OF SOCIETY ACTUARIES/CASUALTY ACTUARIAL SOCIETY
EXAM P PROBABILITY
EXAM P SAMPLE SOLUTIONS
Copyright 2008 by the Society of Actuaries and the Casualty Actuarial Society
Some of the questions in this study note are taken from past SOA/CAS examinations.
P-09-08
PRINTED IN U.S.A.
Page 1 of 60
1. Solution: D Let
G = event that a viewer watched gymnastics
B = event that a viewer watched baseball S = event that a viewer watched soccer Then we want to find c Pr ( G B S ) = 1 Pr ( G B S )
= 1 Pr ( G ) + Pr ( B ) + Pr ( S ) Pr ( G B ) Pr ( G S ) Pr ( B S ) + Pr ( G B S ) = 1 ( 0.28 + 0.29 + 0.19 0.14 0.10 0.12 + 0.08 ) = 1 0.48 = 0.52
-------------------------------------------------------------------------------------------------------2. Solution: A Let R = event of referral to a specialist L = event of lab work We want to find P[RL] = P[R] + P[L] P[RL] = P[R] + P[L] 1 + P[~(RL)] = P[R] + P[L] 1 + P[~R~L] = 0.30 + 0.40 1 + 0.35 = 0.05 .
-------------------------------------------------------------------------------------------------------3. Solution: D First note P [ A B ] = P [ A] + P [ B ] P [ A B ]
P [ A B '] = P [ A] + P [ B '] P [ A B '] Then add these two equations to get P [ A B ] + P [ A B '] = 2 P [ A] + ( P [ B ] + P [ B '] ) ( P [ A B ] + P [ A B '] ) 1.6 = 2 P [ A] + 1 P [ A]
P [ A] = 0.6
0.7 + 0.9 = 2 P [ A] + 1 P ( A B ) ( A B ')
Page 2 of 60
4.
Solution: A For i = 1, 2, let Ri = event that a red ball is drawn form urn i
Bi = event that a blue ball is drawn from urn i .
Then if x is the number of blue balls in urn 2, = Pr [ R1 ] Pr [ R2 ] + Pr [ B1 ] Pr [ B2 ] 4 16 6 x + 10 x + 16 10 x + 16
0.44 = Pr[( R1 R2 ) ( B1 B2 )] = Pr[ R1 R2 ] + Pr [ B1 B2 ]
= Therefore,
32 3x 3x + 32 + = x + 16 x + 16 x + 16 2.2 x + 35.2 = 3 x + 32 0.8 x = 3.2 x=4 2.2 = -------------------------------------------------------------------------------------------------------5. Solution: D Let N(C) denote the number of policyholders in classification C . Then N(Young Female Single) = N(Young Female) N(Young Female Married) = N(Young) N(Young Male) [N(Young Married) N(Young Married Male)] = 3000 1320 (1400 600) = 880 .
-------------------------------------------------------------------------------------------------------6. Solution: B Let H = event that a death is due to heart disease F = event that at least one parent suffered from heart disease Then based on the medical records, 210 102 108 P H F c = = 937 937 937 312 625 P F c = = 937 937 c P H F 108 625 108 = = = 0.173 and P H | F c = c 937 937 625 P F
Page 3 of 60
7.
Solution: D Let A = event that a policyholder has an auto policy H = event that a policyholder has a homeowners policy Then based on the information given, Pr ( A H ) = 0.15
Pr ( A H c ) = Pr ( A ) Pr ( A H ) = 0.65 0.15 = 0.50 Pr ( Ac H ) = Pr ( H ) Pr ( A H ) = 0.50 0.15 = 0.35
and the portion of policyholders that will renew at least one policy is given by 0.4 Pr ( A H c ) + 0.6 Pr ( Ac H ) + 0.8 Pr ( A H ) = ( 0.4 )( 0.5 ) + ( 0.6 )( 0.35 ) + ( 0.8 )( 0.15 ) = 0.53
( = 53% )
-------------------------------------------------------------------------------------------------------100292 01B-9 8. Solution: D Let C = event that patient visits a chiropractor T = event that patient visits a physical therapist We are given that Pr [C ] = Pr [T ] + 0.14 Pr ( C T ) = 0.22 Pr ( C c T c ) = 0.12 Therefore, 0.88 = 1 Pr C c T c = Pr [C T ] = Pr [C ] + Pr [T ] Pr [C T ] = Pr [T ] + 0.14 + Pr [T ] 0.22 = 2 Pr [T ] 0.08 or
Pr [T ] = ( 0.88 + 0.08 ) 2 = 0.48
Page 4 of 60
9.
Solution: B Let M = event that customer insures more than one car S = event that customer insures a sports car Then applying DeMorgans Law, we may compute the desired probability as follows: c Pr ( M c S c ) = Pr ( M S ) = 1 Pr ( M S ) = 1 Pr ( M ) + Pr ( S ) Pr ( M S ) = 1 Pr ( M ) Pr ( S ) + Pr ( S M ) Pr ( M ) = 1 0.70 0.20 + ( 0.15 )( 0.70 ) = 0.205
-------------------------------------------------------------------------------------------------------10. Solution: C Consider the following events about a randomly selected auto insurance customer: A = customer insures more than one car B = customer insures a sports car We want to find the probability of the complement of A intersecting the complement of B (exactly one car, non-sports). But P ( Ac Bc) = 1 P (A B) And, by the Additive Law, P ( A B ) = P ( A) + P ( B ) P ( A B ). By the Multiplicative Law, P ( A B ) = P ( B | A ) P (A) = 0.15 * 0.64 = 0.096 It follows that P ( A B ) = 0.64 + 0.20 0.096 = 0.744 and P (Ac Bc ) = 0.744 = 0.256
-------------------------------------------------------------------------------------------------------11. Solution: B Let C = Event that a policyholder buys collision coverage D = Event that a policyholder buys disability coverage Then we are given that P[C] = 2P[D] and P[C D] = 0.15 . By the independence of C and D, it therefore follows that 0.15 = P[C D] = P[C] P[D] = 2P[D] P[D] = 2(P[D])2 (P[D])2 = 0.15/2 = 0.075 P[D] = 0.075 and P[C] = 2P[D] = 2 0.075 Now the independence of C and D also implies the independence of CC and DC . As a result, we see that P[CC DC] = P[CC] P[DC] = (1 P[C]) (1 P[D]) = (1 2 0.075 ) (1 0.075 ) = 0.33 .
Page 5 of 60
12.
Solution: E Boxed numbers in the table below were computed. High BP Low BP Norm BP
Total
Regular heartbeat 0.09 0.20 0.56 0.85 Irregular heartbeat 0.05 0.02 0.08 0.15 Total 0.14 0.22 0.64 1.00 From the table, we can see that 20% of patients have a regular heartbeat and low blood pressure. -------------------------------------------------------------------------------------------------------13. Solution: C The Venn diagram below summarizes the unconditional probabilities described in the problem.
In addition, we are told that P[ A B C] 1 x = P [ A B C | A B] = = P [ A B] x + 0.12 3 It follows that 1 1 x = ( x + 0.12 ) = x + 0.04 3 3 2 x = 0.04 3 x = 0.06 Now we want to find c P ( A B C ) c P ( A B C ) | Ac = c P A 1 P[ A B C] = 1 P [ A]
= = 1 3 ( 0.10 ) 3 ( 0.12 ) 0.06 1 0.10 2 ( 0.12 ) 0.06 0.28 = 0.467 0.60
Page 6 of 60
14.
Solution: A 1 11 111 1 pk 2 = pk 3 = ... = p0 pk = pk 1 = 5 55 555 5
k
k0
p 5 1 pk = p0 = 0 = p0 k =0 5 14 k =0 1 5 p0 = 4/5 . Therefore, P[N > 1] = 1 P[N 1] = 1 (4/5 + 4/5 1/5) = 1 24/25 = 1/25 = 0.04 .
1=
k
-------------------------------------------------------------------------------------------------------15. Solution: C A Venn diagram for this situation looks like:
We want to find w = 1 ( x + y + z )
1 1 5 We have x + y = , x + z = , y + z = 4 3 12 Adding these three equations gives 115 ( x + y) + ( x + z) + ( y + z) = + + 4 3 12 2( x + y + z) = 1
x+ y+ z = 1 2
11 = 22 Alternatively the three equations can be solved to give x = 1/12, y = 1/6, z =1/4 1 1 1 1 again leading to w = 1 + + = 12 6 4 2 w = 1 ( x + y + z ) = 1
Page 7 of 60
16.
Solution: D Let N1 and N 2 denote the number of claims during weeks one and two, respectively. Then since N1 and N 2 are independent,
Pr [ N1 + N 2 = 7 ] = n =0 Pr [ N1 = n ] Pr [ N 2 = 7 n ]
7 7 1 1 = n =0 n +1 8 n 2 2 7 1 = n =0 9 2 8 1 1 = 9= 6= 2 2 64
-------------------------------------------------------------------------------------------------------17. Solution: D Let O = Event of operating room charges E = Event of emergency room charges Then 0.85 = Pr ( O E ) = Pr ( O ) + Pr ( E ) Pr ( O E ) Since
So
( Independence ) Pr ( E c ) = 0.25 = 1 Pr ( E ) , it follows Pr ( E ) = 0.75 . 0.85 = Pr ( O ) + 0.75 Pr ( O ) ( 0.75 ) Pr ( O )(1 0.75 ) = 0.10 Pr ( O ) = 0.40
= Pr ( O ) + Pr ( E ) Pr ( O ) Pr ( E )
-------------------------------------------------------------------------------------------------------18. Solution: D Let X1 and X2 denote the measurement errors of the less and more accurate instruments, respectively. If N(,) denotes a normal random variable with mean and standard deviation , then we are given X1 is N(0, 0.0056h), X2 is N(0, 0.0044h) and X1, X2 are X1 + X 2 0.00562 h 2 + 0.00442 h 2 is N (0, ) = N(0, 2 4 0.00356h) . Therefore, P[0.005h Y 0.005h] = P[Y 0.005h] P[Y 0.005h] = P[Y 0.005h] P[Y 0.005h] 0.005h = 2P[Y 0.005h] 1 = 2P Z 1 = 2P[Z 1.4] 1 = 2(0.9192) 1 = 0.84. 0.00356h independent. It follows that Y =
Page 8 of 60
19.
Solution: B Apply Bayes Formula. Let A = Event of an accident B1 = Event the drivers age is in the range 16-20 B2 = Event the drivers age is in the range 21-30 B3 = Event the drivers age is in the range 30-65 B4 = Event the drivers age is in the range 66-99 Then Pr ( A B1 ) Pr ( B1 ) Pr ( B1 A ) = Pr ( A B1 ) Pr ( B1 ) + Pr ( A B2 ) Pr ( B2 ) + Pr ( A B3 ) Pr ( B3 ) + Pr ( A B4 ) Pr ( B4 ) =
( 0.06 )( 0.08) = 0.1584 ( 0.06 )( 0.08) + ( 0.03)( 0.15) + ( 0.02 )( 0.49 ) + ( 0.04 )( 0.28)
--------------------------------------------------------------------------------------------------------
20.
Solution: D Let S = Event of a standard policy F = Event of a preferred policy U = Event of an ultra-preferred policy D = Event that a policyholder dies Then P [ D | U ] P [U ] P [U | D ] = P [ D | S ] P [ S ] + P [ D | F ] P [ F ] + P [ D | U ] P [U ]
=
( 0.001)( 0.10 ) ( 0.01)( 0.50 ) + ( 0.005 )( 0.40 ) + ( 0.001)( 0.10 )
= 0.0141 -------------------------------------------------------------------------------------------------------21. Solution: B Apply Bayes Formula: Pr Seri. Surv.
= =
( 0.9 )( 0.3) = 0.29 ( 0.6 )( 0.1) + ( 0.9 )( 0.3) + ( 0.99 )( 0.6 )
Pr Surv. Crit. Pr [ Crit.] + Pr Surv. Seri. Pr [Seri.] + Pr Surv. Stab. Pr [Stab.]
Pr Surv. Seri. Pr [Seri.]
Page 9 of 60
22.
Solution: D Let H = Event of a heavy smoker L = Event of a light smoker
N = Event of a non-smoker D = Event of a death within five-year period 1 Now we are given that Pr D L = 2 Pr D N and Pr D L = Pr D H 2 Therefore, upon applying Bayes Formula, we find that Pr D H Pr [ H ] Pr H D = Pr D N Pr N + Pr D L Pr L + Pr D H Pr H [] [] [] 2 Pr D L ( 0.2 ) 0.4 = = = 0.42 1 0.25 + 0.3 + 0.4 Pr D L ( 0.5 ) + Pr D L ( 0.3) + 2 Pr D L ( 0.2 ) 2 -------------------------------------------------------------------------------------------------------23. Solution: D Let C = Event of a collision T = Event of a teen driver Y = Event of a young adult driver M = Event of a midlife driver S = Event of a senior driver Then using Bayes Theorem, we see that P[C Y ]P[Y ] P[YC] = P[C T ]P[T ] + P[C Y ]P[Y ] + P[C M ]P[ M ] + P[C S ]P[ S ] = (0.08)(0.16) = 0.22 . (0.15)(0.08) + (0.08)(0.16) + (0.04)(0.45) + (0.05)(0.31)
-------------------------------------------------------------------------------------------------------24. Solution: B Observe
Pr [1 N 4] 1 1 1 1 1 1 1 1 1 = + + + + + + + Pr N 1 N 4 = Pr [ N 4] 6 12 20 30 2 6 12 20 30 = 10 + 5 + 3 + 2 20 2 = = 30 + 10 + 5 + 3 + 2 50 5
Page 10 of 60
25.
Solution: B Let Y = positive test result D = disease is present (and ~D = not D) Using Bayes theorem: P[Y | D]P[ D] (0.95)(0.01) = = 0.657 . P[D|Y] = P[Y | D]P[ D] + P[Y |~ D]P[~ D] (0.95)(0.01) + (0.005)(0.99)
-------------------------------------------------------------------------------------------------------26. Solution: C Let: S = Event of a smoker C = Event of a circulation problem Then we are given that P[C] = 0.25 and P[SC] = 2 P[SCC] Now applying Bayes Theorem, we find that P[CS] = = 2 P[ S C C ]P[C ] 2 P[ S C ]P[C ] + P[ S C ](1 P[C ])
C C
P[ S C ]P[C ]
P[ S C ]P[C ] + P[ S C C ]( P[C C ])
=
2(0.25) 2 2 = =. 2(0.25) + 0.75 2 + 3 5
-------------------------------------------------------------------------------------------------------27. Solution: D Use Bayes Theorem with A = the event of an accident in one of the years 1997, 1998 or 1999. P[ A 1997]P[1997] P[1997|A] = P[ A 1997][ P[1997] + P[ A 1998]P[1998] + P[ A 1999]P[1999] = (0.05)(0.16) = 0.45 . (0.05)(0.16) + (0.02)(0.18) + (0.03)(0.20)
--------------------------------------------------------------------------------------------------------
Page 11 of 60
28.
Solution: A Let C = Event that shipment came from Company X I1 = Event that one of the vaccine vials tested is ineffective P [ I1 | C ] P [ C ] Then by Bayes Formula, P [ C | I1 ] = P [ I1 | C ] P [ C ] + P I1 | C c P C c Now 1 P [C ] = 5 14 P C c = 1 P [ C ] = 1 = 55
30 P [ I1 | C ] = ( 1 ) ( 0.10 )( 0.90 ) = 0.141 29 29 30 P I1 | C c = ( 1 ) ( 0.02 )( 0.98 ) = 0.334
Therefore,
P [ C | I1 ] =
( 0.141) (1/ 5) = 0.096 ( 0.141)(1/ 5) + ( 0.334 )( 4 / 5)
-------------------------------------------------------------------------------------------------------29. Solution: C Let T denote the number of days that elapse before a high-risk driver is involved in an accident. Then T is exponentially distributed with unknown parameter . Now we are given that 0.3 = P[T 50] =
50
e
0
t
dt = e t
80
50 0
= 1 e50
Therefore, e50 = 0.7 or = (1/50) ln(0.7) It follows that P[T 80] = =1e
(80/50) ln(0.7)
e
0
t
dt = e t
80 0
= 1 e80
= 1 (0.7)80/50 = 0.435 .
-------------------------------------------------------------------------------------------------------30. e 2 e 4 Let N be the number of claims filed. We are given P[N = 2] = =3 = 3 P[N 2! 4! = 4]24 2 = 6 4 2 = 4 = 2 Therefore, Var[N] = = 2 . Solution: D
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31.
Solution: D Let X denote the number of employees that achieve the high performance level. Then X follows a binomial distribution with parameters n = 20 and p = 0.02 . Now we want to determine x such that Pr [ X > x ] 0.01
or, equivalently, x k 20 k 0.99 Pr [ X x ] = k =0 ( 20 ) ( 0.02 ) ( 0.98 ) k
The following table summarizes the selection process for x: x Pr [ X = x ] Pr [ X x ] 0 1 2
( 0.98) = 0.668 19 20 ( 0.02 )( 0.98 ) = 0.272 2 18 190 ( 0.02 ) ( 0.98 ) = 0.053
20
0.668 0.940 0.993
Consequently, there is less than a 1% chance that more than two employees will achieve the high performance level. We conclude that we should choose the payment amount C such that 2C = 120, 000 or C = 60, 000 -------------------------------------------------------------------------------------------------------32. Solution: D Let X = number of low-risk drivers insured Y = number of moderate-risk drivers insured Z = number of high-risk drivers insured f(x, y, z) = probability function of X, Y, and Z Then f is a trinomial probability function, so Pr [ z x + 2] = f ( 0, 0, 4 ) + f (1, 0,3) + f ( 0,1,3) + f ( 0, 2, 2 )
= ( 0.20 ) + 4 ( 0.50 )( 0.20 ) + 4 ( 0.30 )( 0.20 ) +
4 3 3
= 0.0488
4! 2 2 ( 0.30 ) ( 0.20 ) 2!2!
Page 13 of 60
33.
1 0.005 ( 20 t ) dt = 0.005 20t t 2 20 x x 2 1 1 = 0.005 400 200 20 x + x 2 = 0.005 200 20 x + x 2 2 2 where 0 < x < 20 . Therefore, 2 Pr [ X > 16] 200 20 (16 ) + 1 2 (16 ) 81 Pr X > 16 X > 8 = = = Pr X > 8 = 1 ( 8)2 72 9 [ ] 200 20 (8) + 2 Pr [ X > x ] =
20
Solution: B Note that
-------------------------------------------------------------------------------------------------------34. Solution: C 2 We know the density has the form C (10 + x ) for 0 < x < 40 (equals zero otherwise).
First, determine the proportionality constant C from the condition
1 = C (10 + x ) dx = C (10 + x) 1
2 0 40 40
40 0
f ( x)dx =1 :
2 CC =C 0 10 50 25 so C = 25 2 , or 12.5 . Then, calculate the probability over the interval (0, 6): 6 2 1 6 1 1 = (12.5 ) = 0.47 . 12.5 (10 + x ) dx = (10 + x ) 0 0 10 16 =
-------------------------------------------------------------------------------------------------------35. Solution: C Let the random variable T be the future lifetime of a 30-year-old. We know that the density of T has the form f (x) = C(10 + x)2 for 0 < x < 40 (and it is equal to zero otherwise). First, determine the proportionality constant C from the condition 4 00 f ( x)dx =1: 40 2 40 1 = f ( x)dx = C (10 + x) 1 |0 = C 0 25 25 so that C = = 12.5. Then, calculate P(T < 5) by integrating f (x) = 12.5 (10 + x)2 2 over the interval (0.5).
Page 14 of 60
36.
Solution: B To determine k, note that 1 k k 4 5 1 = k (1 y ) dy = (1 y ) 1 = 0 5 5 0 k=5 We next need to find P[V > 10,000] = P[100,000 Y > 10,000] = P[Y > 0.1] =
0.1
5 (1 y ) dy = (1 y )
4
1
51 0.1
= (0.9)5 = 0.59 and P[V > 40,000]
= P[100,000 Y > 40,000] = P[Y > 0.4] =
0.4
5 (1 y ) dy = (1 y )
4
1
51 0.4
= (0.6)5 = 0.078 .
It now follows that P[V > 40,000V > 10,000] P[V > 40, 000 V > 10, 000] P[V > 40, 000] 0.078 = = = 0.132 . = P[V > 10, 000] P[V > 10, 000] 0.590 -------------------------------------------------------------------------------------------------------37. Solution: D Let T denote printer lifetime. Then f(t) = et/2, 0 t Note that 1 1 P[T 1] = e t / 2 dt = e t / 2 1 = 1 e1/2 = 0.393 0 2 0 P[1 T 2] =
2e
1
2
1
t / 2
dt = e t / 2
2 1
= e 1/2 e 1 = 0.239
Next, denote refunds for the 100 printers sold by independent and identically distributed random variables Y1, . . . , Y100 where with probability 0.393 200 Yi = 100 with probability 0.239 i = 1, . . . , 100 0 with probability 0.368 Now E[Yi] = 200(0.393) + 100(0.239) = 102.56 Therefore, Expected Refunds =
E [Y ] = 100(102.56) = 10,256 .
i =1 i
100
Page 15 of 60
38.
Solution: A Let F denote the distribution function of f. Then Using this result, we see
F ( x ) = Pr [ X x ] = 3t 4 dt = t 3 = 1 x 3
x x 1 1
Pr [ X < 2| X 1.5] =
Pr ( X < 2 ) ( X 1.5 ) Pr [ X 1.5]
3
=
Pr [ X < 2] Pr [ X 1.5] Pr [ X 1.5]
3 3
F ( 2 ) F (1.5 ) (1.5 ) ( 2 ) = = 3 1 F (1.5 ) (1.5 )
3 = 1 4
= 0.578
-------------------------------------------------------------------------------------------------------39. Solution: E Let X be the number of hurricanes over the 20-year period. The conditions of the problem give x is a binomial distribution with n = 20 and p = 0.05 . It follows that P[X < 2] = (0.95)20(0.05)0 + 20(0.95)19(0.05) + 190(0.95)18(0.05)2 = 0.358 + 0.377 + 0.189 = 0.925 .
-------------------------------------------------------------------------------------------------------40. Solution: B Denote the insurance payment by the random variable Y. Then if 0 < X C 0 Y = X C if C < X < 1 Now we are given that 0.64 = Pr (Y < 0.5 ) = Pr ( 0 < X < 0.5 + C ) = Therefore, solving for C, we find C = 0.8 0.5 Finally, since 0 < C < 1 , we conclude that C = 0.3
0.5+ C 0
2 x dx = x
2
0.5 + C 0
= ( 0.5 + C )
2
Page 16 of 60
41.
Solution: E Let X = number of group 1 participants that complete the study. Y = number of group 2 participants that complete the study. Now we are given that X and Y are independent. Therefore, P ( X 9 ) ( Y < 9 ) ( X < 9 ) ( Y 9 )
{
}
= P ( X 9 ) ( Y < 9 ) + P ( X < 9 ) ( Y 9 ) = 2 P ( X 9 ) ( Y < 9 ) = 2 P [ X 9 ] P [Y < 9 ] = 2 P [ X 9] P [ X < 9] = 2 P [ X 9] (1 P [ X 9] )
9 10 9 10 = 2 ( 10 ) ( 0.2 )( 0.8 ) + ( 10 ) ( 0.8 ) 1 ( 10 ) ( 0.2 )( 0.8 ) ( 10 ) ( 0.8 ) 10 9 10 9 = 2 [ 0.376][1 0.376] = 0.469
(due to symmetry) (again due to symmetry)
-------------------------------------------------------------------------------------------------------42. Solution: D Let IA = Event that Company A makes a claim IB = Event that Company B makes a claim XA = Expense paid to Company A if claims are made XB = Expense paid to Company B if claims are made Then we want to find C Pr I A I B ( I A I B ) ( X A < X B )
{
}
C = Pr I A I B + Pr ( I A I B ) ( X A < X B ) C = Pr I A Pr [ I B ] + Pr [ I A ] Pr [ I B ] Pr [ X A < X B ]
(independence)
= ( 0.60 )( 0.30 ) + ( 0.40 )( 0.30 ) Pr [ X B X A 0] = 0.18 + 0.12 Pr [ X B X A 0] Now X B X A is a linear combination of independent normal random variables. Therefore, X B X A is also a normal random variable with mean M = E [ X B X A ] = E [ X B ] E [ X A ] = 9, 000 10, 000 = 1, 000 and standard deviation = Var ( X B ) + Var ( X A ) = It follows that
( 2000 ) + ( 2000 )
2
2
= 2000 2
Page 17 of 60
1000 Pr [ X B X A 0] = Pr Z 2000 2 1 = Pr Z 2 2 1 = 1 Pr Z < 2 2 = 1 Pr [ Z < 0.354]
(Z is standard normal)
Finally,
= 1 0.638 = 0.362
C Pr I A I B ( I A I B ) ( X A < X B ) = 0.18 + ( 0.12 )( 0.362 ) = 0.223
{
}
-------------------------------------------------------------------------------------------------------43. Solution: D If a month with one or more accidents is regarded as success and k = the number of failures before the fourth success, then k follows a negative binomial distribution and the requested probability is 4 k 3 3+ k 3 2 Pr [ k 4] = 1 Pr [ k 3] = 1 ( k ) 5 5 k =0 3 = 1 5
4
3 2 0 4 2 1 5 2 2 6 2 3 ( 0 ) + ( 1 ) + ( 2 ) + ( 3 ) 5 5 5 5
3 8 8 32 = 1 1 + + + 5 5 5 25 = 0.2898 Alternatively the solution is 2 4 2 3 5 2 3 6 2 3 + ( 1 ) + ( 2 ) + ( 3 ) = 0.2898 5 5 5 5 5 5 5 which can be derived directly or by regarding the problem as a negative binomial distribution with i) success taken as a month with no accidents ii) k = the number of failures before the fourth success, and iii) calculating Pr [ k 3]
4 4 4 2 4 3
4
Page 18 of 60
44.
Solution: C If k is the number of days of hospitalization, then the insurance payment g(k) is 100k for k =1, 2, 3 g( k ) = 300 + 50 (k 3) for k = 4, 5.
{
5
Thus, the expected payment is
g (k ) p
k =1
k
= 100 p1 + 200 p2 + 300 p3 + 350 p4 + 400 p5 =
1 (100 5 + 200 4 + 300 3 + 350 2 + 400 1) =220 15 -------------------------------------------------------------------------------------------------------45. Solution: D
0 0 4
2 4x x2 x3 x3 8 64 56 28 + = + = = dx = Note that E ( X ) = dx + 0 10 2 10 30 2 30 0 30 30 30 15
-------------------------------------------------------------------------------------------------------46. Solution: D The density function of T is 1 f ( t ) = et / 3 , 0 < t < 3 Therefore, E [ X ] = E max ( T , 2 ) =
2 0 t 2 t / 3 e dt + e t / 3 dt 23 3 2
2 = 2e t / 3 | 0 te t / 3 | + e t / 3 dt 2
= 2e
2 / 3
+ 2 + 2e
2 / 3
3e t / 3 | 2
= 2 + 3e 2 / 3
Page 19 of 60
47.
Solution: D Let T be the time from purchase until failure of the equipment. We are given that T is exponentially distributed with parameter = 10 since 10 = E[T] = . Next define the payment for 0 T 1 x x for 1 < T 3 P under the insurance contract by P = 2 for T > 3 0 We want to find x such that 1 x t/10 e dt + 1000 = E[P] = 10 0
1/10 3/10
x 1 t/10 t /10 2 10 e dt = xe 1
1/10
3
1 0
x e t /10 2
3 1
= x e + x (x/2) e + (x/2) e We conclude that x = 5644 .
= x(1 e
1/10
e3/10) = 0.1772x .
-------------------------------------------------------------------------------------------------------48. Solution: E Let X and Y denote the year the device fails and the benefit amount, respectively. Then the density function of X is given by x 1 f ( x ) = ( 0.6 ) ( 0.4 ) , x = 1, 2,3... and 1000 ( 5 x ) if x = 1, 2,3, 4 y= if x > 4 0 It follows that 2 3 E [Y ] = 4000 ( 0.4 ) + 3000 ( 0.6 )( 0.4 ) + 2000 ( 0.6 ) ( 0.4 ) + 1000 ( 0.6 ) ( 0.4 ) = 2694 -------------------------------------------------------------------------------------------------------49. Solution: D Define f ( X ) to be hospitalization payments made by the insurance policy. Then 100 X f (X ) = 300 + 25 ( X 3) and if X = 1, 2,3 if X = 4,5
Page 20 of 60
E f ( X ) = f ( k ) Pr [ X = k ]
k =1
5
5 4 3 2 1 = 100 + 200 + 300 + 325 + 350 15 15 15 15 15 1 640 = [100 + 160 + 180 + 130 + 70] = = 213.33 3 3
-------------------------------------------------------------------------------------------------------50. Solution: C Let N be the number of major snowstorms per year, and let P be the amount paid to (3 / 2) n e 3 / 2 , n = 0, 1, 2, . . . and the company under the policy. Then Pr[N = n] = n! for N = 0 0 P= . 10, 000( N 1) for N 1 Now observe that E[P] = = 10,000 e3/2 +
10, 000(n 1)
n =1
(3 / 2) n e 3 / 2 n!
(3 / 2) n e 3 / 2 = 10,000 e3/2 + E[10,000 (N 1)] n! n=0 3/2 = 10,000 e + E[10,000N] E[10,000] = 10,000 e3/2 + 10,000 (3/2) 10,000 = 7,231 .
10, 000(n 1)
-------------------------------------------------------------------------------------------------------51. Solution: C Let Y denote the manufacturers retained annual losses. for 0.6 < x 2 x Then Y = for x > 2 2 and E[Y] = =
2 2.5(0.6) 2.5 2.5(0.6) 2.5 2.5(0.6) 2.5 2(0.6) 2.5 x dx + 2 dx = dx x3.5 x 3.5 x 2.5 x 2.5 0.6 2 0.6
2 0.6
2
2
2.5(0.6) 2.5 1.5 x1.5
+
2(0.6) 2.5 2.5(0.6) 2.5 2.5(0.6) 2.5 (0.6) 2.5 = + + 1.5 = 0.9343 . (2) 2.5 1.5(2)1.5 1.5(0.6)1.5 2
Page 21 of 60
52.
Solution: A Let us first determine K. Observe that 1 1 1 1 60 + 30 + 20 + 15 + 12 137 1 = K 1 + + + + = K = K 60 2 3 4 5 60 60 K= 137 It then follows that Pr [ N = n ] = Pr N = n Insured Suffers a Loss Pr [ Insured Suffers a Loss ] 60 3 , N = 1,...,5 ( 0.05) = 137 N 137 N Now because of the deductible of 2, the net annual premium P = E [ X ] where = 0 X = N 2 , if N 2 , if N > 2
5
Then,
P = E [ X ] = N =3 ( N 2 )
3 3 3 1 = (1) + 3 = 0.0314 + 2 137 N 137 137 ( 4 ) 137 ( 5 )
-------------------------------------------------------------------------------------------------------53. for 1 < y 10 y Let W denote claim payments. Then W = for y 10 10 10 2 2 2 10 10 2 = 2 2/10 + 1/10 = 1.9 . It follows that E[W] = y 3 dy + 10 3 dy = y y y1 y 10 1 10 Solution: D
Page 22 of 60
54.
Solution: B Let Y denote the claim payment made by the insurance company. Then with probability 0.94 0 Y = Max ( 0, x 1) with probability 0.04 14 with probability 0.02 and
E [Y ] = ( 0.94 )( 0 ) + ( 0.04 )( 0.5003)
15 1
( x 1) e x / 2 dx + ( 0.02 )(14 )
15 15 = ( 0.020012 ) xe x / 2 dx e x / 2 dx + 0.28 1 1 15 15 = 0.28 + ( 0.020012 ) 2 xe x / 2 | 15 +2 e x / 2 dx e x / 2 dx 1 1 1 15 = 0.28 + ( 0.020012 ) 30e 7.5 + 2e 0.5 + e x / 2 dx 1 7.5 0.5 x / 2 15 = 0.28 + ( 0.020012 ) 30e + 2e 2e | 1
= 0.28 + ( 0.020012 ) ( 30e 7.5 + 2e 0.5 2e7.5 + 2e0.5 ) = 0.28 + ( 0.020012 ) ( 32e 7.5 + 4e 0.5 ) = 0.28 + ( 0.020012 )( 2.408 )
(in thousands) = 0.328 It follows that the expected claim payment is 328 . -------------------------------------------------------------------------------------------------------55. Solution: C The pdf of x is given by f(x) = k k1 (1 + x)4 dx = 3 (1 + x)3 0 k=3 1=
0
k , 0 < x < . To find k, note (1 + x) 4 = k 3 dx and substituting u = 1 + x, du = dx, we see
It then follows that E[x] =
(1 + x)
0
3x
4
u 2 u 3 3(u 1) 1 1 E[x] = du = 3 (u 3 u 4 )du = 3 = 3 = 3/2 1 = . 4 u 2 3 2 3 1 1 1
Page 23 of 60
56.
Solution: C Let Y represent the payment made to the policyholder for a loss subject to a deductible D. for 0 X D 0 That is Y = x D for D < X 1 Then since E[X] = 500, we want to choose D so that 1000 1 1 1 ( x D) 2 1000 (1000 D) 2 500 = ( x D)dx = = D 4 1000 1000 2 2000 D (1000 D)2 = 2000/4 500 = 5002 1000 D = 500 D = 500 (or D = 1500 which is extraneous).
-------------------------------------------------------------------------------------------------------57. Solution: B 1 for the claim size X in a certain class of accidents. (1 2500t ) 4 (4)(2500) 10, 000 First, compute Mx(t) = = 5 (1 2500t ) (1 2500t )5 (10, 000)(5)(2500) 125, 000, 000 Mx(t) = = (1 2500t )6 (1 2500t )6 Then E[X] = Mx (0) = 10,000 E[X2] = Mx (0) = 125,000,000 Var[X] = E[X2] {E[X]}2 = 125,000,000 (10,000)2 = 25,000,000 Var[ X ] = 5,000 .
We are given that Mx(t) =
-------------------------------------------------------------------------------------------------------58. Solution: E Let XJ, XK, and XL represent annual losses for cities J, K, and L, respectively. Then X = XJ + XK + XL and due to independence x +x +x t M(t) = E e xt = E e( J K L ) = E e xJ t E e xK t E e xLt = MJ(t) MK(t) ML(t) = (1 2t)3 (1 2t)2.5 (1 2t)4.5 = (1 2t)10 Therefore, M(t) = 20(1 2t)11 M(t) = 440(1 2t)12 M(t) = 10,560(1 2t)13 E[X3] = M(0) = 10,560
Page 24 of 60
59.
Solution: B The distribution function of X is given by 2.5 ( 200 ) F ( x) = 200 t 3.5
x
2.5
( 200 ) dt = t 2.5
2.5
2.5 x
( 200 ) = 1
x 2.5
2.5
,
x > 200
200
Therefore, the p percentile x p of X is given by
th
( 200 ) p = F ( xp ) = 1 2.5 100 xp ( 200 ) 1 0.01 p =
xp
=
2.5 2.5
(1 0.01 p )
xp =
25
200 xp
25
(1 0.01 p )
200
It follows that x 70 x 30 =
( 0.30 )
200
25
( 0.70 )
200
25
= 93.06
-------------------------------------------------------------------------------------------------------60. Solution: E Let X and Y denote the annual cost of maintaining and repairing a car before and after the 20% tax, respectively. Then Y = 1.2X and Var[Y] = Var[1.2X] = (1.2)2 Var[X] = (1.2)2(260) = 374 .
-------------------------------------------------------------------------------------------------------61. Solution: A The first quartile, Q1, is found by =
Q1 = 200 (4/3)0.4 = 224.4 . Similarly, the third quartile, Q3, is given by Q3 = 200 (4)0.4 = 348.2 . The interquartile range is the difference Q3 Q1 .
z
Q1
f(x) dx . That is, = (200/Q1)2.5 or
Page 25 of 60
62.
Solution: C First note that the density function of X is given by 1 if x =1 2 1< x < 2 f ( x ) = x 1 if otherwise 0 Then
2 2 1 1 1 1 1 E ( X ) = + x ( x 1) dx = + x 2 x dx = + x3 x 2 1 1 2 2 2 3 2 1
(
)
2
=
184117 4 + + = 1 = 232323 3
EX
()
2
2 2 1 1 1 1 1 = + x 2 ( x 1) dx = + x3 x 2 dx = + x 4 x3 21 21 2 4 3 1
(
)
2
=
1 16 8 1 1 17 7 23 + += = 2 4 3 4 3 4 3 12
Var ( X ) = E X
()
2
23 4 23 16 5 E ( X ) = 12 3 = 12 9 = 36
2
2
-------------------------------------------------------------------------------------------------------63. Solution: C X if 0 X 4 Note Y = 4 if 4 < X 5 Therefore, 41 54 1 4 4 E [Y ] = xdx + dx = x 2 | 0 + x| 5 4 05 45 10 5 16 20 16 8 4 12 = + =+= 10 5 5 5 5 5 41 5 16 1 16 4 E Y 2 = x 2 dx + dx = x 3 | 0 + x| 5 4 05 45 15 5 64 80 64 64 16 64 48 112 = + = += + = 15 5 5 15 5 15 15 15
2 112 12 Var [Y ] = E Y 2 ( E [Y ] ) = = 1.71 15 5 2
Page 26 of 60
64.
Solution: A Let X denote claim size. Then E[X] = [20(0.15) + 30(0.10) + 40(0.05) + 50(0.20) + 60(0.10) + 70(0.10) + 80(0.30)] = (3 + 3 + 2 + 10 + 6 + 7 + 24) = 55 E[X2] = 400(0.15) + 900(0.10) + 1600(0.05) + 2500(0.20) + 3600(0.10) + 4900(0.10) + 6400(0.30) = 60 + 90 + 80 + 500 + 360 + 490 + 1920 = 3500 Var[X] = E[X2] (E[X])2 = 3500 3025 = 475 and Var[ X ] = 21.79 . Now the range of claims within one standard deviation of the mean is given by [55.00 21.79, 55.00 + 21.79] = [33.21, 76.79] Therefore, the proportion of claims within one standard deviation is 0.05 + 0.20 + 0.10 + 0.10 = 0.45 .
-------------------------------------------------------------------------------------------------------65. Solution: B Let X and Y denote repair cost and insurance payment, respectively, in the event the auto is damaged. Then if x 250 0 Y = x 250 if x > 250 and 1500 1 1 12502 2 E [Y ] = x 250 )dx = x 250 ) 1500 = = 521 ( ( 250 250 1500 3000 3000 1500 1 1 12503 2 3 E Y 2 = = 434, 028 ( x 250 ) dx = ( x 250 ) 1500 = 250 250 1500 4500 4500
Var [Y ] = E Y 2 { E [Y ]} = 434, 028 ( 521)
2 2
Var [Y ] = 403
-------------------------------------------------------------------------------------------------------66. Solution: E Let X1, X2, X3, and X4 denote the four independent bids with common distribution function F. Then if we define Y = max (X1, X2, X3, X4), the distribution function G of Y is given by G ( y ) = Pr [Y y ]
= Pr ( X 1 y ) ( X 2 y ) ( X 3 y ) ( X 4 y ) = Pr [ X 1 y ] Pr [ X 2 y ] Pr [ X 3 y ] Pr [ X 4 y ] = F ( y ) =
4
1 3 5 4 y (1 + sin y ) , 16 2 2 It then follows that the density function g of Y is given by
Page 27 of 60
g ( y ) = G '( y ) = = 1 3 (1 + sin y ) ( cos y ) 4 4 cos y (1 + sin y )
5/ 2 3
,
Finally,
3 5 y 2 2
E [Y ] = =
3/ 2 5/ 2
yg ( y ) dy
4
3/ 2
ycos y (1 + sin y ) dy
3
-------------------------------------------------------------------------------------------------------67. Solution: B The amount of money the insurance company will have to pay is defined by the random variable 1000 x if x < 2 Y = if x 2 2000 where x is a Poisson random variable with mean 0.6 . The probability function for X is k e 0.6 ( 0.6 ) p ( x) = k = 0,1, 2,3 and k! k 0.6 0.6 0.6 E [Y ] = 0 + 1000 ( 0.6 ) e + 2000e k = 2 k! k 0.6 = 1000 ( 0.6 ) e 0.6 + 2000 e 0.6 k =0 e 0.6 ( 0.6 ) e0.6 k! = 2000e = 573 E Y 2 = (1000 ) ( 0.6 ) e 0.6 + ( 2000 ) e 0.6 k = 2
2 2
0.6
k =0
( 0.6 )
k!
k
2000e 0.6 1000 ( 0.6 )e 0.6 = 2000 2000e 0.6 600e0.6 0.6k k!
= ( 2000 ) e 0.6 k =0
2
= ( 2000 )
2
0.6k 2 2 2 ( 2000 ) e 0.6 ( 2000 ) (1000 ) ( 0.6 ) e 0.6 k! 2 2 2 ( 2000 ) e 0.6 ( 2000 ) (1000 ) ( 0.6 ) e 0.6
2 2
= 816,893 Var [Y ] = E Y 2 { E [Y ]} = 816,893 ( 573) = 488,564 Var [Y ] = 699
Page 28 of 60
68.
Solution: C Note that X has an exponential distribution. Therefore, c = 0.004 . Now let Y denote the for x < 250 x claim benefits paid. Then Y = and we want to find m such that 0.50 for x 250 250 = 0.004e 0.004 x dx = e 0.004 x
0
m m
0
= 1 e0.004m
This condition implies e0.004m = 0.5 m = 250 ln 2 = 173.29 . -------------------------------------------------------------------------------------------------------69. Solution: D The distribution function of an exponential random variable T with parameter is given by F ( t ) = 1 e t , t > 0 Since we are told that T has a median of four hours, we may determine as follows: 1 = F ( 4 ) = 1 e4 2 1 = e 4 2 4 ln ( 2 ) =
=
4 ln ( 2 )
Therefore, Pr (T 5 ) = 1 F ( 5 ) = e 5 = e
5ln ( 2 ) 4
= 25 4 = 0.42
-------------------------------------------------------------------------------------------------------70. Solution: E Let X denote actual losses incurred. We are given that X follows an exponential distribution with mean 300, and we are asked to find the 95th percentile of all claims that exceed 100 . Consequently, we want to find p95 such that Pr[100 < x < p95 ] F ( p95 ) F (100) 0.95 = = where F(x) is the distribution function of X . P[ X > 100] 1 F (100) Now F(x) = 1 ex/300 . 1 e p95 / 300 (1 e 100 / 300 ) e 1/ 3 e p95 / 300 = = 1 e1/ 3e p95 / 300 Therefore, 0.95 = 1 (1 e 100 / 300 ) e 1/ 3 e p95 / 300 = 0.05 e 1/3 p95 = 300 ln(0.05 e1/3) = 999
Page 29 of 60
71.
Solution: A The distribution function of Y is given by G ( y ) = Pr (T 2 y ) = Pr T y = F y = 1 4 y
(
)
()
for y > 4 . Differentiate to obtain the density function g ( y ) = 4 y 2 Alternate solution: Differentiate F ( t ) to obtain f ( t ) = 8t 3 and set y = t 2 . Then t = g ( y ) = f ( t ( y ) ) dt dy = f d ( y ) dt ( y ) = 8 y
3 2
y and
1 1 2 2 y = 4y 2
-------------------------------------------------------------------------------------------------------72. Solution: E We are given that R is uniform on the interval ( 0.04, 0.08 ) and V = 10, 000e R Therefore, the distribution function of V is given by F ( v ) = Pr [V v ] = Pr 10, 000e R v = Pr R ln ( v ) ln (10, 000 )
1 ln ( v )ln (10,000 ) 1 dr = r = 0.04 0.04 0.04 v = 25 ln 0.04 10, 000
ln ( v ) ln (10,000 )
= 25ln ( v ) 25ln (10, 000 ) 1
0.04
-------------------------------------------------------------------------------------------------------73. Solution: E F ( y ) = Pr [Y y ] = Pr 10 X 0.8 y = Pr X Y 10
1 Y 4 Y 10 5 4 Therefore, f ( y ) = F ( y ) = e ( ) 8 10
1
()
10
8
(Y ) = 1 e 10
10
8
Page 30 of 60
74.
Solution: E First note R = 10/T . Then 10 10 10 FR(r) = P[R r] = P r = P T = 1 FT . Differentiating with respect to r T 10 r d 10 r fR(r) = FR(r) = d/dr 1 FT = FT ( t ) 2 r dt r d 1 FT (t ) = fT (t ) = since T is uniformly distributed on [8, 12] . dt 4 1 10 5 Therefore fR(r) = 2 = 2 . 4 r 2r
-------------------------------------------------------------------------------------------------------75. Solution: A Let X and Y be the monthly profits of Company I and Company II, respectively. We are given that the pdf of X is f . Let us also take g to be the pdf of Y and take F and G to be the distribution functions corresponding to f and g . Then G(y) = Pr[Y y] = P[2X y] = P[X y/2] = F(y/2) and g(y) = G(y) = d/dy F(y/2) = F(y/2) = f(y/2) .
-------------------------------------------------------------------------------------------------------76. Solution: A First, observe that the distribution function of X is given by x3 1x 1 F ( x ) = 4 dt = 3 | 1 = 1 3 , x > 1 1t t x Next, let X1, X2, and X3 denote the three claims made that have this distribution. Then if Y denotes the largest of these three claims, it follows that the distribution function of Y is given by G ( y ) = Pr [ X 1 y ] Pr [ X 2 y ] Pr [ X 3 y ] 1 , y>1 = 1 3 y while the density function of Y is given by 1 3 9 1 g ( y ) = G ' ( y ) = 3 1 3 4 = 4 1 3 y y y y Therefore,
2 2 3
, y>1
Page 31 of 60
E [Y ] =
1
9 9 1 2 1 1 3 dy = 3 1 3 + 6 dy 3 1y y y y y
2
9 18 9 9 18 9 = 3 6 + 9 dy = 2 + 5 8 1 y y y 2 y 5 y 8 y 1 1 2 1 = 9 + = 2.025 (in thousands) 2 5 8
-------------------------------------------------------------------------------------------------------77. Solution: D Prob. = 1 Note
2 2
1
1
1 ( x + y )dxdy = 0.625 8
Pr ( X 1) (Y 1) = Pr ( X > 1) (Y > 1) = 1 Pr ( X > 1) (Y > 1) 1 2 2 2 1 ( y + 2 ) ( y + 1) dy 16 18 30 = 1 = = 0.625 48 48 = 1 = 1 = 1
2 1
{
c
}
(De Morgan's Law) = 1
2 1
1 3 3 ( y + 2 ) ( y + 1) 48
1 8 ( x + y ) dxdy
2 1
1 21 22 1 2 ( x + y ) 1 dy 8 1 = 1 ( 64 27 27 + 8 ) 48
-------------------------------------------------------------------------------------------------------78. Solution: B That the device fails within the first hour means the joint density function must be integrated over the shaded region shown below.
This evaluation is more easily performed by integrating over the unshaded region and subtracting from 1.
Page 32 of 60
Pr ( X < 1) (Y < 1) = 1 = 1
3 1
3
1
2 3 x + 2 xy x+ y 13 dx dy = 1 dy = 1 ( 9 + 6 y 1 2 y ) dy 1 27 54 1 54 1 3
3
13 1 1 32 11 2 1 (8 + 4 y ) dy = 1 54 (8 y + 2 y ) 1 = 1 54 ( 24 + 18 8 2 ) = 1 54 = 27 = 0.41 54
-------------------------------------------------------------------------------------------------------79. Solution: E The domain of s and t is pictured below.
Note that the shaded region is the portion of the domain of s and t over which the device fails sometime during the first half hour. Therefore, 1/ 2 1 1 1/ 2 1 1 Pr S T = f ( s, t ) dsdt + f ( s, t ) dsdt 00 2 2 0 1/ 2 (where the first integral covers A and the second integral covers B). -------------------------------------------------------------------------------------------------------80. Solution: C By the central limit theorem, the total contributions are approximately normally distributed with mean n = ( 2025 ) ( 3125 ) = 6,328,125 and standard deviation
random variable is 1.282 . Letting p be the 90th percentile for total contributions, p n = 1.282, and so p = n + 1.282 n = 6,328,125 + (1.282 ) (11, 250 ) = 6,342,548 . n
n = 250 2025 = 11, 250 . From the tables, the 90th percentile for a standard normal
Page 33 of 60
-------------------------------------------------------------------------------------------------------81. Solution: C 1 Let X1, . . . , X25 denote the 25 collision claims, and let X = (X1 + . . . +X25) . We are 25 given that each Xi (i = 1, . . . , 25) follows a normal distribution with mean 19,400 and standard deviation 5000 . As a result X also follows a normal distribution with mean 1 19,400 and standard deviation (5000) = 1000 . We conclude that P[ X > 20,000] 25 X 19, 400 20, 000 19, 400 X 19, 400 > > 0.6 = 1 (0.6) = 1 0.7257 = P = P 1000 1000 1000 = 0.2743 . -------------------------------------------------------------------------------------------------------82. Solution: B Let X1, . . . , X1250 be the number of claims filed by each of the 1250 policyholders. We are given that each Xi follows a Poisson distribution with mean 2 . It follows that E[Xi] = Var[Xi] = 2 . Now we are interested in the random variable S = X1 + . . . + X1250 . Assuming that the random variables are independent, we may conclude that S has an approximate normal distribution with E[S] = Var[S] = (2)(1250) = 2500 . Therefore P[2450 < S < 2600] = S 2500 2450 2500 S 2500 2600 2500 < < < 2 P = P 1 < 50 2500 2500 2500 S 2500 S 2500 = P < 2 P < 1 50 50 S 2500 , we have P[2450 < S < 2600] Then using the normal approximation with Z = 50 P[Z < 2] P[Z > 1] = P[Z < 2] + P[Z < 1] 1 0.9773 + 0.8413 1 = 0.8186 .
-------------------------------------------------------------------------------------------------------83. Solution: B Let X1,, Xn denote the life spans of the n light bulbs purchased. Since these random variables are independent and normally distributed with mean 3 and variance 1, the random variable S = X1 + + Xn is also normally distributed with mean = 3n and standard deviation = n Now we want to choose the smallest value for n such that S 3n 40 3n 0.9772 Pr [ S > 40] = Pr > n n This implies that n should satisfy the following inequality:
Page 34 of 60
40 3n n To find such an n, lets solve the corresponding equation for n: 40 3n 2 = n 2
2 n = 40 3n 3n 2 n 40 = 0
(3
n + 10
)(
n 4 =0 n =4 n = 16
)
-------------------------------------------------------------------------------------------------------84. Solution: B Observe that E [ X + Y ] = E [ X ] + E [Y ] = 50 + 20 = 70 Var [ X + Y ] = Var [ X ] + Var [Y ] + 2 Cov [ X , Y ] = 50 + 30 + 20 = 100 for a randomly selected person. It then follows from the Central Limit Theorem that T is approximately normal with mean E [T ] = 100 ( 70 ) = 7000 and variance Var [T ] = 100 (100 ) = 1002 T 7000 7100 7000 Pr [T < 7100] = Pr < 100 100 = Pr [ Z < 1] = 0.8413 where Z is a standard normal random variable. Therefore,
Page 35 of 60
-------------------------------------------------------------------------------------------------------85. Solution: B Denote the policy premium by P . Since x is exponential with parameter 1000, it follows from what we are given that E[X] = 1000, Var[X] = 1,000,000, Var[ X ] = 1000 and P = 100 + E[X] = 1,100 . Now if 100 policies are sold, then Total Premium Collected = 100(1,100) = 110,000 Moreover, if we denote total claims by S, and assume the claims of each policy are independent of the others then E[S] = 100 E[X] = (100)(1000) and Var[S] = 100 Var[X] = (100)(1,000,000) . It follows from the Central Limit Theorem that S is approximately normally distributed with mean 100,000 and standard deviation = 10,000 . Therefore, 110, 000 100, 000 P[S 110,000] = 1 P[S 110,000] = 1 P Z = 1 P[Z 1] = 1 10, 000 0.841 0.159 . -------------------------------------------------------------------------------------------------------86. Solution: E Let X 1 ,..., X 100 denote the number of pensions that will be provided to each new recruit. Now under the assumptions given, 0 with probability 1 0.4 = 0.6 X i = 1 with probability ( 0.4 )( 0.25 ) = 0.1 2 with probability ( 0.4 )( 0.75 ) = 0.3 for i = 1,...,100 . Therefore,
E [ X i ] = ( 0 )( 0.6 ) + (1) ( 0.1) + ( 2 ) ( 0.3) = 0.7 ,
2 2 2 2 2 E X i = ( 0 ) ( 0.6 ) + (1) ( 0.1) + ( 2 ) ( 0.3) = 1.3 , and 2 2 Var [ X i ] = E X i { E [ X i ]} = 1.3 ( 0.7 ) = 0.81 Since X 1 ,..., X 100 are assumed by the consulting actuary to be independent, the Central
Limit Theorem then implies that S = X 1 + ... + X 100 is approximately normally distributed with mean E [ S ] = E [ X 1 ] + ... + E [ X 100 ] = 100 ( 0.7 ) = 70 and variance Var [ S ] = Var [ X 1 ] + ... + Var [ X 100 ] = 100 ( 0.81) = 81 Consequently, S 70 90.5 70 Pr [ S 90.5] = Pr 9 9 = Pr [ Z 2.28] = 0.99
Page 36 of 60
-------------------------------------------------------------------------------------------------------87. Solution: D Let X denote the difference between true and reported age. We are given X is uniformly distributed on (2.5,2.5) . That is, X has pdf f(x) = 1/5, 2.5 < x < 2.5 . It follows that x = E[X] = 0 x2 = Var[X] = E[X2] = x2 x3 dx = 5 15 2.5
2.5 2.5
2.5
=
2(2.5)3 =2.083 15
x =1.443 Now X 48 , the difference between the means of the true and rounded ages, has a 1.443 = distribution that is approximately normal with mean 0 and standard deviation 48 0.2083 . Therefore, 1 0.25 1 0.25 P X 48 = P Z = P[1.2 Z 1.2] = P[Z 1.2] P[Z 4 0.2083 4 0.2083 1.2] = P[Z 1.2] 1 + P[Z 1.2] = 2P[Z 1.2] 1 = 2(0.8849) 1 = 0.77 . -------------------------------------------------------------------------------------------------------88. Solution: C 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 problem statement implies that the respective distribution functions for X and Y are F ( x ) = 1 e x / 6 , x > 0 and
G ( y ) = 1 e y / 3 , y > 0
Therefore, Pr ( X 3) (Y 2 ) = Pr [ X 3] Pr [Y 2]
= F ( 3) G ( 2 ) = (1 e 1/ 2 )(1 e 2 / 3 ) = 1 e2 / 3 e1/ 2 + e7 / 6
Page 37 of 60
89.
Solution: B 6 (50 x y ) We are given that f ( x, y ) = 125, 000 0 for 0 < x < 50 y < 50 otherwise
and we want to determine P[X > 20 Y > 20] . In order to determine integration limits, consider the following diagram: y
50
x>20 y>20 (20, 30) (30, 20)
50
x
6 We conclude that P[X > 20 Y > 20] = 125, 000 20
30 50 x
20
(50 x y ) dy dx .
-------------------------------------------------------------------------------------------------------90. Solution: C Let T1 be the time until the next Basic Policy claim, and let T2 be the time until the next Deluxe policy claim. Then the joint pdf of T1 and T2 is 1 1 1 f (t1 , t2 ) = e t1 / 2 e t2 / 3 = e t1 / 2 e t2 / 3 , 0 < t1 < , 0 < t2 < and we need to find 2 3 6 P[T2 < T1] =
t1
1 t / 2 t / 3 1 t / 2 t / 3 t1 6e 1 e 2 dt2 dt1 = 2 e 1 e 2 0 dt1 00 0
32 1 t1 / 2 1 t1 / 2 t1 / 3 1 t1 / 2 1 5t1 / 6 t1 / 2 3 5t1 / 6 +e = e e e dt1 = e e = 1 5 = 5 dt1 = e 5 2 2 2 2 0 0 0 = 0.4 .
-------------------------------------------------------------------------------------------------------91. Solution: D We want to find P[X + Y > 1] . To this end, note that P[X + Y > 1] 1 1 2 2x + 2 y 1 = dydx = 2 xy + 2 y 8 y 1 x dx 4 0 1 x 0 = = 11 1 1 2 x + 1 2 2 x(1 x) 2 (1 x) + 8 (1 x) dx = 0
1 12 1 2
x + 2 x
0
1
1
2
11 1 + x + x 2 dx 84 8
8 x
0
1
5
2
+
3 1 3 1 5 3 1 17 5 x + dx = x 3 + x 2 + x = ++= 8 8 0 24 8 8 24 4 8 24
1
Page 38 of 60
92.
Solution: B Let X and Y denote the two bids. Then the graph below illustrates the region over which X and Y differ by less than 20:
Based on the graph and the uniform distribution: Pr X Y < 20 = Shaded Region Area
( 2200 2000 )
2
=
2002 2
1 2 (180 ) 2 2002
1802 2 = 1 ( 0.9 ) = 0.19 2 200 More formally (still using symmetry) Pr X Y < 20 = 1 Pr X Y 20 = 1 2 Pr [ X Y 20] = 1
2200 1 1 x 20 dydx = 1 2 y 2000 dx 2 2020 2000 200 2020 200 2 2200 2 1 2 = 1 x 20 2000 ) dx = 1 x 2020 ) 2 2020 ( 2( 200 200
= 1 2
2200
2
x 20
2200 2020
180 = 1 = 0.19 200
Page 39 of 60
-------------------------------------------------------------------------------------------------------93. Solution: C Define X and Y to be loss amounts covered by the policies having deductibles of 1 and 2, respectively. The shaded portion of the graph below shows the region over which the total benefit paid to the family does not exceed 5:
We can also infer from the graph that the uniform random variables X and Y have joint 1 density function f ( x, y ) = , 0 < x < 10 , 0 < y < 10 100 We could integrate f over the shaded region in order to determine the desired probability. However, since X and Y are uniform random variables, it is simpler to determine the portion of the 10 x 10 square that is shaded in the graph above. That is, Pr ( Total Benefit Paid Does not Exceed 5)
= Pr ( 0 < X < 6, 0 < Y < 2 ) + Pr ( 0 < X < 1, 2 < Y < 7 ) + Pr (1 < X < 6, 2 < Y < 8 X ) =
( 6 )( 2 ) + (1)( 5 ) + (1 2 )( 5)( 5 ) =
100 100 100
12 5 12.5 + + = 0.295 100 100 100
-------------------------------------------------------------------------------------------------------94. Solution: C Let f ( t1 , t2 ) denote the joint density function of T1 and T2 . The domain of f is pictured below:
Now the area of this domain is given by 1 2 A = 62 ( 6 4 ) = 36 2 = 34 2
Page 40 of 60
1 , 0 < t1 < 6 , 0 < t2 < 6 , t1 + t2 < 10 Consequently, f ( t1 , t2 ) = 34 0 elsewhere and E [T1 + T2 ] = E [T1 ] + E [T2 ] = 2 E [T1 ] (due to symmetry)
61 6 10 t1 1 6 4 t t 4 = 2 t1 dt2 dt1 + t1 dt2 dt1 = 2 t1 2 6 dt1 + t1 2 0 0 34 4 0 4 34 0 34 0 34 61 3t12 4 3t1 = 2 dt1 + (10t1 t12 ) dt1 = 2 34 4 34 0 17 4 0 10 t1 0
dt1
+
1 2 1 3 6 5t1 t1 4 34 3
64 24 1 = 2 + 180 72 80 + = 5.7 3 17 34
-------------------------------------------------------------------------------------------------------95. Solution: E t X +Y + t Y X t t X t +t Y M ( t1 , t2 ) = E et1W +t2 Z = E e 1 ( ) 2 ( ) = E e( 1 2 ) e( 1 2 )
= E e( 1
t t2 ) X
E e( t1 +t2 )Y = e 2
1
( t1 t2 )2
1
e2
( t1 + t2 )2
= e2
1
(t
2 2 1 2 t1t2 + t2
) 1 (t 2
e
2 2 1 + 2 t1t2 + t2
)
= et1 +t2
2
2
-------------------------------------------------------------------------------------------------------96. Solution: E Observe that the bus driver collect 21x50 = 1050 for the 21 tickets he sells. However, he may be required to refund 100 to one passenger if all 21 ticket holders show up. Since passengers show up or do not show up independently of one another, the probability that 21 21 all 21 passengers will show up is (1 0.02 ) = ( 0.98 ) = 0.65 . Therefore, the tour operators expected revenue is 1050 (100 ) ( 0.65 ) = 985 .
Page 41 of 60
97.
Solution: C We are given f(t1, t2) = 2/L2, 0 t1 t2 L . L t2 2 22 2 2 Therefore, E[T1 + T2 ] = (t1 + t2 ) 2 dt1dt2 = L 00
t2 L 3 L 3 t1 2 t2 2 3 + t2 t1 dt1 = 2 + t2 dt2 0 3 L 0 3 0 4L L 2 43 2 t 2 = 2 t2 dt2 = 2 2 = L2 3 L 03 L 3 0 t2
2 L2
(L, L)
t1
-------------------------------------------------------------------------------------------------------98. Solution: A Let g(y) be the probability function for Y = X1X2X3 . Note that Y = 1 if and only if X1 = X2 = X3 = 1 . Otherwise, Y = 0 . Since P[Y = 1] = P[X1 = 1 X2 = 1 X3 = 1] = P[X1 = 1] P[X2 = 1] P[X3 = 1] = (2/3)3 = 8/27 . 19 for y = 0 27 8 We conclude that g ( y ) = for y = 1 27 otherwise 0 19 8 t and M(t) = E e yt = 27 + 27 e
Page 42 of 60
99.
Solution: C We use the relationships Var ( aX + b ) = a 2 Var ( X ) , Cov ( aX , bY ) = ab Cov ( X , Y ) , and Var ( X + Y ) = Var ( X ) + Var (Y ) + 2 Cov ( X , Y ) . First we observe 17, 000 = Var ( X + Y ) = 5000 + 10, 000 + 2 Cov ( X , Y ) , and so Cov ( X , Y ) = 1000. We want to find Var ( X + 100 ) + 1.1Y = Var ( X + 1.1Y ) + 100 = Var [ X + 1.1Y ] = Var X + Var (1.1) Y + 2 Cov ( X ,1.1Y ) = Var X + (1.1) Var Y + 2 (1.1) Cov ( X , Y ) = 5000 + 12,100 + 2200 = 19,300.
2
-------------------------------------------------------------------------------------------------------100. Solution: B Note P(X = 0) = 1/6 P(X = 1) = 1/12 + 1/6 = 3/12 P(X = 2) = 1/12 + 1/3 + 1/6 = 7/12 . E[X] = (0)(1/6) + (1)(3/12) + (2)(7/12) = 17/12 E[X2] = (0)2(1/6) + (1)2(3/12) + (2)2(7/12) = 31/12 Var[X] = 31/12 (17/12)2 = 0.58 .
-------------------------------------------------------------------------------------------------------101. Solution: D Note that due to the independence of X and Y Var(Z) = Var(3X Y 5) = Var(3X) + Var(Y) = 32 Var(X) + Var(Y) = 9(1) + 2 = 11 .
-------------------------------------------------------------------------------------------------------102. Solution: E Let X and Y denote the times that the two backup generators can operate. Now the variance of an exponential random variable with mean is 2 . Therefore, Var [ X ] = Var [Y ] = 102 = 100 Then assuming that X and Y are independent, we see Var [ X+Y ] = Var [ X ] + Var [ Y ] = 100 + 100 = 200
Page 43 of 60
103.
addition, let Y = Max ( X 1 , X 2 , X 3 ) . Then
Solution: E Let X 1 , X 2 , and X 3 denote annual loss due to storm, fire, and theft, respectively. In
Pr [Y > 3] = 1 Pr [Y 3] = 1 Pr [ X 1 3] Pr [ X 2 3] Pr [ X 3 3] = 1 (1 e 3 ) 1 e
(
3 1.5
= 1 (1 e 3 )(1 e 2
) (1 e ) ) (1 e )
3 2.4 5 4
*
= 0.414 * Uses that if X has an exponential distribution with mean
Pr ( X x ) = 1 Pr ( X x ) = 1
x
1
e t dt = 1 ( e t ) = 1 e x x
-------------------------------------------------------------------------------------------------------104. Solution: B Let us first determine k: 1= Then
1 0
1 0
kxdxdy =
k =2
11
1k 1 21 k kx | 0 dy = dy = 02 02 2 1
E [ X ] = 2 x 2 dydx = 2 x 2 dx =
00 0
1
2 31 2 x |0= 3 3 1 21 1 y |0= 2 2
E [Y ] = y 2 x dxdy = ydy =
00 0
11
1
E [ XY ] = =
1 0
1 0
2x 2 ydxdy =
2 21 2 1 y |0 = = 6 63
12 231 x y | 0 dy = ydy 03 03 1
1 2 1 1 1 Cov [ X , Y ] = E [ XY ] E [ X ] E [Y ] = = = 0 3 3 2 3 3 (Alternative Solution) Define g(x) = kx and h(y) = 1 . Then f(x,y) = g(x)h(x) In other words, f(x,y) can be written as the product of a function of x alone and a function of y alone. It follows that X and Y are independent. Therefore, Cov[X, Y] = 0 .
Page 44 of 60
105.
Solution: A The calculation requires integrating over the indicated region.
E(X ) = E (Y ) =
1
0
2x
x
14 82 x y dy dx = x 2 y 2 03 3 2x
2x
x
1 4 4 4 dx = x 2 ( 4 x 2 x 2 ) dx = 4 x 4 dx = x5 = 03 0 505 1 1 1
1
1
0
2x
x
18 82 xy dy dx = xy 3 09 3 2x
x
1 56 8 56 5 56 dy dx = x ( 8 x3 x3 ) dx = x 4 dx = x= 09 09 45 0 45 2x
E ( XY ) =
1
0
x
18 822 x y dy dx = x 2 y 3 09 3
x
dx =
1 56 56 28 82 x ( 8 x3 x3 ) dx = x5 dx = = 09 09 54 27 1
Cov ( X , Y ) = E ( XY ) E ( X ) E (Y ) =
28 56 4 = 0.04 27 45 5
-------------------------------------------------------------------------------------------------------106. Solution: C The joint pdf of X and Y is f(x,y) = f2(y|x) f1(x) = (1/x)(1/12), 0 < y < x, 0 < x < 12 . Therefore, 12 x 12 12 1 yx x x 2 12 E[X] = x =6 dydx = dx = dx = 12 x 12 0 12 24 0 00 0 0
12 12 y2 y x x 2 12 144 E[Y] = =3 dydx = dx = dx = = 12 x 24 x 0 24 48 0 48 00 0 0 x 12 x
E[XY] =
12 x
12 12 2 y2 y x x3 12 (12)3 = 24 dydx = dx = dx = = 12 24 72 0 72 24 0 00 0 0 x
Cov(X,Y) = E[XY] E[X]E[Y] = 24 (3)(6) = 24 18 = 6 .
Page 45 of 60
107.
Solution: A Cov ( C1 , C2 ) = Cov ( X + Y , X + 1.2Y )
= Cov ( X , X ) + Cov (Y , X ) + Cov ( X ,1.2Y ) + Cov ( Y,1.2Y ) = Var X + Cov ( X , Y ) + 1.2Cov ( X , Y ) + 1.2VarY = Var X + 2.2 Cov ( X , Y ) + 1.2VarY Var X = E ( X 2 ) ( E ( X ) ) = 27.4 52 = 2.4
2
Var Y = E (Y 2 ) ( E (Y ) ) = 51.4 7 2 = 2.4
2
Var ( X + Y ) = Var X + Var Y + 2 Cov ( X , Y ) 1 ( Var ( X + Y ) Var X Var Y ) = 1 ( 8 2.4 2.4) = 1.6 2 2 Cov ( C1 , C2 ) = 2.4 + 2.2 (1.6 ) + 1.2 ( 2.4 ) = 8.8 Cov ( X , Y ) =
-------------------------------------------------------------------------------------------------------107. Alternate solution: We are given the following information: C1 = X + Y
C2 = X + 1.2Y E[X ] = 5 E [Y ] = 7 E X 2 = 27.4 E Y 2 = 51.4
Var [ X + Y ] = 8 Now we want to calculate Cov ( C1 , C2 ) = Cov ( X + Y , X + 1.2Y ) = E ( X + Y )( X + 1.2Y ) E [ X + Y ]i E [ X + 1.2Y ] = E X 2 + 2.2 XY + 1.2Y 2 ( E [ X ] + E [Y ]) ( E [ X ] + 1.2 E [Y ]) = E X 2 + 2.2 E [ XY ] + 1.2 E Y 2 ( 5 + 7 ) ( 5 + (1.2 ) 7 ) = 27.4 + 2.2 E [ XY ] + 1.2 ( 51.4 ) (12 )(13.4 ) = 2.2 E [ XY ] 71.72
Therefore, we need to calculate E [ XY ] first. To this end, observe
Page 46 of 60
2 2 8 = Var [ X + Y ] = E ( X + Y ) ( E [ X + Y ])
= E X 2 + 2 XY + Y 2 ( E [ X ] + E [Y ]) = E X 2 + 2 E [ XY ] + E Y 2 ( 5 + 7 ) = 27.4 + 2 E [ XY ] + 51.4 144 = 2 E [ XY ] 65.2 Finally, Cov ( C1,C2 ) = 2.2 ( 36.6 ) 71.72 = 8.8 E [ XY ] = ( 8 + 65.2 ) 2 = 36.6
2
2
-------------------------------------------------------------------------------------------------------108. Solution: A The joint density of T1 and T2 is given by f ( t1 , t2 ) = e t1 e t2 , t1 > 0 , t2 > 0 Therefore, Pr [ X x ] = Pr [ 2T1 + T2 x ]
=
x 0
1 ( x t2 ) 2 0
x e t1 e t2 dt1dt2 = e t2 e t1 0
1 ( x t2 ) 2 0
dt2
11 1 1 x + t2 x t2 x x = e t2 1 e 2 2 dt2 = e t2 e 2 e 2 dt2 0 0 1 1 1 1 1 x t2 xx x x = e t 2 + 2e 2 e 2 0 = e x + 2 e 2 e 2 + 1 2 e 2
= 1 e x + 2e x 2e 2 = 1 2e 2 + e x , x > 0 It follows that the density of X is given by 1 1 x x d g ( x) = 1 2e 2 + e x = e 2 e x , x > 0 dx
1
x
1
x
Page 47 of 60
109.
Solution: B Let u be annual claims, v be annual premiums, g(u, v) be the joint density function of U and V, f(x) be the density function of X, and F(x) be the distribution function of X. Then since U and V are independent, 1 1 g ( u, v ) = ( eu ) e v / 2 = eu e v / 2 , 0 < u < , 0 < v < 2 2 and u F ( x ) = Pr [ X x ] = Pr x = Pr [U Vx ] v vx vx 1 e u e v / 2 dudv = g ( u , v )dudv = 00 0 02 1 1 1 vx = e u e v / 2 | 0 dv = e vx e v / 2 + e v / 2 dv 0 0 2 2 2
1 1 = e v( x +1/ 2) + e v / 2 dv 0 2 2
1 1 e v( x +1/ 2) e v / 2 = = +1 2x + 1 2x + 1 0 2 Finally, f ( x ) = F ' ( x ) = 2 ( 2 x + 1)
-------------------------------------------------------------------------------------------------------110. Solution: C Note that the conditional density function 1 f (1 3, y ) 2 f y x= = , 0< y< , 3 f x (1 3) 3 23 23 23 16 1 f x = 24 (1 3) y dy = 8 y dy = 4 y 2 = 0 9 0 3 0 1 9 9 2 f (1 3, y ) = y , 0 < y < It follows that f y x = = 3 16 2 3
139 9 y dy = y 2 Consequently, Pr Y < X X = 1 3 = 02 4 13
=
0
1 4
Page 48 of 60
111.
Solution: E
3 f ( 2, y ) Pr 1 < Y < 3 X = 2 = dy 1 f x ( 2)
f ( 2, y ) =
2 1 4 1 2 1 y ( ) = y 3 4 ( 2 1) 2
1 1 1 f x ( 2 ) = y 3 dy = y 2 = 2 4 4 1 1 Finally, Pr 1 < Y < 3 X = 2 =
3 1
1 3 y dy 3 18 2 = y 2 = 1 = 1 1 99 4
-------------------------------------------------------------------------------------------------------112. Solution: D We are given that the joint pdf of X and Y is f(x,y) = 2(x+y), 0 < y < x < 1 . Now fx(x) = (2 x + 2 y )dy = 2 xy + y 2
0 x x 0
= 2x2 + x2 = 3x2, 0 < x < 1
f ( x, y ) 2( x + y ) 2 1 y = = + 2 , 0 < y < x 3x 2 3 x x f x ( x) 2 1 y 2 + = [10 + 100 y ] , 0 < y < 0.10 f(y|x = 0.10) = 3 0.1 0.01 3 0.05 2 100 2 0.05 1 1 5 20 =+ = = 0.4167 . y P[Y < 0.05|X = 0.10] = [10 + 100 y ] dy = y + 3 3 3 12 12 3 0 0
so f(y|x) =
-------------------------------------------------------------------------------------------------------113. Solution: E Let W = event that wife survives at least 10 years H = event that husband survives at least 10 years B = benefit paid P = profit from selling policies Then Pr [ H ] = P [ H W ] + Pr H W c = 0.96 + 0.01 = 0.97 and Pr [W | H ] = Pr [W H ] 0.96 = = 0.9897 Pr [ H ] 0.97 Pr H W c Pr [ H ] = 0.01 = 0.0103 0.97
Pr W c | H =
Page 49 of 60
It follows that E [ P ] = E [1000 B ] = 1000 E [ B ] = 1000 ( 0 ) Pr [W | H ] + (10, 000 ) Pr W c | H = 1000 10, 000 ( 0.0103) = 1000 103 = 897
{
}
-------------------------------------------------------------------------------------------------------114. Solution: C Note that P(Y = 0X = 1) =
= 0.286 P(Y = 1X=1) = 1 P(Y = 0 X = 1) = 1 0.286 = 0.714 Therefore, E(YX = 1) = (0) P(Y = 0X = 1) + (1) P(Y = 1X = 1) = (1)(0.714) = 0.714 E(Y2X = 1) = (0)2 P(Y = 0X = 1) + (1)2 P(Y = 1X = 1) = 0.714 Var(YX = 1) = E(Y2X = 1) [E(YX = 1)]2 = 0.714 (0.714)2 = 0.20 -------------------------------------------------------------------------------------------------------115. Solution: A Let f1(x) denote the marginal density function of X. Then Consequently,
f1 ( x ) =
x +1 x
P( X = 1, Y = 0) P( X = 1, Y = 0) 0.05 = = P( X = 1) P( X = 1, Y = 0) + P ( X = 1, Y = 1) 0.05 + 0.125
2 xdy = 2 xy | x +1 = 2 x ( x + 1 x ) = 2 x x
,
0< x<1
f ( x, y ) 1 if: x < y < x + 1 = f1 ( x ) 0 otherwise x [phone] E [Y | X ] = ydy = y 2 | x +1 = ( x + 1) x 2 = x 2 + x + x 2 = x + x x 2 2 2 2 22 2 x +1 1 1 1 3 E Y 2 | X = y 2 dy = y 3 | x +1 = ( x + 1) x3 x x 3 3 3 1 11 1 = x3 + x 2 + x + x3 = x 2 + x + 3 33 3 f ( y| x ) =
2 2 2 2
1 1 Var [Y | X ] = E Y | X { E [Y | X ]} = x + x + x + 3 2 1 11 = x2 + x + x2 x = 3 4 12
Page 50 of 60
116.
Solution: D Denote the number of tornadoes in counties P and Q by NP and NQ, respectively. Then E[NQ|NP = 0] = [(0)(0.12) + (1)(0.06) + (2)(0.05) + 3(0.02)] / [0.12 + 0.06 + 0.05 + 0.02] = 0.88 E[NQ2|NP = 0] = [(0)2(0.12) + (1)2(0.06) + (2)2(0.05) + (3)2(0.02)] / [0.12 + 0.06 + 0.05 + 0.02] = 1.76 and Var[NQ|NP = 0] = E[NQ2|NP = 0] {E[NQ|NP = 0]}2 = 1.76 (0.88)2 = 0.9856 .
-------------------------------------------------------------------------------------------------------117. Solution: C The domain of X and Y is pictured below. The shaded region is the portion of the domain over which X<0.2 .
Now observe Pr [ X < 0.2] =
0.2 0
1 x
0
6 1 ( x + y ) dydx = 6 0
0.2
1 2 y xy 2 y dx 0
1 x
0.2 0.2 1 1 2 2 2 = 6 1 x x (1 x ) (1 x ) dx = 6 (1 x ) (1 x ) dx 0 0 2 2 0.2 1 2 3 3 = 6 (1 x ) dx = (1 x ) | 0.2 = ( 0.8) + 1 = 0.488 0 02
-------------------------------------------------------------------------------------------------------118. Solution: E The shaded portion of the graph below shows the region over which f ( x, y ) is nonzero:
We can infer from the graph that the marginal density function of Y is given by
g ( y ) = 15 y dx = 15 xy
y y y
= 15 y
y
(
y y = 15 y 3 2 (1 y1 2 ) , 0 < y < 1
)
Page 51 of 60
12 32 15 y (1 y ) , 0 < y < 1 or more precisely, g ( y ) = otherwise 0
-------------------------------------------------------------------------------------------------------119. Solution: D The diagram below illustrates the domain of the joint density f ( x, y ) of X and Y .
We are told that the marginal density function of X is f x ( x ) = 1 , 0 < x < 1 while f y x ( y x ) = 1 , x < y < x + 1 1 It follows that f ( x, y ) = f x ( x ) f y x ( y x ) = 0 Therefore, Pr [Y > 0.5] = 1 Pr [Y 0.5] = 1 = 1
1 2 1 2 0
if 0 < x < 1 , x < y < x + 1 otherwise
1 x
2
dydx
1 2 12 117 1 1 x dx = 1 x x 0 = 1 + = 0 0 2 488 2 2 [Note since the density is constant over the shaded parallelogram in the figure the solution is also obtained as the ratio of the area of the portion of the parallelogram above y = 0.5 to the entire shaded area.] y
1 2 x
dx = 1
1
2
Page 52 of 60
120.
Solution: A We are given that X denotes loss. In addition, denote the time required to process a claim by T. 3 2 1 3 x = x , x < t < 2 x, 0 x 2 Then the joint pdf of X and T is f ( x, t ) = 8 x8 0, otherwise. Now we can find P[T 3] = 4 4 2 4 3 12 3 12 3 12 1 36 27 xdxdt = x 2 dt = t 2 dt = t 3 = 1 t2 8 16 t / 2 16 64 4 16 64 3 16 64 3/ 3 3 = 11/64 = 0.17 . t t = 2x
42
4 3 2 1 12
t=x
x
-------------------------------------------------------------------------------------------------------121. Solution: C The marginal density of X is given by
1
1 1 xy 3 1 x fx ( x) = 10 xy 2 ) dy = 10 y ( = 10 0 64 64 3 0 64 3 Then E ( X ) = x f x ( x)dx =
2 10 10 2
1
1 2 x3 1 x2 5x 10 x dx = 64 9 2 64 3
10
=
1 1000 8 500 9 20 9 = 5.778 64
Page 53 of 60
122.
Solution: D The marginal distribution of Y is given by f2(y) = 6 e e
x
y
2y
dx = 6 e
2y
= 6 e2y ey + 6e2y = 6 e2y 6 e3y, 0 < y <
0
e
0
y
x
dx
Therefore, E(Y) =
6 6 2y 3y 2 ye dy 3 3 y e dy 20 0 But 2 y e2y dy and 3 y e3y dy are equivalent to the means of exponential random
0 0
0
y
f2(y) dy = (6 ye
0
2 y
6 ye
3 y
) dy = 6
ye
0
2 y
dy 6
y
0
e3y dy =
variables with parameters 1/2 and 1/3, respectively. In other words, 2 y e2y dy = 1/2
0
and 3 y e3y dy = 1/3 . We conclude that E(Y) = (6/2) (1/2) (6/3) (1/3) = 3/2 2/3 =
0
9/6 4/6 = 5/6 = 0.83 . -------------------------------------------------------------------------------------------------------123. Solution: C Observe Pr [ 4 < S < 8] = Pr 4 < S < 8 N = 1 Pr [ N = 1] + Pr 4 < S < 8 N > 1 Pr [ N > 1] 8 1 4 1 1 = e 5 e 5 + e 2 e 1 * 3 6 = 0.122 *Uses that if X has an exponential distribution with mean
(
)(
)
Pr ( a X b ) = Pr ( X a ) Pr ( X b ) =
a
1
e
t
dt
b
1
e
t
dt = e
a
e
b
Page 54 of 60
124.
Solution: A Because f(x,y) can be written as f ( x) f ( y ) = e x 2e 2 y and the support of f(x,y) is a cross product, X and Y are independent. Thus, the condition on X can be ignored and it suffices to just consider f ( y ) = 2e 2 y . Because of the memoryless property of the exponential distribution, the conditional density of Y is the same as the unconditional density of Y+3. Because a location shift does not affect the variance, the conditional variance of Y is equal to the unconditional variance of Y. Because the mean of Y is 0.5 and the variance of an exponential distribution is always equal to the square of its mean, the requested variance is 0.25.
--------------------------------------------------------------------------------------------------------------------125. Solution: E The support of (X,Y) is 0 < y < x < 1. f X ,Y ( x, y ) = f ( y | x) f X ( x) = 2 on that support. It is clear geometrically (a flat joint density over the triangular region 0 < y < x < 1) that when Y = y we have X ~ U(y, 1) so that f ( x | y ) = By computation: f Y ( y ) = 2dx = 2 2 y f ( x | y ) =
y 1
1 for y < x < 1 . 1 y
f X ,Y ( x, y ) f Y ( y)
=
2 1 = for y < x < 1 2 2y 1 y
Page 55 of 60
126.
Solution: C Using the notation of the problem, we know that p0 + p1 =
p0 + p1 + p2 + p3 + p4 + p5 = 1 . Let pn pn +1 = c for all n 4 . Then pn = p0 nc for 1 n 5 . Thus p 0 + ( p 0 c ) + ( p 0 2c ) + ... + ( p 0 5c ) = 6 p 0 15c = 1.
2 and 5
2 Also p0 + p1 = p0 + ( p0 c ) = 2 p0 c = . Solving simultaneously 5
6 5 1 2 1 25 25 . Thus p0 = . 6 p0 + 15c = 1 . So c = and 2 p0 = + = 60 5 60 60 120 1 12c = 5 17 15 32 We want p4 + p5 = ( p0 4c ) + ( p0 5c ) = + = = 0.267 . 120 120 120 6 p0 3c =
6 p0 15c = 1 2 2 p0 c = 5
-----------------------------------------------------------------------------------------------------------
127.
Solution: D Because the number of payouts (including payouts of zero when the loss is below the deductible) is large, we can apply the central limit theorem and assume the total payout S is normal. For one loss there is no payout with probability 0.25 and otherwise the payout is U(0, 15000). So, E[ X ] = 0.25 * 0 + 0.75 * 7500 = 5625 , 15000 2 ) = 56,250,000 , so the variance of one claim is 12 Var ( X ) = E[ X 2 ] E[ X ]2 = 24,609,375 . E[ X 2 ] = 0.25 * 0 + 0.75 * (7500 2 + Applying the CLT, S (200)(5625) P[1,000,000 < S < 1,200,000] = P 1.781741613 < < 1.069044968 (200)(24,609,375) which interpolates to 0.8575-(1-0.9626)=0.8201 from the provided table.
Page 56 of 60
128.
Key: B Let H be the percentage of clients with homeowners insurance and R be the percentage of clients with renters insurance. Because 36% of clients do not have auto insurance and none have both homeowners and renters insurance, we calculate that 8% (36% 17% 11%) must have renters insurance, but not auto insurance. (H 11)% have both homeowners and auto insurance, (R 8)% have both renters and auto insurance, and none have both homeowners and renters insurance, so (H + R 19)% must equal 35%. Because H = 2R, R must be 18%, which implies that 10% have both renters and auto insurance.
129.
Key: B The reimbursement is positive if health care costs are greater than 20, and because of the memoryless property of the exponential distribution, the conditional distribution of health care costs greater than 20 is the same as the unconditional distribution of health care costs. We observe that a reimbursement of 115 corresponds to health care costs of 150 (100% x (120 20) + 50% x (150 120)), which is 130 greater than the deductible of 20. Therefore, G (115) = F (130) = 1 e
130 100
= 0.727 .
130.
Key: C E 100(0.5)
[
X
] = 100E[(0.5) ] = 100E[e (
X
ln 0.5 ) X
] = 100M
X
(ln 0.5) = 100
1 = 41.9 1 2 ln 0.5
Page 57 of 60
131. Solution: E
p(n1 , n2 ) , p1 (n1 ) where p1 (n1 ) is the marginal probability function of N 1 . To find the latter, sum the joint probability function over all possible values of N 2 obtaining
First, find the conditional probability function of N 2 given N 1 = n1 : p 2|1 (n 2 | n1 ) =
p1 (n1 ) =
since
n2 =1
p(n1 , n2 ) =
31 44
n1 1
n2 =1
e n1 1 e n1
(
)
n2 1
=
31 44
n1 1
,
n2 =1
e (1 e )
n1
n1 n2 1
= 1 as the sum of the probabilities of a geometric random variable. The
p (n1 , n 2 ) = e n1 1 e n1 p1 (n1 )
conditional probability function is
p 2|1 (n 2 | n1 ) =
(
)
n2 1
,
which is the probability function of a geometric random variable with parameter p = e n1 . The mean of this distribution is 1 / p = 1 / e n1 = e n1 , and becomes e 2 when n1 = 2 . 132. Solution: C
The number of defective modems is 20% x 30 + 8% x 50 = 10.
10 70 2 3 The probability that exactly two of a random sample of five are defective is = 0.102 . 80 5 133. Solution: B Pr(man dies before age 50) = Pr(T < 50 | T > 40) = Pr(40 < T < 50) F (50) F (40) = Pr(T > 40) 1 F (40) e
11.140 1000
=
e
11.150 1000
e
11.140 1000
= 1 e
(1.140 1.150 ) 1000
= 0.0696 Expected Benefit = 5000 Pr(man dies before age 50) = (5000) (0.0696) = 347.96
Page 58 of 60
134.
Solutions: C Letting t denote the relative frequency with which twin-sized mattresses are sold, we have that the relative frequency with which king-sized mattresses are sold is 3t and the relative frequency with which queen-sized mattresses are sold is (3t+t)/4, or t. Thus, t = 0.2 since t + 3t + t = 1. The probability we seek is 3t + t = 0.80.
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135.
Key: E Var (N) = E [ Var ( N | )] + Var [ E ( N | )] = E () + Var () = 1.50 + 0.75 = 2.25
136.
Key: D X follows a geometric distribution with p =
1 . Y = 2 implies the first roll is not a 6 and the 6 second roll is a 6. This means a 5 is obtained for the first time on the first roll (probability = 20%) or a 5 is obtained for the first time on the third or later roll (probability = 80%). E [X | X 3] = 1 + 2 = 6 + 2 = 8 , so E [X Y = 2] = 0.2(1) + 0.8(8) = 6.6 p
137.
Key: E
Because X and Y are independent and identically distributed, the moment generating function of X + Y equals K2(t), where K(t) is the moment generating function common to X and Y. Thus, K(t) = -t t 0.30e + 0.40 + 0.30e . This is the moment generating function of a discrete random variable that assumes the values -1, 0, and 1 with respective probabilities 0.30, 0.40, and 0.30. The value we seek is thus 0.70.
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