAn(E97 U E93 U E99))'
Using the deﬁnition of conditional probability
M159, n(A n (E,7 u 13,, u E,,))) Pr(E97Aﬂ(E97UE98UE99))= Pr(An(E UE UE 97 98 99 Pr[A n E97 n [E97 u E93 u E99
____—J Mutually exclusive Mutually exclusive
_ Pr(AnE97)
Pr(AnE,,)+Pr(AnE,,)+Pr(AnE,,) Pr(AE97) ' Pr(E97)+ Pr(AE9s)'Pr(E9s)+ FAME») ' Pr(E99) = 0.050.16 = 0.008 20.4545. 0.05~0.16+0.020.18+0.030.20 0.0176 ' We could also do this problem by drawing a probability tree, and using only a portion of it, excluding other years’ models: Consider only
these three
possibilities, N0 acoldent instead of full
D Bayes
Acoldent Theorem,
setup No accident Accident No accident
Answer D. “5
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof OstaszeWSki  156  "w PRACTICE EXAMINATION N0. 2
11. May 2000 Course 1 Examination, Problem No. 3, also Study Note P0908, Problem No.
37
The lifetime of a printer costing 200 is exponentially distributed with mean 2 years. The
manufacturer agrees to pay a full refund to a buyer if the printer fails during the ﬁrst year
following its purchase, and a onehalf refund if it fails during the second year. If the
manufacturer sells 100 printers, how much should it expect to pay in refunds? A. 6,321 B. 7,358 C. 7,869 D. 10,256 E. 12,642 Solution Z The exponential distribution with a mean of 2 has density fx (x) = 5e 2 , for x > 0, and the l
x cumulative distribution function Fx (1:) = l — e 2 , for x > 0. The probability that a printer will
fail in the ﬁrst year is l
Pr(X s 1) = Fx(1)=1— e 2 = 0.39347,
so that the expected number of failures in the ﬁrst year out of 100 printers is 39.347. The probability that a printer will fail in the second year is
2 l __
Pr(l < X: 2)=Fx(2)Fx(1)= sx(1)sx(2)= e 2 —e 2 =0.23865,
so that the expected number of failures in the second year out of 100 printers is 23.865. The
expected amount the manufacturer will pay in refunds is 200  39.347 + 100 ~ 23.865 = 10256.
Answer D. 12. May 2000 Course 1 Examination, Problem No. 4, also Study Note P0908, Problem No.
74 Let T denote the time in minutes for a customer service representative to respond to 10 telephone
inquiries. T is uniformly distributed on the interval with endpoints 8 minutes and 12 minutes.
Let R denote the average rate, in customers per minute, at which the representative responds to
inquiries. Which of the following is the density function of the random variable R on the interval l2 8 12 5 51n(r) 10 5
A.— B. 3— C.3  D. E.— 5 2r r 2 r2 2r2
Solution. Since Thas a uniform distribution on the interval from 8 to 12, its PDF is f, (t) =% for ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  157   SSI  msmazsmso 301255291 Kq soozvooz luﬁuﬁdoa ©'“°!19“!“‘3¥El lam!an [Id 931:10:) 10.1 [mum Rpms wsv 01 [12an pm; uogﬁax 91.11 .1er wusuoo s! (‘1‘ ‘1)! uopoun; ﬁnsqu mgog‘ __..——__——____—__ ——____..—___——_ 'soun panop {mm paxrem Sugaq uogﬁox 12m
‘11 pm ‘1 go uopnqynsgp zugof 9m .10; Knsuop omz—uou J0 uogﬁeu am samogpu; molaq emﬁg sq; 'uonnlos
7 a l— a i '3 3 a — v
‘ {19 {1z {I {I .21 pure {L J0 5912an sq: JO was 9111 go 9mm p9139dx9 an; aunmmaq jumsuoo aAmsod e 317 91mm
7 s ‘1 s ‘1 s 0 Sq paugap LIOIBQI 9m JaAo unogyun s; {L pun [L .10; uogoun; Kusuap 1x110! sq;
'aoyxap oguonoap m: u; swauodmoo paxuu om J0 smoq u; sawpam 9m masoldel {L pue [L 191 L6
'ON mammd ‘sosod 91°51 ﬁrms 0818 ‘5 ‘ON memoId ‘uoneuyumxa I asmoo oooz Raw '91 '3 .IOMSUV
.Z'lz [I1 J 17 JP .1 y
——= ———=—.11 =.l ‘aloploql '1: I sagde 5—: g ‘osw 'asgmamo 0 = (1)1] put; ‘31 s 15 g OI 01 Z 'ON NOILVNIWVXH HDILOVHJ PRACTICE EXAMINATION NO. 2
l l 2 Areaofregion = 1.1} — L2
2
Therefore, the expected value is L" 2 2
“(tf + 133mm, = Z;
0 0 I012 +t§)dtl]dt2 =
0 ‘ .‘ Answer C. 14. May 2000 Course 1 Examination, Problem No. 6, also Study Note P0908, Problem No. 18 Two instruments are used to measure the height, h, of a tower. The error made by the less
accurate instrument is normally distributed with mean 0 and standard deviation 0.0056h. The
error made by the more accurate instrument is normally distributed with mean 0 and standard
deviation 0.0044h. Assuming the two measurements are independent random variables, what is
the probability that their average value is within 0.005h of the height of the tower? . A. 0.38 B.0.47 C. 0.68 D. 0.84 E. 0.90 Solution.
Let us denote by El and E2 the errors of the first and second instrument, respectively. We are given that 15(5.) = 15(52): 0, Var(E, ) = (0.0056h)2, and Var(E,) = (0.0044h)2. We also
know that E, +132 is normal with mean zero, and since El and E2 are independent,
Var(E, + 5,) = Var(E,)+ Var(E,) = (0.005%)2 +(o.oo44h)2 = (0.00712h)2. The average of El and E2 is a random variable A = %(E1 + E2), which is also normal with mean 0 and variance
Var(A) = ivmzz, + 15,) = (0.00356h)2. The average height is within 0.005h of the height of the tower if the absolute error in the average
is less than 0.005h. Thus, we want to ﬁnd Pr(A < 0005):) = Pr(0.005h < A < 0.005h).
We standardize A to get Pr(—0.005h < A < 0.005h) = r M < Varm‘ varm) =Pr(—0.005h—0 Z 0.005h0 P[—0.005hE(A) AE(A) 0.005h—E(A)] < < —
0.00356h 0.0035611
ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  159  )=Pr(—1.4<Z<l.4), PRACTICE EXAMINATION NO. 2 , .
where Z is standard normal. From the standard normal table, Pr(Z < 1.4)= 0.9192, so that Pr(Z < —1.4) = Pr(Z > 1.4) = 1— Pr(Z $1.4): 0.0808
and ﬁnally Pr(1.4 < Z <1.4)= 0.9192 — 0.0808 = 0.8384.
Answer D. 15. May 2000 Course 1 Examination, Problem No. 7, also Study Note P0908, Problem No.
55 An insurance company’s monthly claims are modeled by a continuous, positive random variable
X , whose probability density function is proportional to (1+ x)" , where 0 < x < oo. Determine the company’s expected monthly claims. 1 l 1 A.  B.  C. — D. 1 E. 3
6 3 2
Solution. The PDF is of the form fx (x) = c(1+ x)'4 where c is an unknown positive constant, for x > 0, and fx (x) = 0 for x S 0. We must have .. xw: 1=£e(1+x)“¢x=[_is(1+x)"] This gives C: 3 and fx (x): 3(11x)’4 for x >0, fx (x): 0 for x 50. Hence, SAX): Tf(t)dt = +f3(l+t)"a1t =(—(1+t)'3) X is a random variable, which is nonnegative almost surely, and therefore
x)°°
l 2, x=0 r—m = (1+x)—3. (=3 +00 E(X) = js, (x)dx = j (1 +x)"¢x = Ema1]
o o 
You could also see that this a Pareto distribution, of the form a0“
fx 0‘)  W: with 05:3 and 9:1, sothat x=0 16. May 2000 Course 1 Examination, Problem No. 8, also Study Note P0908, Problem No. 64 ,
A probability distribution of the claim sizes for an auto insurance policy is given in the table ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostaszewski  160  PRACTICE EXAMINATION NO. 2
below:
Claim Probability
Size
20 0.15 0 .05 What percentage of the claims are within one standard deviation of the mean claim size?
A. 45% B. 55% C. 68% D. 85% E. 100% Solution.
The mean claim size is: E(X)= 200.15+30~0.1+400.05+500.20+600.10+700.10+800.30 =55.
The second moment of the claim size is: E(X2)= 202 0.15 + 302 0.1+4o’ o.05 +502 0.20+ + 602 0.10 + 702 0.10 + 802 030 = 3500.
Therefore, the variance of the claim size is:
Var(X) = E(X2)— (E(X))2 = 3500 — 552 = 475.
The standard deviation is .[Var(x) = Jﬁ = 21.79. The claim sizes within one standard deviation of the mean claim size of 55 are those claim sizes
between 55 — 21.79 = 33.21 and 55 + 21.79 = 76.79. Those claim sizes are 40, 50, 60 and 70.
The total probability of those claim sizes is 0.05 + 0.2 + 0.1+ 0.1: 0.45.
Answer A. 17. May 2000 Course 1 Examination, Problem No. 9, also Study Note P0908, Problem No.
85 The total claim amount for a health insurance policy follows a distribution with density function
1 L
f (x) = We '°°° forx > 0. The premium for the policy is set at 100 over the expected total claim amount. If 100 policies are sold, what is the approximate probability that the insurance
company will have claims exceeding the premiums collected? A. 0.001 B. 0.159 C. 0.333 D. 0.407 E. 0.460 Solution.
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  161  PRACTICE EXAMINATION N0. 2
While the problem does not say so directly, you are expected to understand that the 100 policies
are independent. This is the PDF of an exponential distribution with a mean of 1000. The expected claim per policy is 1000, and the variance is 10002. The premium collected is 1100 per
policy. For 100 policies, a total of 110,000 is collected in premium. The total claim is W=x,+x,+~X,om and
E(W)=100E(X)=100,000,
while, due to independence of the 100 policies,
Var(W) = 100Var(X) = 100  10002.
By the Central Limit Theorem W has an approximately normal distribution. Thus (as always, Z
denotes a standard normal random variable) Pr(W>110,000)=pr[W_‘§(l)>Mﬂl]= JVar W) Var(W)
110,000 — 100,000 .P.[z>
410010002 J: Pr(Z >1)=l—0.8413= 0.1587. Answer B. 18. May 2000 Course 1 Examination, Problem No. 10, also Study Note P0908, Problem No. 90
An insurance company sells two types of auto insurance policies: Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean two days. The
time until the next Deluxe Policy claim is an independent exponential random variable with
mean three days. What is the probability that the next claim will be a Deluxe Policy claim? A. 0.172 B. 0.223 C. 0.400 D. 0.487 E. 0.500 Solution.
Let TB be the random waiting time till next Basic policy claim, and TD be the waiting time till next Deluxe policy claim. We know that TB has the exponential distribution with mean 2, and thus its PDF is f3 (s) = £27. Similarly, TD has the exponential distribution with mean 3 and its I
PDF is f0 (t) = ge—R Since TB and TD are independent, their joint density is ms.»=f.(s)~f.(r)=[ge‘i][gaé} Therefore, ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  162  PRACTICE EXAMINATION NO. 2
You can also argue intuitively as follows. In the next 6 days we expect 3 Basic claims (one every 2 days) and 2 Deluxe claims (one every 3 days). Thus, of the next 5 claims, there is a :—= 0.4 chance that it is from a Deluxe policy on average.
Answer C. 19. May 2000 Course 1 Examination, Problem No. 11, also Study Note P0908, Problem No. 112
A company offers a basic life insurance policy to its employees, as well as a supplemental life insurance policy. To purchase the supplemental policy, an employee must ﬁrst purchase the
basic policy. Let X denote the proportion of employees who purchase the basic policy, and Y the
proportion of employees who purchase the supplemental policy. Let X and Y have the joint density function f (x, y) = 2(x + y) on the region where the density is positive. Given that 10% of the employees buy the basic policy, what is the probability that fewer than 5% buy the
supplemental policy? A.0.010 B. 0.013 C.0.108 D. 0.417 E. 0.500 Solution.
We must have Y S X, as no more than proportion X buy the supplementary policy. The region of positive joint density is the triangular region 0 S y S x S1 (below the line y = x, inside the unit
square). We want to ﬁnd the conditional probability Pr(Y < 0.05X = 0.1).
The conditional density for Y given X = 0.1 is , X=0.1 = —' where fx (0.1) is the marginal density ofX at 0.1. We ﬁnd it as (note that 0 S y S x = 0.10 in
the integral) ! 0.l fx (01)= If” (01.y)dy = j2(0.1+ y)dy= 0.03. 0 Therefore,
01. 2 0.1+
fy(yX=0.l)=M=(_Y),
fx(0'1) 0.03
and
0052 , 2 >=ons
Pr(Y<0‘05X=0'1)= i—L l+y)dy=2—0O.05+ my =0.4l67.
0 0.03 3 F0 Answer D. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  163  PRACTICE EXAMINATION N O. 2
20. May 2000 Course 1 Examination, Problem No. 18, Problem No. 19, also Study Note P 0908, Problem No. 68 An insurance policy reimburses dental expense, X, up to a maximum beneﬁt of 250. The
probability density function for X is: ﬁx) _ ce'm“, forx 2 0,
0, otherwise,
where c is a constant. Calculate the median beneﬁt for this policy. A. 161 B. 165 C. 173 D. 182 E. 250 Solution.
We can see immediately that the form of the density is exactly as for the exponential distribution, and this requires that c = 0.004. Let R be the reimbursed amount and X be the actual expense.
Then R = {X, X S 250,
250, X > 250.
The median beneﬁt is the amount c such that Pr(R S c) = 0.5. The median of the exponential
distribution is alwaysan times its mean. Since
Pr(X > 250) = e'°‘°°“25° = e‘l < 0.50, “‘2 = 173.29.
04 the median of R is the same as the median of X, i.e., Answer C. 21. May 2000 Course 1 Examination, Problem No. 19, also Study Note P0908, Problem
No. 87 In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The
difference between the true age and the rounded age is assumed to be uniformly distributed on
the interval from —2.5 years to 2.5 years. The healthcare data are based on a random sample of
48 people. What is the approximate probability that the mean of the rounded ages is within 0.25
years of the mean of the true ages? A. 0.14 B. 0.38 C. 0.57 D. 0.77 E. 0.88 Solution.
For any given rounded age X, ,its error E, is uniform between —2.5 and 2.5. Therefore, E(E,.)=o and 48
The total error in 48 independent rounded ages 1S 2E" which has a mean of 0, and variance ASM Study Manual for Course Pl] Actuarial Examination. © Copyright 20044008 by Knysztof Ostaszewski  164  PRACTICE EXAMINATION NO. 2 48 48  g = 100. The mean error of the 48 rounded ages, E = aIEZE, has expected value 0 and
i=1 variance
48
V E.
Var(E)  ———ar[§ ') — ﬂ). — 482 482 48 The question asked in the problem is equivalent to asking for the probability of the mean error being within 0.25 years from 0. Using the normal approximation for the distribution of E , as
justiﬁed by the Central Limit Theorem, we obtain h(§ < 0.25) = Pr(—0.25 < E < 0.25) = =Pr _'___<___<'—— =Pr(—l.2<Z<l.2),. 48
where Z is standard normal. Using the data from the standard normal distribition table,
Pr(l.2<Z <1.2)=Pr(Z <l.2)Pr(Z s1.2)=Pr(Z< l.2)—Pr(Z >1.2)=
= 2Pr(Z < 1.2)— l = 2 0.8849  l = 0.7698.
Answer D. 22. May 2000 Course 1 Examination, Problem No. 20, also Study Note P0908, Problem No. 106
Let X and Y denote the values of two stocks at the end of a ﬁveyear period. X is uniformly distributed on the interval [0, 12]. Given X = x, Y is uniformly distributed on the interval
(0,x). Determine Cov(X,Y) according to this model. A.0 B.4 C.6 D.12 E.24 Solution. 2
First, E(X)= 6 and Var(X)=%= 12, ains uniform on [0, 12]. Furthermore, 15(X2 ) = Var(X) + (1r5(x))2 = 48. Since conditional Y given X = x is uniform on [0, x], we have E(Y IX) = We conclude that E(Y) = Ex (E. (YX)) = Ex = 3, and 2 2
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  165  E(XY) = Ex (E, (XYX)) = Ex (XE, (YX)) = Ex = ﬂ = 24. PRACTICE EXAMINATION NO. 2
Note that in the expression E, (XYIX) , X is given, so it can be treated like a constant. Finally, Cov(X,Y) = E(XY) — E(X) . E(Y) = 24 — 6 3 = 6.
Answer C. 23. May 2000 Course 1 Examination, Problem No. 22, also Study Note P0908, Problem
No. 116 An actuary determines that the annual numbers of tornadoes in counties P and Q are jointly
distributed as follows: Annual number of Tornadoes in county Q 1 2 Annual number 0.12 0.06 0.05 0.02
of tornadoes l 0.13 0.15 0.12 0.03 '
in county P 2 0.05 0.15 0.10 0.02 Calculate the conditional variance of the annual number of tornadoes in county Q, given that
there are no tornadoes in county P. A. 0.51 B. 0.84 C. 0.88 D. 0.99 E. 1.76 Solution.
Let Q be the random number of tornadoes in county Q, and P be the total number of tornadoes in County P. Then we have this conditional probability:
use = raw) = 0})
Pr(P = 0) for n = 0,1,2,3. The denominator is
Pr(P= 0): Pr((Q =0)n(P = 0))+Pr((Q =1)n(P =0))+
+Pr((Q = 2)n(P =0))+Pr((Q= 3)n(P = 0)):
=0.12+0.06+0.05 +0.02 = 0.25. Pr(Q=nP=0)= 9 Then, n(Q=0P=0)=EH—Q——:£I}:;g—§iO—D=g%§=o.4s,
and similarly Pr(Q =1P = o) = 0.24, Pr(Q = 2P =0)=0.20, Pr(Q = 3P =O)=0.08. Hence,
E(QP=0)=O0.48+10.24+20.2+30.08 =0.88, ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by szysztof Ostaszewski  166  PRACTICE EXAMINATION N0. 2 15(Q2 [P = 0) = o2 o.4s +12 o.24+ 22 o.2 + 32 0.08 = 1.76,
and
Var(QP = 0) = E(Q2 P = o) (E(QP = 0))2 = 1.76 —0.882 = 0.9856.
Answer D. 24. May 2000 Course 1 Examination, Problem No. 23, also Study Note P0908, Problem No. 120
An insurance policy is written to cover a loss X where X has density function
3 2
—x , forOSxSZ,
f(x) = 8 0, otherwise. The time (in hours) to process a claim of size x, where 0 S x S 2, is uniformly distributed on the interval from x to 2x. Calculate the probability that a randomly chosen claim on this policy is
processed in three hours or more. A. 0.17 B. 0.25 C. 0.32 D. 0.58 E. 0.83 Solution.
Consider the area where the joint density is positive (marked with dotted lines below): We see that the joint density is: 3213 =—x fr.x("x)=fx(x)'fr(tlx=x)=§x '2x_x s for 0 S x S 2 and x < t < 2):. Thus, for any tbetween 0 and 2, as x varies between 1»t and t ' 3
mn=1§w=éza a. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostaszewski  167  PRACTICE EXAMINATION NO. 2 while for t between 2 and 4, x varies between %t and 2 and we have 23 3 3::2 t= —dx=———. f7” lisx 4 64
2 In fact, we only need that second part of density. Based on it 4 2 3r=4
h(T>3)=j[3§‘—]dt=3—’— =3—[1—ﬂj 48 37 11~0.17.
3 4 64 4 64M 4 64 Answer A. 25. May 2000 Course 1 Examination, Problem No. 24, also Study Note P0908, Problem
No. 30 An actuary has discovered that policyholders are three times as likely to ﬁle two claims as to ﬁle
four claims. If the number of claims ﬁled has a Poisson distribution, what is the variance of the
number of claims filed? l
A. — B. 1 C. J2 D. 2 E. 4
J3
Solution.
Let N be the random number of claims filed. We are given that
elAZ €4.14
Pr(N=2)= 2' =3 Pr(N=4)=3 4 Therefore, 2.2 = 4 and A = 2. The variance is the same as the mean and the parameter A = 2.
Answer D. 26. May 2000 Course 1 Examination, Problem No. 27, also Study Note P0908, Problem
No. 100 A car dealership sells 0, l, or 2 luxury cars on any day. When selling a car, the dealer also tries
to persuade the customer to buy an extended warranty for the car. Let X denote the number of
luxury cars sold in a given day, and let Y denote the number of extended warranties sold. 1
Pr(X=0,Y=0)=%, Pr(X=1,Y=0)=%, Pr(X=2,Y=0)=i—2,
Pr(X=1,Y=1)=%, Pr(X=2,Y= )=%, P(X=2,Y=2)=—;. What is the variance of X? A. 0.47 B. 0.58 c. 0.83 D. 1.42 E. 2.58 Solution.
ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  168  PRACTICE EXAMINATION N O. 2
The marginal distribution of X is found by summing probabilities over all possible values of Y. 0
1
fx(0)=Pr(X=0)=ZP(X=0,Y=y)=6,
y=0
f (1)=P(x=1)=iP(x=1,Y=y)=i+l=l,
X F0 12 6 4
and
f(2)P(X—2)§:P(X2Y i+l+l——7—
X F0 ’ y 12 3 6 12‘
Therefore, the ﬁrst moment of X equals
2 1 1 7 17
EX: .Px= =0—+1—+2—=—,
()onx( x) 6 41212
and the second moment equals
E(X=)=ix2P(X=x)=02.l+12l+22l=§i
6 4 1212’ y=0
so that the variance equals Var(X) = E(X2) — (1300)2 = 3—21— = 0.576. Answer B. 27. May 2000 Course 1 Examination, Problem No. 33, also Study Note P0908, Problem
No. 25 A blood test indicates the presence of a particular disease 95% of the time when the disease is
actually present. The same test indicates the presence of the disease 05% of the time when the
disease is not present. One percent of the population actually has the disease. Calculate the
probability that a person has the disease given that the test indicates the presence of the disease. A. 0.324 B. 0.657 C. 0.945 D. 0.950 E. 0.995 Solution.
Start by labeling the events. Let D be the event that a person has the disease, and P be the event that a person tests positive for the disease. We are given
Pr(D)=0.01, Pr(DC)=0.99, Pr(PD_)=0.95, Pr(PCD)=0.05, Pr(PDC)=0.005, Pr(PCDC)=O.995. We want to find Pr(DP) . Any time you have this kind of “ﬂipflop” of conditional
probabilities, you use Bayes’ Theorem: ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  169  PRACTICE EXAMINATION NO. 2 _ Pr(I’lID)Pr(D)
HDIP) — Pr(PD)Pr(D)+Pr(PlDC)Pr(DC)
0.95.001 _ 0.0095 = _ =0.657.
0.95 '0.01+0.005 O.99 0.01445 Answer B. 28. May 2000 Course 1 Examination, Problem No. 34, also Study Note P0908, Problem
No. 53 An insurance policy reimburses a loss up to a beneﬁt limit of 10. The policyholder’s loss, Y,
follows a distribution with density function: 0, otherwise.
What is the expected value of the beneﬁt paid under the insurance policy? A. 1.0 B. 1.3 C. 1.8 D. 1.9 E.2.0 Solution.
The amount paid under this policy is Y, 1< Y S 10,
W =
10, Y >10.
The expected amount paid is 4‘5 =2—0.2+0.1=1.9. IO ‘° 2 *°° 2 [2]
EW = y~—dy+ 10—dy= ——
( ) y. y. y 10
{El
2
l y 29. May 2000 Course 1 Examination, Problem No. 35, also Study Note P0908, Problem
No. 58 A company insures homes in three cities, J, K, and L. Since sufﬁcient distance separates the
cities, it is reasonable to assume that the losses occurring in these cities are independent. The
moment generating functions for the loss distributions of the cities are: M, (r) = (1 — 2r)”, MK (t) = (1  2:)“, ML (t) = (1  2:)“.
Let X represent the combined losses from the three cities. Calculate E (X3). Answer D. A. 1,320 B. 2,082 C. 5,760 D. 8,000 B. 10,560 Solution.
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  170  PRACTICE EXAMINATION N0 . 7.
Write X = J + K + L, where J, K, L represent the losses in cities J, K, and L, re5pectively. Because of independence of J, K and L we have
MK (1) = M, (t) MK (I)  ML(t) = (1  2t)”3 .(1 2:)‘25 (1— 2:)“*5 = (1 2:)"°. From the properties moment generating functions we know that M2 M, = E (2").
Therefore,
E(X3) = 5—ng (z) o = (—10)(—2)(11)(2)(—12)(—2)(1 — 20%”o = 10,560.
.=
Answer E. 30. May 2000 Course 1 Examination, Problem No. 36, also Study Note P0908, Problem
No. 14
In modeling the number of claims filed by an individual under an automobile policy during a threeyear period, an actuary makes the simplifying assumption that for all integers n 2 0,
pMl = % p", where p" represents the probability that the policyholder ﬁles n claims during the period. Under this assumption, what is the probability that a policyholder ﬁles more than one
claim during the period? A. 0.04 B. 0.16 C. 0.20 D. 0.80 E. 0.96 Solution. In order for this to be a probability distribution we must have 2 p, = 1. Since
k=0 1 _l.l .1.1. .1
I’ll spnl 5 spa2 5 5 ' ' 5P0!
it follows that
°° °° I" l
l=zp"=z[[§) 'p°]=”°'—1=125Po'
n=0 n=0 1——
5
Consequently, p0 = i and p” = . Finally,
5 5 5
4 14 l
PrN>1 =1PrN=0 —PrN=1=1 — = ———— .—=—.
() ()()P0P155525 Answer A. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  171  ...
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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
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