Actuary PS3

5/19/2018 Actuary PS3

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Math 370/408, Spring 2008

Prof. A.J. Hildebrand

Actuarial Exam Practice Problem Set 3Solutions

About this problem set: These are problems from Course 1/P actuarial exams that Ihave collected over the years, grouped by topic, and ordered by difficulty. All of the exams haveappeared on the SOA website http://www.soa.org/at some point in the past, though most of

them are no longer there. The exams are copy-righted by the SOA.

Note on the topics covered: This problem set covers problems on continuous randomvariables (Chapter 3 of Hogg/Tanis).

Note on the ordering of the problems: The problems are loosely grouped by topic, andvery roughly ordered by difficulty (though that ordering is very subjective). The last group ofproblems (those with triple digit numbers) are harder, either because they are conceptually moredifficult, or simply because they require lengthy computations and may take two or three timesas long to solve than an average problem.

Answers and solutions: I will post an answer key and hints/solutions on the coursewebpage, www.math.uiuc.edu/hildebr/370.

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Math 370/408 Spring 2008 Actuarial Exam Practice Problem Set 3 Solutions

1. [3-1]

An insurance company issued insurance policies to 32 independent risks. For each policy,the probability of a claim is 1/6. The benefit given that there is a claim has probabilitydensity function

f(y) =2(1 y), 0< y


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4. [3-4]

The loss amount, X, for a medical insurance policy has cumulative distribution function

F(x) =

0, x 3.

Calculate the mode of the distribution.

(A) 23 (B) 1 (C) 32 (D) 2 (E) 3

Answer: D: 2

Hint/Solution: The mode of a distribution is the point wheref(x) is maximal. Withthe given distribution we have f(x) = F(x) = (4/9)x (1/9)x2 for 0 < x < 3, andf(x) = 0 outside the interval [0, 3]. Differentiating, we get f(x) = (4/9) (2/9)x, andsetting this equal to 0, we see that x = 2 is the only critical point off. Hence the mode

must occur at x = 2.

5. [3-5]

An insurance companys monthly claims are modeled by a continuous, positive randomvariable X, whose probability density function is proportional to (1 + x)4, where 0< x 0. Since the median is 4 hours, we have F(4) = 1/2. This allows usto determine the parameter : 1 e4/ = 1/2, so = 4/ ln 2. The probability that thecomponent will work for at least 5 hours is given by 1 F(5) =e5/ =e(5/4)ln2 = 0.42.

8. [3-8]

The number of days that elapse between the beginning of a calendar year and the momenta high-risk driver is involved in an accident is exponentially distributed. An insurancecompany expects that 30% of high-risk drivers will be involved in an accident during thefirst 50 days of a calendar year. What portion of high-risk 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

Answer: C: 0.43

Hint/Solution: IfXdenotes the number of days elapsed until an accident occurs, thenwe are given that X is exponentially distributed, and that P(X 50) = 0.3, and we haveto compute P(X 80). By the general of the c.d.f. of an exponential distribution, wehave F(x) =P(X x) = 1 ex/ forx >0. By the given information, F(50) =P(X50) = 0.3, which allows us to determine : We have 1 e50/ = 0.3, so50/ = ln 0.7,or =50/ ln 0.7. Substituting this back intoF(x) and setting x = 80, we get F(80) =P(X 80) = 1 e80/ = 1 e(80/50)ln0.7 = 0.43

9. [3-9]

The monthly profit of Company I can be modeled by a continuous random variable withdensity function f. Company I I has a monthly profit that is twice that of Company I.Determine the probability density function of the monthly profit of Company II.

(A) 12

fx2

(B)f

x2

(C) 2f

x2

(D) 2f(x) (E) 2f(2x)

Answer: A

Hint/Solution: This is an easy exercise in changing variables in density functions ifyou apply to correct procedure for that. Let X denote the profit for Company I and Ythat for Company II. Then Y = 2X. Let f(x) and F(x) denote the p.d.f. and c.d.f.

of X, and let g(x) and G(x) denote the corresponding functions for Y. Then we haveG(x) = P(Y x) = P(2X x) = P(X x/2) = F(x/2). Differentiating both sides withrespect tox we getg(x) = G(x) = (1/2)F(x/2) = (1/2)f(x/2) sinceF(x) =f(x). Thus,the density ofY is (1/2)f(x/2).

10. [3-10]

An insurance company insures a large number of homes. The insured value, X, of arandomly selected home is assumed to follow a distribution with density function

f(x) =

3x4 forx >1,

0 otherwise.

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Given that a randomly selected home is insured for at least 1.5, what is the probabilitythat it is insured for less than 2?

(A) 0.578 (B) 0.684 (C) 0.704 (D) 0.829 (E) 0.875

Answer: A: 0.578

Hint/Solution: We need to compute

P(X 2|X 1.5) = P(1.5 X 2)P(X >1.5)

.

Now,

P(X >1.5) =

1.5

3x4 = 1.53 = 1.53 = 0.296,

andP(1.5 X 2) = P(X >1.5) P(X >2) = 1.53 23 = 0.171,

so P(X

2|X

1.5) = 0.171/0.296 = 0.578.

11. [3-11]

An insurance policy reimburses a loss up to a benefit limit of 10. The policyholders loss,Y, follows a distribution with density function

f(y) =

2y3 fory >1,

0 otherwise.

What is the expected value of the benefit paid under the insurance policy?

(A) 1.0 (B) 1.3 (C) 1.8 (D) 1.9 (E) 2.0

Answer: D: 1.9

Hint/Solution: The benefit, X, is given by X=Y ifY 10, and byX= 10 ifY >10.Thus,

E(X) =

101

y 2y3dy+ 10

10 2y3dy = 2

1 110

+ 10 102 = 1.9

12. [3-12]

The lifetime of a machine part has a continuous distribution on the interval (0 , 40) withprobability density function f, where f(x) is proportional to (10 + x)2. Calculate theprobability that the lifetime of the machine part is less than 6.

(A) 0.04 (B) 0.15 (C) 0.47 (D) 0.53 (E) 0.94

Answer: C: 0.47

Hint/Solution: We are given that f(x) = c(10 +x)2 for 0 x 40, with someconstant c. To determine c, compute the integral of f(x) over the full interval (0, 40)

and set the integral equal to 1: 1 =400

c(10 +x)xdx = c((1/10) (1/50)) = c0.08,so c = 1/0.08 = 12.5. The probability to compute is then equal to

60

c(10 +x)2dx =c((1/10) (1/16)) =c(3/80) = 0.47.

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13. [3-13]

The time, T, that a manufacturing system is out of operation has cumulative distributionfunction

F(t) = 1

2t

2for t >2,

0 otherwise.

The resulting cost to the company is Y = T2. Determine the density function ofY, fory >4.

(A) 4y2 (B) 8y3/2 (C) 8y3 (D) 16y1 (E) 1024y5

Answer: A

Hint/Solution: This is a routine application of the change of variables techniques; asalways in these problems, you need to take a detour via the corresponding c.d.f.s to getthe p.d.f. of the new variable.

14. [3-14]

Under a group insurance policy, an insurer agrees to pay 100% of the medical bills incurredduring the year by employees of a small company, up to a maximum total of one milliondollars. The total amount of bills incurred, X, has probability density function

f(x) =

x(4x)

9 for 0< x


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Hint/Solution: Let X denote the lifetime of the printer. Then the refund for eachprinter, Y, is given by Y = 200 ifX 1, Y = 100 if 1 X 2, and Y = 0 otherwise.Thus, E(Y) = 200P(X 1) + 100P(1 < X 2). The probabilities P(. . . ) here areeasily computed using that fact X has exponential distribution with mean 2, and thusc.d.f. F(x) = P(X

x) = 1

ex/2 for x

0. Thus, P(X

1) = 1

e1/2, P(1 100,

P(X x|X 100) = P(X x)P(X 100 =

ex/300

e100/300 =e(x100)/300.

Setting this equal to 0.05 and solving for x, we get (x 100)/300 = ln(0.05), so x =300 ln(0.05) + 100 = 1000.

18. [3-18]

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A piece of equipment is being insured against early failure. The time from purchase untilfailure of the equipment is exponentially distributed with mean 10 years. The insurancewill pay an amount x if the equipment fails during the first year, and it will pay 0 .5x iffailure occurs during the second or third year. If failure occurs after the first three years,no payment will be made. At what level must x be set if the expected payment made

under this insurance is to be 1000?

(A) 3858 (B) 4449 (C) 5382 (D) 5644 (E) 7235

Answer: D: 5644

Hint/Solution: Let T be the time until failure and X the payout. Then X = x ifT 1, X = (1/2)x if 1 T 3, and X = 0 otherwise. (Note that x here is notthe variable in a p.d.f. or c.d.f., but the unknown given in the problem that we need todetermine.) Thus, E(X) = xP(T 1)+(1/2)xP(1 T 3). NowP(T t) = 1 et/10by the given exponential distribution for T. Thus, P(T 1) = 1 e1/10 = 0.096 andP(1 T 3) = e1/10 e3/10 = 0.163. Substituting these values into E(X) givesE(X) = 0.096x + 0.5

0.163x= 0.177x. Setting this equal to 1000 and solving for x we get

the answer, x = 1000/0.1777 = 5644.

19. [3-19]

A manufacturers annual losses follow a distribution with density function

f(x) =

2.5(0.6)2.5

x3.5 for x >0.6,

0 otherwise.

To cover its losses, the manufacturer purchases an insurance policy with an annual de-ductible of 2. What is the mean of the manufacturers annual losses not paid by theinsurance policy?

(A) 0.84 (B) 0.88 (C) 0.93 (D) 0.95 (E) 1.00

Answer: C: 0.93

Hint/Solution: Let Xdenote the annual loss, and Y the part of the loss not coveredby insurance. ThenY = X ifX 2, and Y = 2 ifX 2. To computeE(Y), split theintegral into intervals [0.6, 2] and [2,].

20. [3-20]

A device that continuously measures and records seismic activity is placed in a remoteregion. The time, T, to failure of this device is exponentially distributed with mean 3years. Since the device will not be monitored during its first two years of service, the time

to discovery of its failure is X= max(T, 2). DetermineE(X).(A) 2 + 1

3e6

(B) 2 2e2/3 + 5e4/3(C) 3

(D) 2 + 3e2/3

(E) 5

Answer: D

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Hint/Solution: Since T is exponentially distributed with mean 3, the density ofT isf(t) = (1/3)et/3 fort >0. SinceX= max(T, 2), we haveX= 2 if 0 T 2 andX=Tif 2< T < .Thus,

E(X) = 2

0

2 13

et/3dt +

2

t 13

et/3dt

= 2(1 e2/3) tet/32

+

2

et/3dt

= 2(1 e2/3) + 2et/3 + 3e2/3 = 2 + 3e2/3

21. [3-21]

An insurance policy pays for a random loss Xsubject to a deductible ofC, where 0< C 200.

SettingF(x) equal to 0.3 and 0.7 and solving for x we get the 30th and 70th percentiles:x= 230.7 and x = 323.7. The answer is the difference between these two numbers, 93.

27. [3-53]

The warranty on a machine specifies that it will be replaced at failure or age 4, whicheveroccurs first. The machines age at failure, X, has density function

f(x) =

1/5 for 0< x


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Thus,

E(Y) =

40

x1

5dt +

54

41

5dt= 2.4,

E(Y2

) = 40 x

2 1

5 dt + 54 4

2 1

5 dt= 7.47

Var(Y) = E(Y2) E(Y)2 = 1.707.

28. [3-54]

The owner of an automobile insures it against damage by purchasing an insurance policywith a deductible of 250. In the event that the automobile is damaged, repair costs can bemodeled by a uniform random variable on the interval (0, 1500). Determine the standarddeviation of the insurance payment in the event that the automobile is damaged.

(A) 361 (B) 403 (C) 433 (D) 464 (E) 521

Answer: B: 403Hint/Solution: LetXdenote the repair costs, and let Ydenote the insurance payment.We need to compute =

Var(Y).

We are given thatX is uniformly distributed on the interval (0, 1500), so the density ofXisf(x) = 1/1500 for 0< x 0, of the random variable Y .

(A) 10y0.8e8y0.2

(B) 8y0.2e10y0.8

(C) 8y0.2e0.1y1.25

(D) (0.1y)1.25e0.125(0.1y)0.25

(E) 0.125(0.1y)0.25e(0.1y)1.25

Answer: E

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Hint/Solution: This is a change of variables exercise. Let g(x) and G(x) denote thedensity and c.d.f. ofX, and let f(y) and F(y) denote the density and c.d.f. ofY . Then,for y >0,

F(y) = P(Y y) = P(10X0.8 y)=P(X (0.1y)1.25) = G((0.1y)1.25).

Differentiating with respect to y we get

f(y) = F(y) =G((0.1y)1.25)1.25(0.1y)0.250.1

=g((0.1y)1.25)0.125(0.1y)0.25 =e(0.1y)1.25

0.125(0.1y)0.25.

sinceg (x) =ex by the exponential distribution ofXwith mean 1.

30. [3-56]

A tour operator has a bus that can accommodate 20 tourists. The operator knows thattourists may not show up, so he sells 21 tickets. The probability that an individual tourist

will not show up is 0.02, independent of all other tourists. Each ticket costs 50, and isnon-refundable if a tourist fails to show up. If a tourist shows up and a seat is not available,the tour operator has to pay 100 (ticket cost + 50 penalty) to the tourist. What is theexpected revenue of the tour operator?

(A) 935 (B) 950 (C) 967 (D) 976 (E) 985

Answer: E:985

Hint/Solution: This is simpler than it might seem since there are only two cases toconsider:

(i) All 21 ticket holders show up. This occurs with probability (1 0.02)21 = 0.654.In this case, the operator earns 21 50 = 1050 from ticket sales, but incurs a costof 100 for the one ticket holder that cannot be accommodated, so the revenue is1050 100 = 950.

(ii) Not all 21 ticket holders show up. This occurs with the complementary probability,1 0.654 = 0.346. In this case, the operator earns 1050 from ticket sales, but incursno penalty, so the revenue is 1050.

Hence the expected revenue is 950 0.654 + 1050 0.346 = 985.

31. [3-57]

An investment account earns an annual interest rate R that follows a uniform distributionon the interval (0.04, 0.08). The value of a 10, 000 initial investment in this account after

one year is given by V = 10, 000eR

. Determine the cumulative distribution function, F(v),ofV for values v that satisfy 0< F(v)< 1.

(A) 10, 000ev/10,000 10, 408

425

(B) 25ev/10,000 0.04(C)

v 10, 40810, 833 10, 408

(D) 25

v

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(E) 25

ln

v

10, 000

0.04

Answer: E

Hint/Solution: This is a change of variables exercise. Let g(r) and G(r) denote thedensity and c.d.f. ofR, and let f(v) and F(v) denote the density and c.d.f. ofY. Then,forv in the appropriate range,

F(v) =P(V v) =P(10000eR v)=P

R ln

v10000

= G

ln v

10000

.

SinceR has uniform distribution on the interval (0.04, 0.08),G(r) is given by (for r in thisinterval)

G(r) = r 0.040.08 0.04= 25(r 0.04).

Substituting this above gives the correct answer, (E):

F(v) = 25

ln

v

10, 000

0.04

.

32. [3-61]

You are given the following information about N, the annual number of claims for arandomly selected insured:

P(N= 0) =1

2, P(N= 1) =

1

3, P(N >1) =

1

6.

LetSdenote the total annual claim amount for an insured. When N= 1,Sis exponentiallydistributed with mean 5. When N > 1, S is exponentially distributed with mean 8.

DetermineP(4< S


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(B) 3 52r(C) 3r 5 ln r2(D) 10r2

(E) 52r2

Answer: E

Hint/Solution: The key observation here is that R and T are related by R = 10/T.Since we know the distribution of T (uniform distribution on [8, 12]), the problem thenbecomes a change of variables problem for density functions. As usual, this requires adetour through the corresponding c.d.f.s. Letf(t) andF(t) denote the p.d.f. and c.d.f. ofT, and letg (r) and G(r) denote the corresponding functions for R. By the given uniformdistribution ofTon [8, 12], we havef(t) = 1/4 for 8 t 12. Hence, for 10/12 r 10/8,

G(r) = P(R r) = P

10

T r

= P

T 10

r

= 1 F

10

r

.

Differentiating, we get

g(r) = G(r) = F

10

r

(10r2) =f

10

r

10r2 =

1

410r2 =

5

2r2.

34. [3-103]

An auto insurance company insures an automobile worth 15, 000 for one year under a policywith a 1, 000 deductible. During the policy year there is a 0.04 chance of partial damageto the car and a 0.02 chance of a total loss of the car. If there is partial damage to the car,the amount X of damage (in thousands) follows a distribution with density function

f(x) =0.5003ex/2 for 0< x


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Thus,

E(Y|D) = 151

(x 1)f(x)dx= 151

(x 1)0.5003ex/2dx

=

(2)(x 1)0.5003ex/215

1 + 2 151 0.5003e

x/2

dx

= (2)(14)0.5003e15/2 + 2 2 0.5003

e1/2 e15/2

= 1.2049

Hence,E(Y) = 1.2049 0.04 + 14 0.02 = 0.328.

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Actuary PS3