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Unformatted text preview: SECTION 12 PRACTICE EXAMINATION NUMBER 8
SOLUTIONS 1. Two types of vehicles, cars or trucks, enter a certain tunnel. The probability that a next vehicle
to enter the tunnel is a car is 10 times the probability that the next vehicle is a truck. Determine
the probability that among the ﬁrst 23 vehicles to enter the tunnel, there are at least 20 cars. A. Less than 72% B. At least 72%, but less than 77%
C. At least 77%, but less than 82%
D. At least 82%, but less than 87%
E. 87% or more Solution. A vehicle entering the tunnel has a % chance of being a car and a ﬁ chance of being a truck. We are asked for the probability that of the first 23 vehicles to enter the tunnel, at least 20 are
cars. That probability is Pr(20 out of 23 vehicles are cars) + Pr(21 out of 23 vehicles are cars) +
+ Pr(22 out of 23 vehicles are cars) + Pr(23 out of 23 vehicles are cars) = iii6%)”(ﬁ—JiiiJ(i—‘JT'(£31
+[::)(a”ia‘+[::Ma”ia°= 20 21 22 23
10 +253lo—+2311(1)7+1(1)—23=0.8490. 1123 1123 =1771 Answer D. 2. May 2005 CAS Course 3 Examination, Problem No. 25
The probability density function of the kth order statistic of size n is: m(F(y>)k“UF<y»”m), where F is the cumulative distribution function of the original distribution, and f is the density of
it. Samples are selected from a uniform distribution on [0, 10]. Determine the expected value of
the fourth order statistic for a sample of size ﬁve. A. Less than 6.5
B. At least 6.5, but less than 7.0 ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  322  PRACTICE EXAMINATION No. 8
C. At least 7.0, but less than 7.5 D. At least 7.5, but less than 8.0
E. 8.0 or more Solution. The density of the uniform distribution is f (y) = %, for 0 S y 510, and 0 otherwise. Its CDF is F(y) = % for 0 S y S 10, and 0 for y < 0, as well as 1 for y > 1. Therefore, the 4th order statistic of a sample of size 55 has the density fx(.,(x x)= 34""?! (F(x)) .‘(1F(x)) .—_f(x)2 0(135 [1— % )' _l__10_ x 2301;; . The expected value 18
1 IO E(x(,,) = Tmm = 35%“ —0.lx5)dx= o 500
__1_ L19: __1000.[1_1]39
.0 500560 556 3' =L [lxuixs]
500 5 60 .— 3. May 2005 CAS Course 3 Examination, Problem No. 36 XYZ Insurance issues 1year policies. ° The probability that a new insured has no accidents last year is 0.70. 0 The probability that an insured who was accidentfree last year will be accidentfree this year is
0.80. 0 The probability that an insured who was not accidentfree last year will be accidentfree this year is 0.60.
What is the probability that a new insured with an unknown accident history will be accidentfree in the second year of coverage? Answer B . A. Less than 71% B. At least 71%, but less than 72%
C. At least 72%, but less than 73%
D. At least 73%, but less than 74%
E. 74% or more Solution. Let E be the event that a new insured will be accidentfree this year, and F be the event that a new
insured was accidentfree last year. Then ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  323  SECTION 12
Pr(E):P1'((Ef‘F)U(EhF"))=Pr(EhF)+Pr(EnFC)= = Pr(EF).Pr(F)+Pr(EFC)Pr(FC)=0.8 o.7 +0.60.3=0.74. Answer E. 4. Let X1,X2 ,X3 be a random sample from a continuous probability distribution. Find
Pr(X, s X2 3 X3). A D . l B. l C. —l . l E. Not enough information to ﬁnd
2 3 4 6 Solution.
Since the individual distributions of X1 ,X2 ,X3 are continuous, so is their joint distribution. Therefore, P1'(Xl = X2) = 0, Pr(Xt = X3) = 0, and Pr(Xz = X3) = 0. The complement of the
event
{X1 = X,}u {X2 = X3}p{XI = X3}
can be partitioned into a union of the following six mutually exclusive events:
{X1<X2<X3}, {X3<X,<X2}, {X2<X3<X,},
{Xl <X2 <X3}, {X3 <X2 <X,}, {X2 < Xl <X3}.
These six events must all have the same probability, because X1, X2, and X3 have the same
distribution. Therefore,
Pr(X1 5 X2 s X3) = Pr(Xl < X2 < X3) = 115.
Answer D. 5. Let X be a discrete random variable with moment generating function
1 1 +o= m—l e
”What“ _, n. for t < 2. Find E(X). l 3 3 l“ e7 ,
A. l B. — C. — D. E. E(X) does not exrst
4 2 ":0 n!
Solution.
1 1 2 . . . . .
Note that —2— = 2 .2—t equals to a half of the MGF of the exponent1a1 dlstnbution Wlth mean
_ t _
ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  324  PRACTICE EXAMINATION NO. 8
0.5 (or hazard rate 11 = 2). Also «0 emd —li(el)n 1 e' _ ¢'l
",0 n! e "=0 n! e
is the MGF of the Poisson distribution with mean 1. The MGF given in this problem is a 50% 50% weighted average mixed distribution of the two, so that 11_1_ 3
X=—+— 1=—.
EU 22 4 Alternatively,2 since we established that «a ml
M(,__I_+1 s__ __1_.1..a—1,
2—t 2n=on! 2 t 2
wehave
I _d 1 1 d J1_ 2 l. e'I r
Mx(t)—E(—(t—2) )+— 5e —(t2) +2 e e,
and
__ l °n. 01 12
E(=X) M’(0)— (_2)+2 e e—4+2 4.
And there is also this third way if you do not want to fold the series immediately:
 l d +""e""l .2 1 *“ em“
'  2l —— =t2 +— .
M x—(t) :t( ”(t ) )+2 dt,,,,o n! ( ) 2 Eh—l)!
W Therefore,
11*""e‘l 11,“1113
EX=M’ = +—. =_+_. . .._=_+_=_.
H "( (—2)2 2.2.:(n—1)! 4 23 “on! 4 2 4
AnswerB. 6. Study Note P0908, Problem No. 125
The distribution of Y, given X, is uniform on the interval [0, X]. The marginal density of X is f( )_ 2x, for0<x<l
x x _ 0, otherwise. Determine the conditional density of X, given Y = y, where positive. A. 1 B. 2 C. 2x D. l E. L
y ly Solution. Because the conditional density of Y, given X, is uniform on the interval [0, X] , and the density of X is positive only for 0 < x < l, the joint density of X and Y is positive only on the region deﬁned by the condition 0 < y < x < 1, shown in the ﬁgure below. (.._ ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  325  SECTION 12 Where positive, that joint density equals fxx (350’) = fy (YIX = x)fx (x) = 2. This, of course, means that the joint density is uniform on the triangle deﬁned by the condition
0 < y < x < 1. This implies that the conditional distribution of X, given Y = y, is also uniform. But X, given Y: y, has positive density only for y < x < 1, so that the density, which is constant on
the interval [y,l] must equal to l
x Y = = — fx(  Y) 1 _ y for y < x < 1. That’s answer E. You can also calculate the conditional density from its deﬁnition
f , (x,y) 2 2 1
fx(xY=Y)=—Xfy‘r;j_=l—=2—_—2';=ﬂ.
Y j 2dx
y Answer E. 7. Let X be a discrete random variable with moment generating function 1 1 +°°t"
M =—1 '°' — —.
X(t) 4( +e )+2 "=0”! for —°o<t<+oc. Find Pr(X23). A B C D. 2:: E. Cannot be determined 1 l l
'8 '4 '2 Solution.
When a random variable X is discrete and assumes values x1 ,x2 ,x3,. .. with probabilities pl , p2 ,p3,. . . then its moment generating function is
Mx(t)=pl e"" +p, .e‘“ +p3oe"’ +....
In this case, we have ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  326  / PRACTICE EXAMINATION N0. 8 Mx(t)= l(1+e'°')+l ’— =le°" +le‘°" +le' = leo“ +le” +1e‘°".
4 2 “,0 n! 4 4 2 4 2 4
Therefore, the random variable whose MGF we are analyzing assumes values x1 = 0, x2 = 1,
l l l and x3 =10, with probabilities p, = 4’ p2 = 2’ and p3 = 4' respectively, so that Pr(X23)=Pr(X=10)=::.
AnswerB. 8. November 1981 Course 110 Examination, Problem No. 11
Mr. Flowers plants 10 rosebushes in a row. Eight of the bushes are white and two are red, and he plants them in a random order. What is the probability that he will consecutively plant seven or
more white bushes? A. —1 B. l C. l D. l E. l
10 9 15 45 5
Solution.
Given n objects, 0f WhiCh "i are 0f tYPe 1. n2 are of type 2,..., n,‘ are of type k, etc., there are
I
—' n'. I different ways of ordering them (if objects of the same type are interchangeable). If
nl'.n2"..nk° the ordering is done randomly, each of those ways is equally likely. In this problem, there are 1 l
i = 45 equally likely ways of ordering the 8 white and 2 red rosebushes. Let ‘R’ denote a 8 ! 2!
red rosebush and let ‘W‘ denote a white rosebush. The following 9 conﬁgurations are the only
orderings in which 7 or more white rosebushes are planted consecutively: RRWWWWWWWW, RWRWWWWWWW,
RWWWWWWWRW, RWWWWWWWWR,
WRRWWWWWWW, WRWWWWWWWR
WWWWWWWRRW, WWWWWWWRWR,
WWWWWWWWRR. 9 I The probability sought is therefore 45 5 .
Answer E. 9. You are given that N has Poisson distribution with mean 4. Find Var(NN 2 4). A. 2.10 B. 3.57 C. 4.00 D. 4.67 E. 5.33 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  327  SECTION 12 Solution. Let X be a Bernoulli Trial deﬁned as
'_ 1 if N 2 4,
_ 0 otherwise. Then the probability of success for X is p= Pr(N>4)= 1— Pr(N= o)— Pr(N= 1)— Pr(N= 2)— Pr(N= 3): l—ge" =O.5665. Also
4: Var(N) = E(V ar(NX))+Var(E(NX)).
Now (NX — 1): (N W. 4) and (NX=O) =(NN < 4). Therefore, v (NX)— Var(NN24) whenX=1,
ar ‘ Var(NN<4) whenX=0, and
E(NX)_ {E(NN 2 4) whenX= 1, } =
E(NN < 4) whenX = 0,
E(NN <4)+1(E(NN24)E(NN <4)) whenX= 1,
_ E(NN <4)+0(E(NN 24)—E(NN <4)) whenX=O,
l whenX=1,
= E(NN < 4)+(E(NN 24)E(NN < 4))~{O whenX=0.
This implies that
E(Var(NX)) = pVar(NN 2 4)+ (1  p)Var(NN < 4),
and Var(E(NX)) = p(1— p)(E(NN 2 4) E(NN < 4))‘.
Combining these two with
4 = Var(N) = E(Var(NX)) + Var(E(NX)),
we et
g 4 =Var(N) = pVar(NN 2 4)+(1— p)Var(NN < 4)+ +p(1p)(E(NN?4)'E(NN<4))2 Also,
= E(N) = E(E(NX)) = pE(NN z 4)+ (1 — p)E(NN < 4) =
Pr(X= 1) Pr(X = 2) Pr(X= 3))
= 2 1— . 1 + 2 + 3 =
pE(NN 4)+( p) [ 1—1) l—p l—p
= pE(NN 2 4)+ 52e'4.
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004—2008 by Knysztof Ostaszewski  328  Therefore,
_ 4
E(NN 2 4) = 4—32e—— = %‘5—~ ~ 5.3794.
p 1 —e"
3
Also,
Pr(N=nN<4)=——————= forn=0,1,2,and3,sothat 0 with probability PRACTICE EXAMINATION NO. 8 71 —e"’
3 3
4e" 0 with probability —,
1 with probability 71 , 71
.4
3" 1 with probability 3
(MN < 4): 8e
2 With. probability 7—1e," 2 with probability a,
3e 32
3_13 e“ 3 with probability7 —
3 with probability
71 —e"’
3
Based on this
3 12 24 32 12+48+96 156
E N N<4 =0—+l—+2 — +3—=—=—==2. 1972,
( I ) 71 71 71 71 71
and
E(~2N<4)=oz.i+12.2+22._ 24 +2.2_m_ﬁ
71 71+ 71 71 71
as well as
24
Var(NN < 4)_3__9_6_[l_56]2_ 281162 336_ 3780 =0 ”7499
71 71 71
so that
(1— p)Var(NN <4)=7—1e“ 3771820— _ 1—260 e" =0 H3250
Recalling that
Var(N) =
= pVar(NN 2 4)+ (1 — p)Var(NN < 4)+ p(l — p)(E(NN 2 4) E(NN < 4))2,
we conclude that
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004.2008 by Knysztof Osmszewski  329  SECTION 12 _ 4
.=(.2.4]v..(N.N..).7_I.«.3783 (12.02.4. _4 5% E
3 3 7l 3 3 714 71 = 0.5665Var(NN 2 4) + 0.3250 + 0.5665 0.4335 . (5.3794 — 2.1972)2 —v——‘
2A868 and based on that
Var(NN 2 4) = W = 2.0973.
0.5665
Answer A. 10. The claim amount on a certain insurance contract has a normal distribution with mean $1,000
and standard deviation $250. Given 10 independent claims, what is the probability that the
number of claims less than $1,050 is less than or equal to 2? A. 0.9026 B. 0.6025 C. 0.5793 D. 0.3356 E. 0.0174 Solution.
Let X be the random individual claim amount, and Z denote the standard normal random
variable. Then P,(X $1050): Pr(x— 1000 S 1050 1000 250 250
If we have 10 claims, we can treat each claim comparison to the 1050 threshold as a Bernoulli
Trial, with being below 1050 treated as a success and being above 1050 treated as failure. Then
the total number of claims among 10 less than $1,050 follows the binomial distribution with
probability of success p = 0.5793 and number of Bernoulli trials n = 10, so that the the
probability that the number of claims less than $1050 is less than or equal to 2 equals 10 0 10 10 l 9 10 2 8
0 0.5793 0.4207 + 1 0.5793 0.4207 + 2 05793 0.4207 =0.0174. J: Pr(Z s 0.2) = 05793. Answer E. l l 11. A random variable has the probability density function fx (x) = a, for E .<. x S e, and 0 otherwise. What is the probability that among of four independent observations of X there are
three less than 1 and one that is greater than 1? A.— B C .l 15.3 3.1
16 4 4 g l
' 2 ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  330  PRACTICE EXAMINATION NO. 8
Solution. We have
I
(PrX < 1) Mid): —
1 2x
Therefore, an observation of this random variable being less than 1 can be treated as four Bernoulli Trials with probability of success of : and the same probability of failure. The probability of exactly three successes is ﬁlerElﬁ Answer B. 12. A machine consists of two components. The lifetimes of the two components are identically distributed with the following common density function fx (1:) = %e‘3’ + ge'z‘ for x > 0, and 0 otherwise. The machine breaks down if any of the two components fails. Find the expected
lifetime of the device. A. 0.1124 B. 0.2260 C. 0.2981 D. 0.3267 E. 0.3999 Solution. The density can be written as fx (x) = % 3e’3" + 2%  2e'2‘ , and therefore this is a mixed distribution, with a weight of 0.25 in an exponential with hazard rate 3 and a weight of 0.75 in an
exponential with hazard rate 2. The survival function of this distribution is an analogous mixture of the survival functions of the two exponential distributions, i.e., sx (x) = i6” + g6”. Let Xl be the lifetime of the ﬁrst component, X2 be the lifetime of the second component, and Y be the
lifetime of the device. Then Y = min(X, ,Xz). Therefore 1 3 s, (y) = Pr(Y > y) = Pr(min(X1,X2) > y) = (sx (y))2 = (Xe3’ + ZeZyjz. The expected value of Y is therefore E(Y)= T(—e‘3’+43 —e 2’2=)dy ![—e’°’+—e‘4’+1—66e‘5’)dy= _9_ £.l_i.[1_° 135 7_2] 1 217 217 l l 1
=——+ —+ +—+ =——=—=
l6 6 l6 4 l6 5 16 60 60 60 16 60 960 02260 Answer B . ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  331  SECTION 12
13. Let X be an exponential random variable with mean 2. Deﬁne a new random variable Y: min(X2,2). Find 17,0). 1 1 1
A.0 B.l C.— D.— E.l—
e J2 J2
Solution.
We have x2, for0<XS~/_2, 2, for X > J2.
By the deﬁnition of the cumulative distribution function 1~}(1)=Pr(rs1)=Pr({X2 Sl}n{0<XS~/2})=Pr(XSI)=l—e_%. Answer E. Y = min(Xz,2) ={ 14. X is a random variable uniformly disn'ibuted on the interval [1, 2]. Find the probability
density function of Y = 1n X. Y
A. i B. 1n2 c. lny D.e— E. e’
1n2 y
Solution.
First note that fx (x) = 1 for 1 .<_ x S 2, and fx (1:) = 0 otherwise. and We have Y = lnX so that X = e’ and %= e” = X. Therefore =le’=e’, fr (Y) = fx (x(y));% where 0 S y S 1n2.
Answer E. 15. You are given that XI and X2 are two independent and identically distributed random
variables with a Poisson distribution with mean 2. Let Y = max(X1 ,Xz). Find Pr(Y =1). A. 0.1201 B. 0.1465 C. 0.2578 D. 0.3381 E. 0.4255 Solution. We have ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski . 332  PRACTICE EXAMINATION NO. 8 16. Let X1 and X2 be a random sample from the uniform distribution on [0, l] and let
Y1 =min(Xl,X2), Y2 = max(Xl,X2). Find firl (yIIYZ =Y2) B.l— Cl D.2 E] A. l
2 23’: 3’2 Solution.
Recalling the formula for the joint PDF of two order statistics, we obtain immediately: meJm (3502) = 2! 'fx (y1)‘fx(y2)= 2! ’1']: 2'
This actually tells us that the joint PDF of the two order statistics is uniform on the region
{0 S y1 S y2 S 1}. We conclude then that the conditional distribution of 1’,le = )22 is uniform on [0,y2] by looking at the figure below
3’2 Area where )21 S y2 Y,r2 = y2 is distributed on this line,
and the distribution is uniform, because the joint distribution is
uniform on the shaded triangle ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  333  SECTION 12
This means that for 0 S y, S y2 1
fr, (3)1le =y2) = _°
3’2
Answer C. 17. Michiko is performing a Bernoulli Trial experiment with probability of success of P, where P
is distributed on the interval [0, 1] according to a probability distribution with density f, (p) = 3 p2 for 0 S p S 1, and 0 otherwise. She performs the experiment 10 times and has two successes. Find the mean of the posterior probability distribution of P, i.e., E (PI X = 2), where
X is the number of successes in 10 Bernoulli Trials with a probability of success P. A B C. i D. — E.
5 l l i
'2 '5 10 14 Solution.
Recall that if the prior distribution of a probability p in a Bernoulli Trial is beta with parameters r
and s, then the posterior distribution for X = n, with m total Bernoulli trials, is also beta, with parameters r+n and s+mn. Here
3! 1‘(3+1)
= 3 2 = _. 2 = ____.
MP) p 211! r(3)r(1)
for 0 S p 51, and 0 otherwise, so that the prior distribution is beta with parameters 3 and 1. The posterior for X = 2, with 10 Bernoulli trials, is beta with parameters 3 + 2 = 5 and l + 10  2 = 9.
Therefore x3'1 (1 —x)l'1 We can also calculate E (PI X = 2) from its deﬁnition. We have 10 101: 2 10 x+2 lot
fx.p(x.p)=fx(XIP=P)'fp(p)=[x]p’°(1p) .3p=3.[x].p (1p) . Therefore, f (p‘X_2)_M_ 135p‘(1—p)‘ _ p‘(1—p)‘
P "  " l ’ l '
f" (2) [135174 (1  P)8 dp I!" (1  p)8 dp
o 0 Recall the beta function '  1‘(r)1“(s) B(r,s) = z"' (1 z)‘ 1 dz = —— J). 1"(r + s)
Here,r=5 ands=9and
ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 20042008 by Knysztof Ostaszewski  334  PRACTICE EXAMINATION NO. 8
r(5)r(9) _ 418! 1 l
4 — 8 = = —— _ —— = —,
{p (I p) d” 3(5’9) r(14) 13! 6435
Therefore,
f,(pX= 2): 6435;:“(1—12)’3 .
We conclude that ‘ 9.101113 518! 5
— = 5 _ a = . =———.—=—.
E(PX_2) 6435!]; (1 p)dp 6435 B(6,9) 2 14! 14 Answer E. 18. November 1981 Course 110 Examination, Problem No. 12
, 2 . 7
LetX have the density function f (x;9) = ii, for 0 < x < 0, and 0 otherWISe. If Pr(X > 1) = g, what is the value of 6? I I
—  1
B. (1)3 c. (E): D. 23 E. 2
8 7 Since f (x;0) = O for x > 0 and since Pr(X > 1) = %, we can conclude that 9 > 1. Therefore, A.1
2 Solution. ’2. Pr(X > 1) = jf(x;9)dx = 1%: {3;} l 1 7 '?=8' x=l This implies that a}; = %, and 6 = 2. Answer: E. 19. Telephone numbers in a certain area all start with 7 and then 6. The third digit can be 4, 5, or
6. Each of the last four digits can be any number from 0 to 9. There are 18,243 telephone
numbers assigned in the area. How many phone numbers are still unassigned? A. 18,243 B. 30,000 C. 321,757 D. 11,757 E. None Solution.
There are 3 choices for the third digit, and 10 choices for each of the last four digits, resulting in
310 101010 = 30,000 numbers available. Since 18,243 numbers are already assigned, there are 30,000 — 18,243 = 11,757 numbers still unassigned.
Answer D. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  335  SECTION 12
20. November 1981 Course 110 Examination, Problem No. 13 A random sample of size 16 is to be taken from a normal population having mean 100 and
variance 4. What is the 90th percentile of the distribution of the sample mean 2? ? A. 97.44 B. 100.08 C. 100.32 D. 100.64 B. 102.56 Solution.
2
Forarandom sample X1....,Xn from N([1,O'2) the sample mean X is N[/,t,o—], and
n
 J; 2
ﬂ = _(_"_t_). .. N(0.1).
£_ 0
J; In this problem, n =16, ,u = 100 and cr2 = 4, so that X ~ N(100%) and JR (2?  100)
2
Let p,,0 be the 90th percentile of the sample mean )7. By definition Pr(f S p90) = 0.90. Let (D
be the CDF of the standard normal distribution. This is equivalent to n(2(2?— 100) s 2(1)» 100))= <I>(2(p90 — 100)) = 0.90.
From the table of the standard normal distribution (I>(l .28) = 0.90 and therefore, 2(1)90 —100) = 1.28, or p90 =100.64.
Answer D. = 2(2—100)~N(0,1). 21....
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