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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 independe...
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 Normal Distribution, Probability theory, probability density function, Krzysztof Ostaszewski, Study Note P0908

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